Search Results

Search found 18347 results on 734 pages for 'generate password'.

Page 129/734 | < Previous Page | 125 126 127 128 129 130 131 132 133 134 135 136  | Next Page >

  • jQuery + Dialog Form Validation

    - by Panther24
    Hi, I have a jQuery Dialog form and on submit I'm trying to validate the fields. I'm using http://docs.jquery.com/Plugins/Validation to validate. In this I'm facing an issue, the validate function is not being called. I'm posting some snippet of my code $("#register-dialog-form").dialog({ autoOpen: false, height: 350, width: 450, modal: true, buttons: { 'Register': function() { $("#registerFrm").validate({ rules: { accountid: "required", name: { required: true, minlength: 5 }, username: { required: true, minlength: 5 }, password: { required: true, minlength: 5 } }, messages: { firstname: "Please enter your firstname", accountid: "Please enter the lastname", name: "Please enter a user friendly name", username: { required: "Please enter a username", minlength: jQuery.format("Enter at least {0} characters") }, password: { required: "Please provide a password", minlength: jQuery.format("Password must be at least {0} characters long") } } }); //****************** //TODO: Need to submit my form here //****************** $(this).dialog('close'); }, Cancel: function() { $(this).dialog('close'); } }, close: function() { //$('registerFrm').clearForm(); } }); Can someone please tell me what I'm doing wrong here. I've also tried to put the validation into $(document).ready(function() {}, but with no success. Here is the html code <div id="register-dialog-form" title="Register Account - Master" align="center" style="display: none"> <s:form name="registerFrm" id="registerFrm" action="registermaster" method="POST"> <table width="90%" border="0" class="ui-widget"> <tr> <td> <s:textfield label="Account Id" name="accountid" id="accountid" cssClass="text ui-widget-content ui-corner-all" /> </td> </tr> <tr> <td> <s:textfield label="Name" name="name" id="name" cssClass="text ui-widget-content ui-corner-all" /> </td> </tr> <tr> <td> <s:textfield label="Username" name="username" id="username" cssClass="text ui-widget-content ui-corner-all" /> </td> </tr> <tr> <td> <s:password label="Password" name="password" id="password" cssClass="text ui-widget-content ui-corner-all" /> </td> </tr> </table> </s:form> </div><!--End of RegisterAcc form-->

    Read the article

  • VB.Net Split A Group Of Text

    - by Ben
    I am looking to split up multiple lines of text to single them out, for example: Url/Host:ftp://server.com/1 Login:Admin1 Password:Password1 Url/Host:ftp://server.com/2 Login:Admin2 Password:Password2 Url/Host:ftp://server.com/3 Login:Admin3 Password:Password3 How can I split each section into a different textbox, so that section one would be put into TextBox1.Text on its own: Url/Host:ftp://server.com/1 Login:Admin1 Password:Password1 Thanks in advance :)!

    Read the article

  • How can I get an Active Directory data code from System.DirectoryServices[.Protocols]?

    - by Alex Waddell
    When using .Protocols, I can run the following pseudocode to authenticate to an AD: try { LdapConnection c = new LdapConnection("User", "Password"); c.Bind(); } catch (LdapException le) { Debug.WriteLine(le.ResultCode); } This code will allow me to get the "Invalid Credentials" error string, and the AD code "49", but I need to get the additional data errors similar to an LDAP Java client : [LDAP: error code 49 - 80090308: LdapErr: DSID-0C09030F, comment: AcceptSecurityContext error, data **525**, vece ] 525 – user not found 52e – invalid credentials (bad password) 530 – logon time restriction 532 – password expired 533 – account disabled 701 – account expired 773 – user must reset password

    Read the article

  • A problem in my windows boot menu

    - by user210332
    Hi, One i had kept a supervisor password to my windows boot screen, but now i forgot that password, Now i am unable to access the boot menu since its asking the password, all menu options are disabled. Is it possible to remove that password and can i get the boot menu default settings back? Processor: Intel Pentium dual core (2) OS : XP Thanks in Advance,

    Read the article

  • getting Cannot identify image file when trying to create thumbnail in django

    - by Mo J. Mughrabi
    Am trying to create a thumbnail in django, am trying to build a custom class specifically to be used for generating thumbnails. As following from StringIO import StringIO from PIL import Image class Thumbnail(object): source = '' size = (50, 50) output = '' def __init__(self): pass @staticmethod def load(src): self = Thumbnail() self.source = src return self def generate(self, size=(50, 50)): if not isinstance(size, tuple): raise Exception('Thumbnail class: The size parameter must be an instance of a tuple.') self.size = size # resize properties box = self.size factor = 1 fit = True image = Image.open(self.source) # Convert to RGB if necessary if image.mode not in ('L', 'RGB'): image = image.convert('RGB') while image.size[0]/factor > 2*box[0] and image.size[1]*2/factor > 2*box[1]: factor *=2 if factor > 1: image.thumbnail((image.size[0]/factor, image.size[1]/factor), Image.NEAREST) #calculate the cropping box and get the cropped part if fit: x1 = y1 = 0 x2, y2 = image.size wRatio = 1.0 * x2/box[0] hRatio = 1.0 * y2/box[1] if hRatio > wRatio: y1 = int(y2/2-box[1]*wRatio/2) y2 = int(y2/2+box[1]*wRatio/2) else: x1 = int(x2/2-box[0]*hRatio/2) x2 = int(x2/2+box[0]*hRatio/2) image = image.crop((x1,y1,x2,y2)) #Resize the image with best quality algorithm ANTI-ALIAS image.thumbnail(box, Image.ANTIALIAS) # save image to memory temp_handle = StringIO() image.save(temp_handle, 'png') temp_handle.seek(0) self.output = temp_handle return self def get_output(self): return self.output.read() the purpose of the class is so i can use it inside different locations to generate thumbnails on the fly. The class works perfectly, I've tested it directly under a view.. I've implemented the thumbnail class inside the save method of the forms to resize the original images on saving. in my design, I have two fields for thumbnails. I was able to generate one thumbnail, if I try to generate two it crashes and I've been stuck for hours not sure whats the problem. Here is my model class Image(models.Model): article = models.ForeignKey(Article) title = models.CharField(max_length=100, null=True, blank=True) src = models.ImageField(upload_to='publication/image/') r128 = models.ImageField(upload_to='publication/image/128/', blank=True, null=True) r200 = models.ImageField(upload_to='publication/image/200/', blank=True, null=True) uploaded_at = models.DateTimeField(auto_now=True) Here is my forms class ImageForm(models.ModelForm): """ """ class Meta: model = Image fields = ('src',) def save(self, commit=True): instance = super(ImageForm, self).save(commit=True) file = Thumbnail.load(instance.src) instance.r128 = SimpleUploadedFile( instance.src.name, file.generate((128, 128)).get_output(), content_type='image/png' ) instance.r200 = SimpleUploadedFile( instance.src.name, file.generate((200, 200)).get_output(), content_type='image/png' ) if commit: instance.save() return instance the strange part is, when i remove the line which contains instance.r200 in the form save. It works fine, and it does the thumbnail and stores it successfully. Once I add the second thumbnail it fails.. Any ideas what am doing wrong here? Thanks Update: I tried earlier doing the following but I still got the same error class ImageForm(models.ModelForm): """ """ class Meta: model = Image fields = ('src',) def save(self, commit=True): instance = super(ImageForm, self).save(commit=True) instance.r128 = SimpleUploadedFile( instance.src.name, Thumbnail.load(instance.src).generate((128, 128)).get_output(), content_type='image/png' ) instance.r200 = SimpleUploadedFile( instance.src.name, Thumbnail.load(instance.src).generate((200, 200)).get_output(), content_type='image/png' ) if commit: instance.save() return instance

    Read the article

  • How to pass parameters for OPENDATASOURCE

    - by Rapunzo
    I can connect to a linked server with this: SELECT testNo, soruTuruId, soruNo, cevap , degerlendirenTcNo, degerlendirilenTcNo FROM OPENDATASOURCE('SQLOLEDB', 'Data Source=192.168.150.42;User ID=readerUser;Password=1').akreditasyon.dbo.tblPerfCevap But I have to pass the password as parameter. and I try like this: SET @connectionString = 'Data Source=192.168.150.42;User ID=readerUser;Password='+@pw SELECT testNo, soruTuruId, soruNo, cevap , degerlendirenTcNo, degerlendirilenTcNo FROM OPENDATASOURCE('SQLOLEDB', @connectionString ).akreditasyon.dbo.tblPerfCevap and SELECT testNo, soruTuruId, soruNo, cevap , degerlendirenTcNo, degerlendirilenTcNo FROM OPENDATASOURCE('SQLOLEDB', 'Data Source=192.168.150.42;User ID=readerUser;Password='+@pw ).akreditasyon.dbo.tblPerfCevap but didnt work:S does anyone have an idea?

