Search Results

Search found 12055 results on 483 pages for 'password complexity'.

Page 89/483 | < Previous Page | 85 86 87 88 89 90 91 92 93 94 95 96  | Next Page >

  • python getelementbyid from string

    - by matthewgall
    Hey, I have the following program, that is trying to upload a file (or files) to an image upload site, however I am struggling to find out how to parse the returned HTML to grab the direct link (contained in a ). I have the code below: #!/usr/bin/python # -*- coding: utf-8 -*- import pycurl import urllib import urlparse import xml.dom.minidom import StringIO import sys import gtk import os import imghdr import locale import gettext try: import pynotify except: print "Please install pynotify." APP="Uploadir Uploader" DIR="locale" locale.setlocale(locale.LC_ALL, '') gettext.bindtextdomain(APP, DIR) gettext.textdomain(APP) _ = gettext.gettext ##STRINGS uploading = _("Uploading image to Uploadir.") oneimage = _("1 image has been successfully uploaded.") multimages = _("images have been successfully uploaded.") uploadfailed = _("Unable to upload to Uploadir.") class Uploadir: def __init__(self, args): self.images = [] self.urls = [] self.broadcasts = [] self.username="" self.password="" if len(args) == 1: return else: for file in args: if file == args[0] or file == "": continue if file.startswith("-u"): self.username = file.split("-u")[1] #print self.username continue if file.startswith("-p"): self.password = file.split("-p")[1] #print self.password continue self.type = imghdr.what(file) self.images.append(file) for file in self.images: self.upload(file) self.setClipBoard() self.broadcast(self.broadcasts) def broadcast(self, l): try: str = '\n'.join(l) n = pynotify.Notification(str) n.set_urgency(pynotify.URGENCY_LOW) n.show() except: for line in l: print line def upload(self, file): #Try to login cookie_file_name = "/tmp/uploadircookie" if ( self.username!="" and self.password!=""): print "Uploadir authentication in progress" l=pycurl.Curl() loginData = [ ("username",self.username),("password", self.password), ("login", "Login") ] l.setopt(l.URL, "http://uploadir.com/user/login") l.setopt(l.HTTPPOST, loginData) l.setopt(l.USERAGENT,"User-Agent: Uploadir (Python Image Uploader)") l.setopt(l.FOLLOWLOCATION,1) l.setopt(l.COOKIEFILE,cookie_file_name) l.setopt(l.COOKIEJAR,cookie_file_name) l.setopt(l.HEADER,1) loginDataReturnedBuffer = StringIO.StringIO() l.setopt( l.WRITEFUNCTION, loginDataReturnedBuffer.write ) if l.perform(): self.broadcasts.append("Login failed. Please check connection.") l.close() return loginDataReturned = loginDataReturnedBuffer.getvalue() l.close() #print loginDataReturned if loginDataReturned.find("<li>Your supplied username or password is invalid.</li>")!=-1: self.broadcasts.append("Uploadir authentication failed. Username/password invalid.") return else: self.broadcasts.append("Uploadir authentication successful.") #cookie = loginDataReturned.split("Set-Cookie: ")[1] #cookie = cookie.split(";",0) #print cookie c = pycurl.Curl() values = [ ("file", (c.FORM_FILE, file)) ] buf = StringIO.StringIO() c.setopt(c.URL, "http://uploadir.com/file/upload") c.setopt(c.HTTPPOST, values) c.setopt(c.COOKIEFILE, cookie_file_name) c.setopt(c.COOKIEJAR, cookie_file_name) c.setopt(c.WRITEFUNCTION, buf.write) if c.perform(): self.broadcasts.append(uploadfailed+" "+file+".") c.close() return self.result = buf.getvalue() #print self.result c.close() doc = urlparse.urlparse(self.result) self.urls.append(doc.getElementsByTagName("download")[0].childNodes[0].nodeValue) def setClipBoard(self): c = gtk.Clipboard() c.set_text('\n'.join(self.urls)) c.store() if len(self.urls) == 1: self.broadcasts.append(oneimage) elif len(self.urls) != 0: self.broadcasts.append(str(len(self.urls))+" "+multimages) if __name__ == '__main__': uploadir = Uploadir(sys.argv) Any help would be gratefully appreciated. Warm regards,