    Read the article

  • Why does MS SQL Mgmt Studio Express keep forgetting my passwords?

    - by Ryan
    I have about had it with this tool, I check the save password box at the login dialogue but it just doesn't work. Sometimes it will for a few days, and then the password will just be gone. Nearly every time I load this thing up I have to track down the password again and type it in. Is there some password rule in the database that would be causing this? This is driving me absolutely crazy.

    Read the article

  • Page not redirecting properly(php)

    - by user225269
    I want to do the login page this way so that I won't be having trouble posting the username in the userpage. But everytime I try to access login.php. I get an error in firefox, that the page is not redirecting properly. What do I do? This works when I separate them into two. Into something like, login.php and verifylogin.php as the form action. But if I do it like this, I get redirection errors: <?php $host="localhost"; $username="root"; $password="nitoryolai123$%^"; $db_name="school"; $tbl_name="users"; mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); $uname = mysql_real_escape_string($_POST['username']); $pword = mysql_real_escape_string($_POST['password']); $SQL = "SELECT * FROM users WHERE username = '$uname' AND password = '$pword'"; $result = mysql_query($SQL); $num_rows = mysql_num_rows($result); if ($result) { if ($num_rows > 0) { session_start(); $_SESSION['login'] = "1"; header ("Location: userpage.php"); } else { session_start(); $_SESSION['login'] = ""; header ("Location: login.php"); } } else { $errorMessage = "Error logging on"; } ?> <tr> <form name="form1" method="post" action="login.php"> <td> <table> <tr> <td><strong><font size="2">Login User</strong></td> </tr> <tr> <td width="30" height="35"><font size="2">Username:</td> <td width="30"><input name="username" type="text" id="username" maxlength="17"></td> </tr> <tr> <td width="30" height="35" ><font size="2">Password:</td> <td width="30"><input name="password" type="password" id="password" maxlength="17"></td> </tr> <td><td align="right" width="30"><input type="submit" name="Submit" value="Submit" /></td> <td><input type="reset" name="Reset" value="Reset"></td></td> </tr> </form> please help, thanks.

    Read the article

  • Java MessageDigest result does not stay constant

    - by user344146
    I've got this function for encrypting passwords in Java, but somehow when I call MessageDigest, it returns a different result every time even though I call it with the same password. I wonder if I am initializing it wrong somehow. public String encrypt (String password) { MessageDigest md = MessageDigest.getInstance("SHA-1"); md.reset(); md.update(password.getBytes(Charset.forName("utf-8")),0,password.length()); String res = md.digest().toString(); }

    Read the article

  • .htaccess authentication from a php script to prevent a browser dialog box

    - by digitalbart
    Using php I authenticate a user, then behind the scenes,they are then again authenticated a second time with a single .htaccess username & password. This would be the same for all users, but I would not want them to have to enter a username and password again and they would now be allowed to enter the password protected directory. I prefer not to use http://username@password:somedomain.com. Any thoughts?

    Read the article

  • ProxyPass or intercept web requests .NET

    - by JerryO
    I have got a .NET winform application that uses a Web Mapping Service that is password protected. Unfortunately I cannot attach a username and password to requests, ( the request are generated from a GIS mapcontrol) I can think of two ways around it Intercept all web requests from my .NET app and add a username/password Set up an Apache webserver and use proxypass to pass along my request adding a username/password Does anyone know how to do either of these?

    Read the article

  • Error in create back up in mysql through java program

    - by Arivu2020
    Runtime.getRuntime().exec("C:\mysql\bin\mysqldump -u root -pmypassword Databasename -r C:/backup.sql"); I am using this code to create back up from my sql. but It creates the empty file in the path.Because it is waiting in the command prompt to get the password. How can i give password to it Using command prompt directly when i press enter after typing, it asks password.After giving password,It creates the backup.Give me any solution for this Thanks in advance

    Read the article

  • Setting up RADIUS + LDAP for WPA2 on Ubuntu

    - by Morten Siebuhr
    I'm setting up a wireless network for ~150 users. In short, I'm looking for a guide to set RADIUS server to authenticate WPA2 against a LDAP. On Ubuntu. I got a working LDAP, but as it is not in production use, it can very easily be adapted to whatever changes this project may require. I've been looking at FreeRADIUS, but any RADIUS server will do. We got a separate physical network just for WiFi, so not too many worries about security on that front. Our AP's are HP's low end enterprise stuff - they seem to support whatever you can think of. All Ubuntu Server, baby! And the bad news: I now somebody less knowledgeable than me will eventually take over administration, so the setup has to be as "trivial" as possible. So far, our setup is based only on software from the Ubuntu repositories, with exception of our LDAP administration web application and a few small special scripts. So no "fetch package X, untar, ./configure"-things if avoidable. UPDATE 2009-08-18: While I found several useful resources, there is one serious obstacle: Ignoring EAP-Type/tls because we do not have OpenSSL support. Ignoring EAP-Type/ttls because we do not have OpenSSL support. Ignoring EAP-Type/peap because we do not have OpenSSL support. Basically the Ubuntu version of FreeRADIUS does not support SSL (bug 183840), which makes all the secure EAP-types useless. Bummer. But some useful documentation for anybody interested: http://vuksan.com/linux/dot1x/802-1x-LDAP.html http://tldp.org/HOWTO/html_single/8021X-HOWTO/#confradius UPDATE 2009-08-19: I ended up compiling my own FreeRADIUS package yesterday evening - there's a really good recipe at http://www.linuxinsight.com/building-debian-freeradius-package-with-eap-tls-ttls-peap-support.html (See the comments to the post for updated instructions). I got a certificate from http://CACert.org (you should probably get a "real" cert if possible) Then I followed the instructions at http://vuksan.com/linux/dot1x/802-1x-LDAP.html. This links to http://tldp.org/HOWTO/html_single/8021X-HOWTO/, which is a very worthwhile read if you want to know how WiFi security works. UPDATE 2009-08-27: After following the above guide, I've managed to get FreeRADIUS to talk to LDAP: I've created a test user in LDAP, with the password mr2Yx36M - this gives an LDAP entry roughly of: uid: testuser sambaLMPassword: CF3D6F8A92967E0FE72C57EF50F76A05 sambaNTPassword: DA44187ECA97B7C14A22F29F52BEBD90 userPassword: {SSHA}Z0SwaKO5tuGxgxtceRDjiDGFy6bRL6ja When using radtest, I can connect fine: > radtest testuser "mr2Yx36N" sbhr.dk 0 radius-private-password Sending Access-Request of id 215 to 130.225.235.6 port 1812 User-Name = "msiebuhr" User-Password = "mr2Yx36N" NAS-IP-Address = 127.0.1.1 NAS-Port = 0 rad_recv: Access-Accept packet from host 130.225.235.6 port 1812, id=215, length=20 > But when I try through the AP, it doesn't fly - while it does confirm that it figures out the NT and LM passwords: ... rlm_ldap: sambaNTPassword -> NT-Password == 0x4441343431383745434139374237433134413232463239463532424542443930 rlm_ldap: sambaLMPassword -> LM-Password == 0x4346334436463841393239363745304645373243353745463530463736413035 [ldap] looking for reply items in directory... WARNING: No "known good" password was found in LDAP. Are you sure that the user is configured correctly? [ldap] user testuser authorized to use remote access rlm_ldap: ldap_release_conn: Release Id: 0 ++[ldap] returns ok ++[expiration] returns noop ++[logintime] returns noop [pap] Normalizing NT-Password from hex encoding [pap] Normalizing LM-Password from hex encoding ... It is clear that the NT and LM passwords differ from the above, yet the message [ldap] user testuser authorized to use remote access - and the user is later rejected...

    Read the article

  • How could I stop ssh offering a wrong key?

    - by Alvaro Maceda
    (This is a problem with ssh, not gitolite) I've configured gitolite on my home server (ubuntu 12.04 server, open-ssh). I want an special identityfile to administer the repositories, so I need to access throught ssh to my own host ussing two different identity keys. This is the content of my .ssh/config file: Host gitadmin.gammu.com User git IdentityFile /home/alvaro/.ssh/id_gitolite_mantra Host git.gammu.com User git IdentityFile /home/alvaro/.ssh/id_alvaro_mantra This is the content of my hosts file: # Git 127.0.0.1 gitadmin.gammu.com 127.0.0.1 git.gammu.com So I should be able to communicate with gitolite this way to access with the "normal" account: $ssh git.gammu.com and this way to access with the administrative account: $ssh gitadmin.gammu.com When I try to access with the normal account, all is ok: alvaro@mantra:~/.ssh$ ssh git.gammu.com PTY allocation request failed on channel 0 hello alvaro, this is gitolite 2.2-1 (Debian) running on git 1.7.9.5 the gitolite config gives you the following access: @R_ @W_ testing Connection to git.gammu.com closed. When I do the same with the administrative account: alvaro@mantra:~$ ssh gitadmin.gammu.com PTY allocation request failed on channel 0 hello alvaro, this is gitolite 2.2-1 (Debian) running on git 1.7.9.5 the gitolite config gives you the following access: @R_ @W_ testing Connection to gitadmin.gammu.com closed. It should show the administrative repository. If I launch ssh with verbose option: ssh -vvv gitadmin.gammu.com ... debug1: SSH2_MSG_SERVICE_REQUEST sent debug2: service_accept: ssh-userauth debug1: SSH2_MSG_SERVICE_ACCEPT received debug2: key: /home/alvaro/.ssh/id_alvaro_mantra (0x7f7cb6c0fbc0) debug2: key: /home/alvaro/.ssh/id_gitolite_mantra (0x7f7cb6c044d0) debug1: Authentications that can continue: publickey,password debug3: start over, passed a different list publickey,password debug3: preferred gssapi-keyex,gssapi-with-mic,publickey,keyboard-interactive,password debug3: authmethod_lookup publickey debug3: remaining preferred: keyboard-interactive,password debug3: authmethod_is_enabled publickey debug1: Next authentication method: publickey debug1: Offering RSA public key: /home/alvaro/.ssh/id_alvaro_mantra debug3: send_pubkey_test debug2: we sent a publickey packet, wait for reply debug1: Server accepts key: pkalg ssh-rsa blen 279 ... It's offering the key id_alvaro_mantra, and it should'nt!! The same happens when I specify the key with the -i option: ssh -i /home/alvaro/.ssh/id_gitolite_mantra -vvv gitadmin.gammu.com ... debug1: SSH2_MSG_SERVICE_REQUEST sent debug2: service_accept: ssh-userauth debug1: SSH2_MSG_SERVICE_ACCEPT received debug2: key: /home/alvaro/.ssh/id_alvaro_mantra (0x7fa365237f90) debug2: key: /home/alvaro/.ssh/id_gitolite_mantra (0x7fa365230550) debug2: key: /home/alvaro/.ssh/id_gitolite_mantra (0x7fa365231050) debug1: Authentications that can continue: publickey,password debug3: start over, passed a different list publickey,password debug3: preferred gssapi-keyex,gssapi-with-mic,publickey,keyboard-interactive,password debug3: authmethod_lookup publickey debug3: remaining preferred: keyboard-interactive,password debug3: authmethod_is_enabled publickey debug1: Next authentication method: publickey debug1: Offering RSA public key: /home/alvaro/.ssh/id_alvaro_mantra debug3: send_pubkey_test debug2: we sent a publickey packet, wait for reply debug1: Server accepts key: pkalg ssh-rsa blen 279 debug2: input_userauth_pk_ok: fp 36:b1:43:36:af:4f:00:e5:e1:39:50:7e:07:80:14:26 debug3: sign_and_send_pubkey: RSA 36:b1:43:36:af:4f:00:e5:e1:39:50:7e:07:80:14:26 debug1: Authentication succeeded (publickey). ... What the hell is happening??? I'm missing something, but I can't find what. These are the contents of my home dir: -rw-rw-r-- 1 alvaro alvaro 395 nov 14 18:00 authorized_keys -rw-rw-r-- 1 alvaro alvaro 326 nov 21 10:21 config -rw------- 1 alvaro alvaro 137 nov 20 20:26 environment -rw------- 1 alvaro alvaro 1766 nov 20 21:41 id_alvaromaceda.es -rw-r--r-- 1 alvaro alvaro 404 nov 20 21:41 id_alvaromaceda.es.pub -rw------- 1 alvaro alvaro 1766 nov 14 17:59 id_alvaro_mantra -rw-r--r-- 1 alvaro alvaro 395 nov 14 17:59 id_alvaro_mantra.pub -rw------- 1 alvaro alvaro 771 nov 14 18:03 id_developer_mantra -rw------- 1 alvaro alvaro 1679 nov 20 12:37 id_dos_pruebasgit -rw-r--r-- 1 alvaro alvaro 395 nov 20 12:37 id_dos_pruebasgit.pub -rw------- 1 alvaro alvaro 1679 nov 20 12:46 id_gitolite_mantra -rw-r--r-- 1 alvaro alvaro 397 nov 20 12:46 id_gitolite_mantra.pub -rw------- 1 alvaro alvaro 1675 nov 20 21:44 id_gitpruebas.es -rw-r--r-- 1 alvaro alvaro 408 nov 20 21:44 id_gitpruebas.es.pub -rw------- 1 alvaro alvaro 1679 nov 20 12:34 id_uno_pruebasgit -rw-r--r-- 1 alvaro alvaro 395 nov 20 12:34 id_uno_pruebasgit.pub -rw-r--r-- 1 alvaro alvaro 2434 nov 21 10:11 known_hosts There are a bunch of other keys which aren't offered... why id_alvaro_mantra is offered and not the other keys? I can't understand. I need some help, don't know where to look....

    Read the article

  • SSH service will not start on fresh Cygwin 1.7.15 install

    - by Coder6841
    OS: Windows 7 x64 Cygwin: 1.7.15-1 OpenSSH: 6.0p1-1 I'm attempting to install an SSH server on Windows 7. The tutorial that I'm following to do this is here: http://www.howtogeek.com/howto/41560/how-to-get-ssh-command-line-access-to-windows-7-using-cygwin/ The issue is that upon executing the net start sshd command I get the following output:The CYGWIN sshd service is starting. The CYGWIN sshd service could not be started. The service did not report an error. More help is available by typing NET HELPMSG 3534. Here is the full output of the setup: AdminUser@ThisComputer ~ $ ssh-host-config *** Info: Generating /etc/ssh_host_key *** Info: Generating /etc/ssh_host_rsa_key *** Info: Generating /etc/ssh_host_dsa_key *** Info: Generating /etc/ssh_host_ecdsa_key *** Info: Creating default /etc/ssh_config file *** Info: Creating default /etc/sshd_config file *** Info: Privilege separation is set to yes by default since OpenSSH 3.3. *** Info: However, this requires a non-privileged account called 'sshd'. *** Info: For more info on privilege separation read /usr/share/doc/openssh/README.privsep. *** Query: Should privilege separation be used? (yes/no) yes *** Info: Note that creating a new user requires that the current account have *** Info: Administrator privileges. Should this script attempt to create a *** Query: new local account 'sshd'? (yes/no) yes *** Info: Updating /etc/sshd_config file *** Query: Do you want to install sshd as a service? *** Query: (Say "no" if it is already installed as a service) (yes/no) yes *** Query: Enter the value of CYGWIN for the daemon: [] *** Info: On Windows Server 2003, Windows Vista, and above, the *** Info: SYSTEM account cannot setuid to other users -- a capability *** Info: sshd requires. You need to have or to create a privileged *** Info: account. This script will help you do so. *** Info: You appear to be running Windows XP 64bit, Windows 2003 Server, *** Info: or later. On these systems, it's not possible to use the LocalSystem *** Info: account for services that can change the user id without an *** Info: explicit password (such as passwordless logins [e.g. public key *** Info: authentication] via sshd). *** Info: If you want to enable that functionality, it's required to create *** Info: a new account with special privileges (unless a similar account *** Info: already exists). This account is then used to run these special *** Info: servers. *** Info: Note that creating a new user requires that the current account *** Info: have Administrator privileges itself. *** Info: No privileged account could be found. *** Info: This script plans to use 'cyg_server'. *** Info: 'cyg_server' will only be used by registered services. *** Query: Do you want to use a different name? (yes/no) no *** Query: Create new privileged user account 'cyg_server'? (yes/no) yes *** Info: Please enter a password for new user cyg_server. Please be sure *** Info: that this password matches the password rules given on your system. *** Info: Entering no password will exit the configuration. *** Query: Please enter the password: *** Query: Reenter: *** Info: User 'cyg_server' has been created with password '[CENSORED]'. *** Info: If you change the password, please remember also to change the *** Info: password for the installed services which use (or will soon use) *** Info: the 'cyg_server' account. *** Info: Also keep in mind that the user 'cyg_server' needs read permissions *** Info: on all users' relevant files for the services running as 'cyg_server'. *** Info: In particular, for the sshd server all users' .ssh/authorized_keys *** Info: files must have appropriate permissions to allow public key *** Info: authentication. (Re-)running ssh-user-config for each user will set *** Info: these permissions correctly. [Similar restrictions apply, for *** Info: instance, for .rhosts files if the rshd server is running, etc]. *** Info: The sshd service has been installed under the 'cyg_server' *** Info: account. To start the service now, call `net start sshd' or *** Info: `cygrunsrv -S sshd'. Otherwise, it will start automatically *** Info: after the next reboot. *** Info: Host configuration finished. Have fun! AdminUser@ThisComputer ~ $ net start sshd The CYGWIN sshd service is starting. The CYGWIN sshd service could not be started. The service did not report an error. More help is available by typing NET HELPMSG 3534. Note that on the line *** Query: Enter the value of CYGWIN for the daemon: [] I haven't entered anything. Tutorials often say to use ntsec or ntsec tty here but those options are removed from the latest version of OpenSSH. I've tried using them anyway and the result is the same. The file /var/log/sshd.log is empty. If I try just running the command /usr/sbin/sshd I get the output /var/empty must be owned by root and not group or world-writable.. The /var/empty directory has the following permissions: drwxr-xr-x+ 1 cyg_server root 0 May 29 15:28 empty. Google searches on this error did not turn up any working fixes. One person seems to have solved it by using the command chown SYSTEM /var/empty but that did not fix it in my case.