    Read the article

  • Having to hit the site login button twice after first turning on computer

    - by John
    Hello, The code below is for a login that I'm using. It works fine, but when I first try logging in after turning on my computer, it only works the second time that I hit the "Login" button. Any idea how I can make it not require hitting the "Login" button twice in this situation? Thanks in advance, John function isLoggedIn() { if (session_is_registered('loginid') && session_is_registered('username')) { return true; // the user is loged in } else { return false; // not logged in } return false; } if (!isLoggedIn()) { if (isset($_POST['cmdlogin'])) { if (checkLogin($_POST['username'], $_POST['password'])) { show_userbox(); } else { echo "Incorrect Login information !"; show_loginform(); } } else { show_loginform(); } } else { show_userbox(); } function show_loginform($disabled = false) { echo '<form name="login-form" id="login-form" method="post" action="./index.php"> <div class="usernameformtext"><label title="Username">Username: </label></div> <div class="usernameformfield"><input tabindex="1" accesskey="u" name="username" type="text" maxlength="30" id="username" /></div> <div class="passwordformtext"><label title="Password">Password: </label></div> <div class="passwordformfield"><input tabindex="2" accesskey="p" name="password" type="password" maxlength="15" id="password" /></div> <div class="registertext"><a href="http://www...com/sandbox/register.php" title="Register">Register</a></div> <div class="lostpasswordtext"><a href="http://www...com/sandbox/lostpassword.php" title="Lost Password">Lost password?</a></div> <p class="loginbutton"><input tabindex="3" accesskey="l" type="submit" name="cmdlogin" value="Login" '; if ($disabled == true) { echo 'disabled="disabled"'; } echo ' /></p></form>'; } EDIT: Here is another function that is used. function isLoggedIn() { if (session_is_registered('loginid') && session_is_registered('username')) { return true; // the user is loged in } else { return false; // not logged in } return false; }

    Read the article

  • Login function runs different between local and server

    - by quangnd
    Here is my check login function: protected bool checkLoginStatus(String email, String password) { bool loginStatus = false; bool status = false; try { Connector.openConn(); String str = "SELECT * FROM [User]"; SqlCommand cmd = new SqlCommand(str, Connector.conn); SqlDataAdapter da = new SqlDataAdapter(cmd); DataSet ds = new DataSet(); da.Fill(ds, "tblUser"); //check valid foreach (DataRow dr in ds.Tables[0].Rows) { if (email == dr["Email"].ToString() && password == Connector.base64Decode(dr["Password"].ToString())) { Session["login_status"] = true; Session["username"] = dr["Name"].ToString(); Session["userId"] = dr["UserId"].ToString(); status = true; break; } } } catch (Exception ex) { } finally { Connector.closeConn(); } return status; } And call it at my aspx page: String email = Login1.UserName.Trim(); String password = Login1.Password.Trim(); if (checkLoginStatus(email, password)) Response.Redirect(homeSite); else lblFailure.Text = "Invalid!"; I ran this page at localhost successful! When I published it to server, this function only can run if email and password correct! Other, error occured: A network-related or instance-specific error occurred while establishing a connection to SQL Server. The server was not found or was not accessible. Verify that the instance name is correct and that SQL Server is configured to allow remote connections. (provider: SQL Network Interfaces, error: 26 - Error Locating Server/Instance Specified) I tried open SQL Server 2008 Configuration Manager and enable SQL Server Browser service (Logon as:NT Authority/Local Service) but it stills error. (note: here is connection string of openConn() at Localhost (run on SQLEXpress 2005) connectionString="Data Source=MYLAPTOP\SQLEXPRESS;Initial Catalog=Spider_Vcms;Integrated Security=True" /> ) At server (run on SQL Server Enterprise 2008) connectionString="Data Source=SVR;Initial Catalog=Spider_Vcms;User Id=abc;password=123456;" /> anyone have an answer for my problem :( thanks a lot!