    Read the article

  • (PHP) User is being forced to RE-LOGIN after trying to do something on an admin page

    - by hatorade
    I have created an admin panel for a client in PHP, which requires a login. Here is the code at the top of the admin page requiring the user to be logged in: admin.php <?php session_start(); require("_lib/session_functions.php"); require("_lib/db.php"); db_connect(); //if the user has not logged in if(!isLoggedIn()) { header('Location: login_form.php'); die(); } ?> Obviously, the if statement is what catches them and forces them to log in. Here is the code on the resulting login page: login_form.php <form name="login" action="login.php" method="post"> Username: <input type="text" name="username" /> Password: <input type="password" name="password" /> <input type="submit" value="Login" /> </form> Which posts info to this controller page: login.php <?php session_start(); //must call session_start before using any $_SESSION variables include '_lib/session_functions.php'; $username = $_POST['username']; $password = $_POST['password']; include '_lib/db.php'; db_connect(); // Connect to the DB $username = mysql_real_escape_string($username); $query = "SELECT password, salt FROM users WHERE username = '$username';"; $result = mysql_query($query); if(mysql_num_rows($result) < 1) //no such user exists { header('Location: login_form.php?login=fail'); die(); } $userData = mysql_fetch_array($result, MYSQL_ASSOC); db_disconnect(); $hash = hash('sha256', $password . $userData['salt']); if($hash != $userData['password']) //incorrect password { header('Location: login_form.php?login=fail'); die(); } else { validateUser(); //sets the session data for this user } header('Location: admin.php'); ?> and the session functions page that provides login functions contains this: session_functions.php <?php function validateUser() { session_regenerate_id (); //this is a security measure $_SESSION['valid'] = 1; $_SESSION['userid'] = $username; } function isLoggedIn() { if($_SESSION['valid']) return true; return false; } function logout() { $_SESSION = array(); //destroy all of the session variables if (ini_get("session.use_cookies")) { $params = session_get_cookie_params(); setcookie(session_name(), '', time() - 42000, $params["path"], $params["domain"], $params["secure"], $params["httponly"] ); } session_destroy(); } ?> I grabbed the sessions_functions.php code of an online tutorial, so it could be suspicious. Any ideas why the user logs in to the admin panel, tries to do something, is forced to re-login, and THEN is allowed to do stuff like normal in the admin panel?

    Read the article

  • Html LogIn form not functioning

    - by Tony C
    Ok, I have a login form that looks like this: <form id="loginForm" name="loginForm" method="post" action="login-exec.php"> <table width="300" border="0" align="center" cellpadding="2" cellspacing="0"> <tr> <td width="112"><b>Login</b></td> <td width="188"><input name="login" type="text" class="textfield" id="login" /></td> </tr> <tr> <td><b>Password</b></td> <td><input name="password" type="password" class="textfield" id="password" /></td> </tr> <tr> <td>&nbsp;</td> <td><input type="submit" name="Submit" value="Login" /></td> </tr> </table> </form> Now, This form is on a page in a directory called members. When i put it on a page in the home directory and change the action to "members/login-exec.php" When I try to logIn it just refreshes the page, but the name of the page in the browser changes to the actions taking place in the form. Any ideas on making this work guys? EDIT, heres the login-exec.php code: <?php //Start session session_start(); //Include database connection details require_once('config.php'); //Array to store validation errors $errmsg_arr = array(); //Validation error flag $errflag = false; //Connect to mysql server $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } //Select database $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } //Function to sanitize values received from the form. Prevents SQL injection function clean($str) { $str = @trim($str); if(get_magic_quotes_gpc()) { $str = stripslashes($str); } return mysql_real_escape_string($str); } //Sanitize the POST values $login = clean($_POST['login']); $password = clean($_POST['password']); //Input Validations if($login == '') { $errmsg_arr[] = 'Login ID missing'; $errflag = true; } if($password == '') { $errmsg_arr[] = 'Password missing'; $errflag = true; } //If there are input validations, redirect back to the login form if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; session_write_close(); header("location: login-form.php"); exit(); } //Create query $qry="SELECT * FROM members WHERE login='$login' AND passwd='".md5($_POST['password'])."'"; $result=mysql_query($qry); //Check whether the query was successful or not if($result) { if(mysql_num_rows($result) == 1) { //Login Successful session_regenerate_id(); $member = mysql_fetch_assoc($result); $_SESSION['SESS_MEMBER_ID'] = $member['member_id']; $_SESSION['SESS_FIRST_NAME'] = $member['firstname']; $_SESSION['SESS_LAST_NAME'] = $member['lastname']; session_write_close(); header("location: members.php"); exit(); }else { //Login failed header("location: login-failed.php"); exit(); } }else { die("Query failed"); } ?>

    Read the article

  • I'm trying to pass a string from my first ViewController to my second ViewController but it returns NULL

    - by Dashony
    In my first view controller I have 3 input fields each of them take the user input into and saves it into a string such as: address, username and password as NSUserDefaults. This part works fine. In my second view controller I'm trying to take the 3 strings from first controller (address, username and password) create a html link based on the 3 strings. I've tried many ways to access the 3 strings with no luck, the result I get is NULL. Here is my code: //.h file - first view controller with the 3 input fields CamSetup.h #import <UIKit/UIKit.h> @interface CamSetup : UIViewController <UITextFieldDelegate> { NSString * address; NSString * username; NSString * password; IBOutlet UITextField * addressField; IBOutlet UITextField * usernameField; IBOutlet UITextField * passwordField; } -(IBAction) saveAddress: (id) sender; -(IBAction) saveUsername: (id) sender; -(IBAction) savePassword: (id) sender; @property(nonatomic, retain) UITextField *addressField; @property(nonatomic, retain) UITextField *usernameField; @property(nonatomic, retain) UITextField *passwordField; @property(nonatomic, retain) NSString *address; @property(nonatomic, retain) NSString *username; @property(nonatomic, retain) NSString *password; @end //.m file - first view controller CamSetup.m #import "CamSetup.h" @interface CamSetup () @end @implementation CamSetup @synthesize addressField, usernameField, passwordField, address, username, password; -(IBAction) saveAddress: (id) sender { address = [[NSString alloc] initWithFormat:addressField.text]; [addressField setText:address]; NSUserDefaults *stringDefaultAddress = [NSUserDefaults standardUserDefaults]; [stringDefaultAddress setObject:address forKey:@"stringKey1"]; NSLog(@"String [%@]", address); } -(IBAction) saveUsername: (id) sender { username = [[NSString alloc] initWithFormat:usernameField.text]; [usernameField setText:username]; NSUserDefaults *stringDefaultUsername = [NSUserDefaults standardUserDefaults]; [stringDefaultUsername setObject:username forKey:@"stringKey2"]; NSLog(@"String [%@]", username); } -(IBAction) savePassword: (id) sender { password = [[NSString alloc] initWithFormat:passwordField.text]; [passwordField setText:password]; NSUserDefaults *stringDefaultPassword = [NSUserDefaults standardUserDefaults]; [stringDefaultPassword setObject:password forKey:@"stringKey3"]; NSLog(@"String [%@]", password); } - (void)viewDidLoad { [addressField setText:[[NSUserDefaults standardUserDefaults] objectForKey:@"stringKey1"]]; [usernameField setText:[[NSUserDefaults standardUserDefaults] objectForKey:@"stringKey2"]]; [passwordField setText:[[NSUserDefaults standardUserDefaults] objectForKey:@"stringKey3"]]; [super viewDidLoad]; } @end //.h second view controller LiveView.h #import <UIKit/UIKit.h> #import "CamSetup.h" @interface LiveView : UIViewController { NSString *theAddress; NSString *theUsername; NSString *thePassword; CamSetup *camsetup; //here is an instance of the first class } @property (nonatomic, retain) NSString *theAddress; @property (nonatomic, retain) NSString *theUsername; @property (nonatomic, retain) NSString *thePassword; @end //.m second view LiveView.m file #import "LiveView.h" @interface LiveView () @end @implementation LiveView @synthesize theAddress, theUsername, thePassword; - (void)viewDidLoad { [super viewDidLoad]; theUsername = camsetup.username; //this is probably not right? NSLog(@"String [%@]", theUsername); //resut here is NULL NSLog(@"String [%@]", camsetup.username); //and here NULL as well } @end

    Read the article

  • How to generate a random unique string with more than 2^30 combination. I also wanted to reverse the process. Is this possible?

    - by Yusuf S
    I have a string which contains 3 elements: a 3 digit code (example: SIN, ABD, SMS, etc) a 1 digit code type (example: 1, 2, 3, etc) a 3 digit number (example: 500, 123, 345) Example string: SIN1500, ABD2123, SMS3345, etc.. I wanted to generate a UNIQUE 10 digit alphanumeric and case sensitive string (only 0-9/a-z/A-Z is allowed), with more than 2^30 (about 1 billion) unique combination per string supplied. The generated code must have a particular algorithm so that I can reverse the process. For example: public static void main(String[] args) { String test = "ABD2123"; String result = generateData(test); System.out.println(generateOutput(test)); //for example, the output of this is: 1jS8g4GDn0 System.out.println(generateOutput(result)); //the output of this will be ABD2123 (the original string supplied) } What I wanted to ask is is there any ideas/examples/libraries in java that can do this? Or at least any hint on what keyword should I put on Google? I tried googling using the keyword java checksum, rng, security, random number, etc and also tried looking at some random number solution (java SecureRandom, xorshift RNG, java.util.zip's checksum, etc) but I can't seem to find one? Thanks! EDIT: My use case for this program is to generate some kind of unique voucher number to be used by specific customers. The string supplied will contains 3 digit code for company ID, 1 digit code for voucher type, and a 3 digit number for the voucher nominal. I also tried adding 3 random alphanumeric (so the final digit is 7 + 3 digit = 10 digit). This is what I've done so far, but the result is not very good (only about 100 thousand combination): public static String in ="somerandomstrings"; public static String out="someotherrandomstrings"; public static String encrypt(String kata) throws Exception { String result=""; String ina=in; String outa=out; Random ran = new Random(); Integer modulus=in.length(); Integer offset= ((Integer.parseInt(Utils.convertDateToString(new Date(), "SS")))+ran.nextInt(60))/2%modulus; result=ina.substring(offset, offset+1); ina=ina+ina; ina=ina.substring(offset, offset+modulus); result=result+translate(kata, ina, outa); return result; } EDIT: I'm sorry I forgot to put the "translate" function : public static String translate(String kata,String seq1, String seq2){ String result=""; if(kata!=null&seq1!=null&seq2!=null){ String[] a=kata.split(""); for (int j = 1; j < a.length; j++) { String b=a[j]; String[]seq1split=seq1.split(""); String[]seq2split=seq2.split(""); int hint=seq1.indexOf(b)+1; String sq=""; if(seq1split.length>hint) sq=seq1split[hint]; String sq1=""; if(seq2split.length>hint) sq1=seq2split[hint]; b=b.replace(sq, sq1); result=result+b; } } return result; }

    Read the article

  • Why can't I reinstall MySQL?

    - by Johannes Nielsen
    I've been looking all around the Internet for an answer but didn't find anything. I hope you can help me now. I have a server with MySQL. From one day to another, MySQL didn't let me enter with my root password anymore (accsess denied for user 'root'@'localhost' using password: 'YES'). So I tried two ways to reset the password: No.1: I typed: shell> /etc/init.d/mysqld stop To stop MySQL. Then I restarted it skipping the grant-tables: shell> mysqld_safe --skip-grant-tables So I was able to log in as root and change the password using: mysql> UPDATE mysql.user SET Password = PASSWORD('MyNewPassword') WHERE User = 'root'; FLUSH PRIVILEGES; I restarted MySQL and tried to log in as root with my new password - didn't work. So I tried the solution that's described here: http://dev.mysql.com/doc/refman/5.0/en/resetting-permissions.html (I don't want to post it here because this post is already pretty long). Didn't work either. Actually it made it worse, because since that day, every time I try to start MySQL, it doesn't even ask me for my password, but I get: shell> ERROR 2002 (HY000): Can't connect to local MySQL server through socket '/var/run/mysqld/mysqld.sock' (111) Well, I've looked up what it means and found that my mysqld.sock is missing. I tried to create it using touch but MySQL can't start with that socket. Now I'm trying to reinstall MySQL but everytime I type in shell> apt-get --purge remove mysql-server mysql-common mysql-client In that or any other order or every one of those three alone, I get: shell> Reading package lists... Done shell> Building dependency tree shell> Reading state information... Done shell> Package mysql-client is not installed, so not removed shell> Package mysql-server is not installed, so not removed shell> You might want to run 'apt-get -f install' to correct these: shell> The following packages have unmet dependencies: shell> libmysqlclient18 : Depends: mysql-common (>= 5.5.28-0ubuntu0.12.04.2) but it is not going to be installed shell> libmysqlclient18:i386 : Depends: mysql-common:i386 (>= 5.5.28-0ubuntu0.12.04.2) shell> mysql-client-5.5 : Depends: mysql-common (>= 5.5.28-0ubuntu0.12.04.2) but it is not going to be installed shell> mysql-server-5.5 : PreDepends: mysql-common (>= 5.5.28-0ubuntu0.12.04.2) but it is not going to be installed shell> psa-firewall : Depends: plesk-core (>= 11.0.9) but it is not installable shell> Depends: mysql-server but it is not going to be installed shell> psa-spamassassin : Depends: plesk-core (>= 11.0.9) but it is not installable shell> psa-vpn : Depends: plesk-core (>= 11.0.9) but it is not installable shell> Depends: plesk-base (>= 11.0.9) but it is not installable shell> Depends: mysql-server but it is not going to be installed shell> E: Unmet dependencies. Try 'apt-get -f install' with no packages (or specify a solution). So I said to my self "let's just remove those files with depenencies, too" (that psa-stuff since plesk is virtual and can't be uninstalled)... Guess what happened: shell> Reading package lists... Done shell> Building dependency tree shell> Reading state information... Done shell> Package mysql-client is not installed, so not removed shell> Package mysql-server is not installed, so not removed shell> You might want to run 'apt-get -f install' to correct these: shell> The following packages have unmet dependencies: shell> libmysqlclient18 : Depends: mysql-common (>= 5.5.28-0ubuntu0.12.04.2) but it is not going to be installed shell> libmysqlclient18:i386 : Depends: mysql-common:i386 (>= 5.5.28-0ubuntu0.12.04.2) shell> mysql-client-5.5 : Depends: mysql-common (>= 5.5.28-0ubuntu0.12.04.2) but it is not going to be installed shell> mysql-server-5.5 : PreDepends: mysql-common (>= 5.5.28-0ubuntu0.12.04.2) but it is not going to be installed shell> E: Unmet dependencies. Try 'apt-get -f install' with no packages (or specify a solution). Of course I tried apt-get -f install, too many times even. What am I doing wrong? No matter, which other packages I include into apt-get --purge remove, I always get new dependencies. Do I have to delete every MySQL-related directory and file manually? Hope there's someone out there who can help me! Cheers! EDIT: After trying apt-get purge mysql-server mysql-common mysql-client libmysqlclient18 libmysqlclient18:i386 mysql-client-5.5 mysql-server-5.5 psa-firewall psa-spamassassin psa-vpn Reading package lists... Done Building dependency tree Reading state information... Done Package mysql-client is not installed, so not removed Package mysql-server is not installed, so not removed You might want to run 'apt-get -f install' to correct these: The following packages have unmet dependencies: libdbd-mysql-perl : Depends: libmysqlclient18 (>= 5.5.13-1) but it is not going to be installed libmyodbc : Depends: libmysqlclient18 (>= 5.5.13-1) but it is not going to be installed libqt4-sql-mysql:i386 : Depends: libmysqlclient18:i386 (>= 5.5.13-1) but it is not going to be installed php5-mysql : Depends: libmysqlclient18 (>= 5.5.13-1) but it is not going to be installed ruby-mysql : Depends: libmysqlclient18 (>= 5.5.13-1) but it is not going to be installed E: Unmet dependencies. Try 'apt-get -f install' with no packages (or specify a solution). So I tried to remove all these and got: Building dependency tree Reading state information... Done Package mysql-client is not installed, so not removed Package mysql-server is not installed, so not removed You might want to run 'apt-get -f install' to correct these:qlclient18:i386 mysql The following packages have unmet dependencies: libmysql-ruby1.8 : Depends: ruby-mysql but it is not going to be installed E: Unmet dependencies. Try 'apt-get -f install' with no packages (or specify a solution). And actually I think removing that file, too solved my problem :-S Next time I'll try everything before asking :D Thank you Eric for keeping me couraged to just go on removing :D

    Read the article

  • Controlar Autentificaci&oacute;n Crystal Reports

    - by Jason Ulloa
    Para todos los que hemos trabajamos con Crystal Reports, no es un secreto que cuando tratamos de conectar nuestro reporte directamente a la base de datos, se nos viene encima el problema de autenticación. Es decir nuestro reporte al momento de iniciar la carga nos solicita autentificarnos en el servidor y sino lo hacemos, simplemente no veremos el reporte. Esto, además de ser tedioso para los usuarios se convierte en un problema de seguridad bastante grande, de ahí que en la mayoría de los casos se recomienda utilizar dataset. Sin embargo, para todos los que aún sabiendo esto no desean utilizar datasets, sino que, quieren conectar su crystal directamente veremos como implementar una pequeña clase que nos ayudará con esa tarea. Generalmente, cuando trabajamos con una aplicación web, nuestra cadena de conexión esta incluida en el web.config y también en muchas ocasiones contiene los datos como el usuario y password para acceder a la base de datos.  De esta cadena de conexión y estos datos es de los que nos ayudaremos para implementar la autentificación en el reporte. Generalmente, la cadena de conexión se vería así <connectionStrings> <remove name="LocalSqlServer"/> <add name="xxx" connectionString="Data Source=.\SqlExpress;Integrated Security=False;Initial Catalog=xxx;user id=myuser;password=mypass" providerName="System.Data.SqlClient"/> </connectionStrings>   Para nuestro ejemplo, nombraremos a nuestra clase CrystalRules (es solo algo que pensé de momento) 1. Primer Paso Creamos una variable de tipo SqlConnectionStringBuilder, a la cual le asignaremos la cadena de conexión que definimos en el web.config, y que luego utilizaremos para obtener los datos del usuario y el password para el crystal report. SqlConnectionStringBuilder builder = new SqlConnectionStringBuilder(ConfigurationManager.ConnectionStrings["xxx"].ConnectionString); 2. Implementación de propiedad Para ser más ordenados crearemos varias propiedad de tipo Privado, que se encargarán de recibir los datos de:   La Base de datos, el password, el usuario y el servidor private string _dbName; private string _serverName; private string _userID; private string _passWord;   private string dataBase { get { return _dbName; } set { _dbName = value; } }   private string serverName { get { return _serverName; } set { _serverName = value; } }   private string userName { get { return _userID; } set { _userID = value; } }   private string dataBasePassword { get { return _passWord; } set { _passWord = value; } } 3. Creación del Método para aplicar los datos de conexión Una vez que ya tenemos las propiedades, asignaremos a las variables los valores que se han recogido en el SqlConnectionStringBuilder. Y crearemos una variable de tipo ConnectionInfo para aplicar los datos de conexión. internal void ApplyInfo(ReportDocument _oRpt) { dataBase = builder.InitialCatalog; serverName = builder.DataSource; userName = builder.UserID; dataBasePassword = builder.Password;   Database oCRDb = _oRpt.Database; Tables oCRTables = oCRDb.Tables; //Table oCRTable = default(Table); TableLogOnInfo oCRTableLogonInfo = default(TableLogOnInfo); ConnectionInfo oCRConnectionInfo = new ConnectionInfo();   oCRConnectionInfo.DatabaseName = _dbName; oCRConnectionInfo.ServerName = _serverName; oCRConnectionInfo.UserID = _userID; oCRConnectionInfo.Password = _passWord;   foreach (Table oCRTable in oCRTables) { oCRTableLogonInfo = oCRTable.LogOnInfo; oCRTableLogonInfo.ConnectionInfo = oCRConnectionInfo; oCRTable.ApplyLogOnInfo(oCRTableLogonInfo);     }   }   4. Creación del report document y aplicación de la seguridad Una vez recogidos los datos y asignados, crearemos un elemento report document al cual le asignaremos el CrystalReportViewer y le aplicaremos los datos de acceso que obtuvimos anteriormente public void loadReport(string repName, CrystalReportViewer viewer) {   // attached our report to viewer and set database login. ReportDocument report = new ReportDocument(); report.Load(HttpContext.Current.Server.MapPath("~/Reports/" + repName)); ApplyInfo(report); viewer.ReportSource = report; } Al final, nuestra clase completa ser vería así public class CrystalRules { SqlConnectionStringBuilder builder = new SqlConnectionStringBuilder(ConfigurationManager.ConnectionStrings["Fatchoy.Data.Properties.Settings.FatchoyConnectionString"].ConnectionString);   private string _dbName; private string _serverName; private string _userID; private string _passWord;   private string dataBase { get { return _dbName; } set { _dbName = value; } }   private string serverName { get { return _serverName; } set { _serverName = value; } }   private string userName { get { return _userID; } set { _userID = value; } }   private string dataBasePassword { get { return _passWord; } set { _passWord = value; } }   internal void ApplyInfo(ReportDocument _oRpt) { dataBase = builder.InitialCatalog; serverName = builder.DataSource; userName = builder.UserID; dataBasePassword = builder.Password;   Database oCRDb = _oRpt.Database; Tables oCRTables = oCRDb.Tables; //Table oCRTable = default(Table); TableLogOnInfo oCRTableLogonInfo = default(TableLogOnInfo); ConnectionInfo oCRConnectionInfo = new ConnectionInfo();   oCRConnectionInfo.DatabaseName = _dbName; oCRConnectionInfo.ServerName = _serverName; oCRConnectionInfo.UserID = _userID; oCRConnectionInfo.Password = _passWord;   foreach (Table oCRTable in oCRTables) { oCRTableLogonInfo = oCRTable.LogOnInfo; oCRTableLogonInfo.ConnectionInfo = oCRConnectionInfo; oCRTable.ApplyLogOnInfo(oCRTableLogonInfo);     }   }   public void loadReport(string repName, CrystalReportViewer viewer) {   // attached our report to viewer and set database login. ReportDocument report = new ReportDocument(); report.Load(HttpContext.Current.Server.MapPath("~/Reports/" + repName)); ApplyInfo(report); viewer.ReportSource = report; }       #region instance   private static CrystalRules m_instance;   // Properties public static CrystalRules Instance { get { if (m_instance == null) { m_instance = new CrystalRules(); } return m_instance; } }   public DataDataContext m_DataContext { get { return DataDataContext.Instance; } }     #endregion instance   }   Si bien, la solución no es robusta y no es la mas segura. En casos de uso como una intranet y cuando estamos contra tiempo, podría ser de gran ayuda.

    Read the article

  • MySQL for Excel 1.1.3 has been released

    - by Javier Treviño
    The MySQL Windows Experience Team is proud to announce the release of MySQL for Excel version 1.1.3, the  latest addition to the MySQL Installer for Windows. MySQL for Excel is an application plug-in enabling data analysts to very easily access and manipulate MySQL data within Microsoft Excel. It enables you to directly work with a MySQL database from within Microsoft Excel so you can easily do tasks such as: Importing MySQL Data into Excel Exporting Excel data directly into MySQL to a new or existing table Editing MySQL data directly within Excel MySQL for Excel is installed using the MySQL Installer for Windows. The MySQL installer comes in 2 versions   Full (150 MB) which includes a complete set of MySQL products with their binaries included in the download Web (1.5 MB - a network install) which will just pull MySQL for Excel over the web and install it when run.   You can download MySQL Installer from our official Downloads page at http://dev.mysql.com/downloads/installer/. MySQL for Excel 1.1.3 introduces the following features:   Upon saving a Workbook containing Worksheets in Edit Mode, the user is asked if he wants to exit the Edit Mode on all Worksheets before their parent Workbook is saved so the Worksheets are saved unprotected, otherwise the Worksheets will remain protected and the users will be able to unprotect them later retrieving the passkeys from the application log after closing MySQL for Excel. Added background coloring to the column names header row of an Import Data operation to have the same look as the one in an Edit Data operation (i.e. gray-ish background). Connection passwords can be stored securely just like MySQL Workbench does and these secured passwords are shared with Workbench in the same way connections are. Changed the way the MySQL for Excel ribbon toggle button works, instead of just showing or hiding the add-in it actually opens and closes it. Added a connection test before any operation against the database (schema creation, data import, append, export or edition) so the operation dialog is not shown and a friendlier error message is shown.   Also this release contains the following bug fixes:   Added a check on every connection test for an expired password, if the password has been expired a dialog is now shown to the user to reset the password. Bug #17354118 - DON'T HANDLE EXPIRED PASSWORDS Added code to escape text values to be imported to an Excel worksheet that start with an equals sign so Excel does not treat those values as formulas that will fail evaluation. This is an option turned on by default that can be turned off by users if they wish to import values to be treated as Excel formulas. Bug #17354102 - ERROR IMPORTING TEXT VALUES TO EXCEL STARTING WITH AN EQUALS SIGN Added code to properly check the reason for a failing connection, if it's a failing password the user gets a dialog to retry the connection with a different password until the connection succeeds, a connection error not related to the password is thrown or the user cancels. If the failing connection is not related to a bad password an error message is shown to the users indicating the reason of the failure. Bug #16239007 - CONNECTIONS TO MYSQL SERVICES NOT RUNNING DISPLAY A WRONG PASSWORD ERROR MESSAGE Added global options dialog that can be accessed from the Schema Selection and DB Object Selection panels where the timeouts for the connection to the DB Server and for the query commands can be changed from their default values (15 seconds for the connection timeout and 30 seconds for the query timeout). MySQL Bug #68732, Bug #17191646 - QUERY TIMEOUT CANNOT BE ADJUSTED IN MYSQL FOR EXCEL Changed the Varchar(65,535) data type shown in the Export Data data type combo box to Text since the maximum row size is 65,535 bytes and any autodetected column data type with a length greater than 4,000 should be set to Text actually for the table to be created successfully. MySQL Bug #69779, Bug #17191633 - EXPORT FAILS FOR EXCEL FILES CONTAINING > 4000 CHARACTERS OF TEXT PER CELL Removed code that was replacing all spaces typed by the user in an overriden data type for a new column in an Export Data operation, also improved the data type detection code to flag as invalid data types with parenthesis but without any text inside or where the contents inside the parenthesis are not valid for the specific data type. Bug #17260260 - EXPORT DATA SET TYPE NOT WORKING WITH MEMBER VALUES CONTAINING SPACES Added support for the year data type with a length of 2 or 4 and a validation that valid values are integers between 1901-2155 (for 4-digit years) or between 0-99 (for 2-digit years). Bug #17259915 - EXPORT DATA YEAR DATA TYPE NOT RECOGNIZED IF DECLARED WITH A DISPLAY WIDTH) Fixed code for Export Data operations where users overrode the data type for columns typing Text in the data type combobox, which is a valid data type but was not recognized as such. Bug #17259490 - EXPORT DATA TEXT DATA TYPE NOT RECOGNIZED AS A VALID DATA TYPE Changed the location of the registry where the MySQL for Excel add-in is installed to HKEY_LOCAL_MACHINE instead of HKEY_CURRENT_USER so the add-in is accessible by all users and not only to the user that installed it. For this to work with Excel 2007 a hotfix may be required (see http://support.microsoft.com/kb/976477). MySQL Bug #68746, Bug #16675992 - EXCEL-ADD-IN IS ONLY INSTALLED FOR USER ACCOUNT THAT THE INSTALLATION RUNS UNDER Added support for Excel 2013 Single Document Interface, now that Excel 2013 creates 1 window per workbook also the Excel Add-In maintains an independent custom task pane in each window. MySQL Bug #68792, Bug #17272087 - MYSQL FOR EXCEL SIDEBAR DOES NOT APPEAR IN EXCEL 2013 (WITH WORKAROUND) Included the latest MySQL Utility with a code fix for the COM exception thrown when attempting to open Workbench in the Manage Connections window. Bug #17258966 - MYSQL WORKBENCH NOT OPENED BY CLICKING MANAGE CONNECTIONS HOTLABEL Fixed code for Append Data operations that was not applying a calculated automatic mapping correctly when the source and target tables had different number of columns, some columns with the same name but some of those lying on column indexes beyond the limit of the other source/target table. MySQL Bug #69220, Bug #17278349 - APPEND DOESN'T AUTOMATICALLY DETECT EXCEL COL HEADER WITH SAME NAME AS SQL FIELD Fixed some code for Edit Data operations that was escaping special characters twice (during edition in Excel and then upon sending the query to the MySQL server). MySQL Bug #68669, Bug #17271693 - A BACKSLASH IS INSERTED BEFORE AN APOSTROPHE EDITING TABLE WITH MYSQL FOR EXCEL Upgraded MySQL Utility with latest version that encapsulates dialog base classes and introduces more classes to handle Workbench connections, and removed these from the Excel project. Bug #16500331 - CAN'T DELETE CONNECTIONS CREATED WITHIN ADDIN You can access the MySQL for Excel documentation at http://dev.mysql.com/doc/refman/5.6/en/mysql-for-excel.html You can find our team’s blog at http://blogs.oracle.com/MySQLOnWindows. You can also post questions on our MySQL for Excel forum found at http://forums.mysql.com/. Enjoy and thanks for the support!

    Read the article

  • Secure Your Wireless Router: 8 Things You Can Do Right Now

    - by Chris Hoffman
    A security researcher recently discovered a backdoor in many D-Link routers, allowing anyone to access the router without knowing the username or password. This isn’t the first router security issue and won’t be the last. To protect yourself, you should ensure that your router is configured securely. This is about more than just enabling Wi-Fi encryption and not hosting an open Wi-Fi network. Disable Remote Access Routers offer a web interface, allowing you to configure them through a browser. The router runs a web server and makes this web page available when you’re on the router’s local network. However, most routers offer a “remote access” feature that allows you to access this web interface from anywhere in the world. Even if you set a username and password, if you have a D-Link router affected by this vulnerability, anyone would be able to log in without any credentials. If you have remote access disabled, you’d be safe from people remotely accessing your router and tampering with it. To do this, open your router’s web interface and look for the “Remote Access,” “Remote Administration,” or “Remote Management” feature. Ensure it’s disabled — it should be disabled by default on most routers, but it’s good to check. Update the Firmware Like our operating systems, web browsers, and every other piece of software we use, router software isn’t perfect. The router’s firmware — essentially the software running on the router — may have security flaws. Router manufacturers may release firmware updates that fix such security holes, although they quickly discontinue support for most routers and move on to the next models. Unfortunately, most routers don’t have an auto-update feature like Windows and our web browsers do — you have to check your router manufacturer’s website for a firmware update and install it manually via the router’s web interface. Check to be sure your router has the latest available firmware installed. Change Default Login Credentials Many routers have default login credentials that are fairly obvious, such as the password “admin”. If someone gained access to your router’s web interface through some sort of vulnerability or just by logging onto your Wi-Fi network, it would be easy to log in and tamper with the router’s settings. To avoid this, change the router’s password to a non-default password that an attacker couldn’t easily guess. Some routers even allow you to change the username you use to log into your router. Lock Down Wi-Fi Access If someone gains access to your Wi-Fi network, they could attempt to tamper with your router — or just do other bad things like snoop on your local file shares or use your connection to downloaded copyrighted content and get you in trouble. Running an open Wi-Fi network can be dangerous. To prevent this, ensure your router’s Wi-Fi is secure. This is pretty simple: Set it to use WPA2 encryption and use a reasonably secure passphrase. Don’t use the weaker WEP encryption or set an obvious passphrase like “password”. Disable UPnP A variety of UPnP flaws have been found in consumer routers. Tens of millions of consumer routers respond to UPnP requests from the Internet, allowing attackers on the Internet to remotely configure your router. Flash applets in your browser could use UPnP to open ports, making your computer more vulnerable. UPnP is fairly insecure for a variety of reasons. To avoid UPnP-based problems, disable UPnP on your router via its web interface. If you use software that needs ports forwarded — such as a BitTorrent client, game server, or communications program — you’ll have to forward ports on your router without relying on UPnP. Log Out of the Router’s Web Interface When You’re Done Configuring It Cross site scripting (XSS) flaws have been found in some routers. A router with such an XSS flaw could be controlled by a malicious web page, allowing the web page to configure settings while you’re logged in. If your router is using its default username and password, it would be easy for the malicious web page to gain access. Even if you changed your router’s password, it would be theoretically possible for a website to use your logged-in session to access your router and modify its settings. To prevent this, just log out of your router when you’re done configuring it — if you can’t do that, you may want to clear your browser cookies. This isn’t something to be too paranoid about, but logging out of your router when you’re done using it is a quick and easy thing to do. Change the Router’s Local IP Address If you’re really paranoid, you may be able to change your router’s local IP address. For example, if its default address is 192.168.0.1, you could change it to 192.168.0.150. If the router itself were vulnerable and some sort of malicious script in your web browser attempted to exploit a cross site scripting vulnerability, accessing known-vulnerable routers at their local IP address and tampering with them, the attack would fail. This step isn’t completely necessary, especially since it wouldn’t protect against local attackers — if someone were on your network or software was running on your PC, they’d be able to determine your router’s IP address and connect to it. Install Third-Party Firmwares If you’re really worried about security, you could also install a third-party firmware such as DD-WRT or OpenWRT. You won’t find obscure back doors added by the router’s manufacturer in these alternative firmwares. Consumer routers are shaping up to be a perfect storm of security problems — they’re not automatically updated with new security patches, they’re connected directly to the Internet, manufacturers quickly stop supporting them, and many consumer routers seem to be full of bad code that leads to UPnP exploits and easy-to-exploit backdoors. It’s smart to take some basic precautions. Image Credit: Nuscreen on Flickr     

    Read the article

  • Trying to configure HWIC-3G-HSPA

    - by user1174838
    I'm trying to configure a couple of Cisco 1941 routes. The are both identical routers. Each as a HWIC-1T (Smart Serial interface) and a HWIC-3G-HSPA 3G interface. These routers are to be sent to remote sites. We have connectivity to one of the sites but if remote site A gors down we lose connectivity to remote site B. The HWIC-1T is the primary WAN interface using frame relay joining the two remote sites We want the HWIC-3G-HSPA to be usable for direct connectivity from head office to remote site B, and also the HWIC-3G-HSPA is do be used for comms between the remote sites when the frame relay is down (happens quite a bit). I initialy tried to do dynamic routing using EIGRP however in my lab setup of laptop - 1941 - 1941 - laptop, I was unable to get end to end connectivity. I later settled on static routing and have got end to end connectivity but only over frame relay, not the HWIC-3G-HSPA. The sanitized running config for remote site A: version 15.1 service tcp-keepalives-in service tcp-keepalives-out service timestamps debug datetime msec service timestamps log datetime msec service password-encryption service udp-small-servers service tcp-small-servers ! hostname remoteA ! boot-start-marker boot-end-marker ! ! logging buffered 51200 warnings enable secret 5 censored ! no aaa new-model clock timezone wst 8 0 ! no ipv6 cef ip source-route ip cef ! ip domain name yourdomain.com multilink bundle-name authenticated ! chat-script gsm "" "ATDT*98*1#" TIMEOUT 30 "CONNECT" ! username admin privilege 15 secret 5 censored ! controller Cellular 0/1 ! interface Embedded-Service-Engine0/0 no ip address shutdown ! interface GigabitEthernet0/0 ip address 192.168.2.5 255.255.255.0 duplex auto speed auto ! interface GigabitEthernet0/1 no ip address shutdown duplex auto speed auto ! interface Serial0/0/0 ip address 10.1.1.2 255.255.255.252 encapsulation frame-relay cdp enable frame-relay interface-dlci 16 frame-relay lmi-type ansi ! interface Cellular0/1/0 ip address negotiated encapsulation ppp dialer in-band dialer idle-timeout 2147483 dialer string gsm dialer-group 1 async mode interactive ppp chap hostname censored ppp chap password 7 censored cdp enable ! interface Cellular0/1/1 no ip address encapsulation ppp ! interface Dialer0 no ip address ! ip forward-protocol nd ! no ip http server no ip http secure-server ! ip route 0.0.0.0 0.0.0.0 Serial0/0/0 210 permanent ip route 0.0.0.0 0.0.0.0 Cellular0/1/0 220 permanent ip route 172.31.2.0 255.255.255.0 Cellular0/1/0 permanent ip route 192.168.3.0 255.255.255.0 10.1.1.1 permanent ip route 192.168.3.0 255.255.255.0 Cellular0/1/0 210 permanent ! access-list 1 permit any dialer-list 1 protocol ip list 1 ! control-plane ! line con 0 logging synchronous login local line aux 0 line 2 no activation-character no exec transport preferred none transport input all transport output pad telnet rlogin lapb-ta mop udptn v120 ssh stopbits 1 line 0/1/0 exec-timeout 0 0 script dialer gsm login modem InOut no exec transport input all rxspeed 7200000 txspeed 5760000 line 0/1/1 no exec rxspeed 7200000 txspeed 5760000 line vty 0 4 access-class 23 in privilege level 15 password 7 censored login local transport input all line vty 5 15 access-class 23 in privilege level 15 password 7 censored login local transport input all line vty 16 1370 password 7 censored login transport input all ! scheduler allocate 20000 1000 end The sanitized running config for remote site B: version 15.1 service tcp-keepalives-in service tcp-keepalives-out service timestamps debug datetime msec service timestamps log datetime msec service password-encryption service udp-small-servers service tcp-small-servers ! hostname remoteB ! boot-start-marker boot-end-marker ! logging buffered 51200 warnings enable secret 5 censored ! no aaa new-model clock timezone wst 8 0 ! no ipv6 cef ip source-route ip cef ! no ip domain lookup ip domain name yourdomain.com multilink bundle-name authenticated ! chat-script gsm "" "ATDT*98*1#" TIMEOUT 30 "CONNECT" username admin privilege 15 secret 5 censored ! controller Cellular 0/1 ! interface Embedded-Service-Engine0/0 no ip address shutdown ! interface GigabitEthernet0/0 ip address 192.168.3.1 255.255.255.0 duplex auto speed auto ! interface GigabitEthernet0/1 no ip address shutdown duplex auto speed auto ! interface Serial0/0/0 ip address 10.1.1.1 255.255.255.252 encapsulation frame-relay clock rate 2000000 cdp enable frame-relay interface-dlci 16 frame-relay lmi-type ansi frame-relay intf-type dce ! interface Cellular0/1/0 ip address negotiated encapsulation ppp dialer in-band dialer idle-timeout 2147483 dialer string gsm dialer-group 1 async mode interactive ppp chap hostname censored ppp chap password 7 censored ppp ipcp dns request cdp enable ! interface Cellular0/1/1 no ip address encapsulation ppp ! interface Dialer0 no ip address ! ip forward-protocol nd ! no ip http server no ip http secure-server ! ip route 0.0.0.0 0.0.0.0 Serial0/0/0 210 permanent ip route 0.0.0.0 0.0.0.0 Cellular0/1/0 220 permanent ip route 172.31.2.0 255.255.255.0 Cellular0/1/0 permanent ip route 192.168.2.0 255.255.255.0 10.1.1.2 permanent ip route 192.168.2.0 255.255.255.0 Cellular0/1/0 210 permanent ! kron occurrence PING in 1 recurring policy-list ICMP ! access-list 1 permit any dialer-list 1 protocol ip list 1 ! control-plane ! line con 0 logging synchronous login local line aux 0 line 2 no activation-character no exec transport preferred none transport input all transport output pad telnet rlogin lapb-ta mop udptn v120 ssh stopbits 1 line 0/1/0 exec-timeout 0 0 script dialer gsm login modem InOut no exec transport input all rxspeed 7200000 txspeed 5760000 line 0/1/1 no exec rxspeed 7200000 txspeed 5760000 line vty 0 4 access-class 23 in privilege level 15 password 7 censored login transport input all line vty 5 15 access-class 23 in privilege level 15 password 7 censored login transport input all line vty 16 1370 password 7 censored login transport input all ! scheduler allocate 20000 1000 end The last problem I'm having is the 3G interfaces go down after only a few minutes of inactivity. I've tried using kron to ping the local HWIC-3G-HSPA interface (cellular 0/1/0) every minute but that hasn't been successful. Manually pinging the IP assigned (by the telco) to ce0/1/0 does bring the interface up. Any ideas? Thanks

    Read the article

  • What is a proper way to pass a parameter to Set-Alias in powershell?

    - by Nick Gorbikoff
    Hello. A little background: I use PowerShell on windows xp at work and I set a bunch of useful shortcuts in Microsoft.PowerShell_profile.ps1 in My Documents, trying to emulate Mac environment inspired by Ryan Bates's shortcuts I have things like: Set-Alias rsc Rails-Console function Rails-Console {Invoke-Expression "ruby script/console"} Which works just fine when in command prompt I say: rsc #it calls the proper command However this doesn't work properly Set-Alias rsg Rails-Generate function Rails-Generate {Invoke-Expression "ruby script/generate"} So when I do : rsg model User which is supposed to call ruby script/generate model User all it calls is ruby script/generate #Dumping my params So how would I properly modify my functions to take params I send to functions? Thank you!!

    Read the article

< Previous Page | 125 126 127 128 129 130 131 132 133 134 135 136  | Next Page >