    Read the article

  • JQuery in ASP.NET - Form Validation Issue

    - by user1026288
    It's not working at all and I'm not sure why. Ideally I'd like to have all the errors pop up in a modal dialog box. But right now I can't even get it to work normally. Any help would be appreciated. Thanks. HTML <html xmlns="http://www.w3.org/1999/xhtml"> <head runat="server"> <title></title> <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.7.0/jquery.min.js" type="text/javascript"></script> <script src="https://ajax.googleapis.com/ajax/libs/jqueryui/1.8.16/jquery-ui.min.js" type="text/javascript"></script> <script src="http://ajax.microsoft.com/ajax/jquery.validate/1.7/jquery.validate.min.js" type="text/javascript"></script> <script src="../Scripts/Frame.js" type="text/javascript"></script> </head> <body runat="server" id="bodyLogin"> <div runat="server" id="frameLogin"> <form runat="server" id="formLogin"> <asp:CheckBox runat="server" ID="checkboxRemember" /> <div><span id="un">Username</span><div id="forgotUsername">?</div><br /> <asp:TextBox runat="server" ID="textUsername" Name="username" /></div> <div><span id="pw">Password</span><div id="forgotPassword">?</div><br /> <asp:TextBox runat="server" ID="textPassword" Name="password" TextMode="Password" /></div> <asp:Button runat="server" ID="buttonLogin" Text="L" /> <asp:Button runat="server" ID="buttonRegister" Text="R" /> </form> </div> <div id="dialog" title="Errors" style="display:none;"><ul></ul></div> </body> </html> JQuery <script type="text/javascript"> $(function () { $("#formLogin").validate({ rules: { username: { required:true, minlength:3, maxlength:15 }, password: { required:true, minlength:6, maxlength:15 }, }, messages: { username: { required: "Username is required.", minlength: "Username minimum length is 3 characters.", maxlength: "Username maxumum length is 15 characters." }, password: { required: "Password is required.", minlength: "Password minumum length is 6 characters.", maxlength: "Password maximum length is 15 characters." } } }); }); </script>

    Read the article

  • Authentication with wget

    - by Llistes Sugra
    I am currently accepting the parameters login and password in my servlet, but logs are storing this info when using wget (as long as it is GET method, and apache is in the middle) Instead of this I want to enhance my servlet's authentication accepting: wget --http-user=login --http-password=password http://myhost/myServlet How can I read, in my servlet, the server side, the login and the password user is sending, in java code?

    Read the article

  • Using GET Method to Maintain Variables After Logging In

    - by John
    Hello, I am using a login system. When I navigate to comments/index.php without logging in, some variables get passed along using the GET method just fine. Then, if I log in while I am on this page, these variables disappear. The variables that disappear are $submission, $submissionid, and $url. I thought I could use the GET method to keep them live after logging in by appending ?submission='.$submission.'&submissionid='.$submissionid.'&url='.$url.' to the URL of the login form action as seen below. But the variables still disappeared after I made this addition. The relevant code I am trying to use is below. Any idea what I can do to make it do what I want? Thanks in advance, John In comments/index.php: require_once "header.php"; include "login.php"; include "comments.php"; In login.php: if (!isLoggedIn()) { if (isset($_POST['cmdlogin'])) { if (checkLogin($_POST['username'], $_POST['password'])) { show_userbox(); } else { echo "Incorrect Login information !"; show_loginform(); } } else { show_loginform(); } } else { show_userbox(); } In comments.php: $url = mysql_real_escape_string($_GET['url']); echo '<div class="subcommenttitle"><a href="http://www.'.$url.'">'.$submission.'</a></div>'; $submission = mysql_real_escape_string($_GET['submission']); $submissionid = mysql_real_escape_string($_GET['submissionid']); The login function: function show_loginform($disabled = false) { echo '<form name="login-form" id="login-form" method="post" action="./index.php?submission='.$submission.'&submissionid='.$submissionid.'&url='.$url.'"> <div class="usernameformtext"><label title="Username">Username: </label></div> <div class="usernameformfield"><input tabindex="1" accesskey="u" name="username" type="text" maxlength="30" id="username" /></div> <div class="passwordformtext"><label title="Password">Password: </label></div> <div class="passwordformfield"><input tabindex="2" accesskey="p" name="password" type="password" maxlength="15" id="password" /></div> <div class="registertext"><a href="http://www...com/sandbox/register.php" title="Register">Register</a></div> <div class="lostpasswordtext"><a href="http://www...com/sandbox/lostpassword.php" title="Lost Password">Lost password?</a></div> <p class="loginbutton"><input tabindex="3" accesskey="l" type="submit" name="cmdlogin" value="Login" '; if ($disabled == true) { echo 'disabled="disabled"'; } echo ' /></p></form>'; }

    Read the article

  • Weird IIS with Windows Authentication + IE problem

    - by Paulius Maruška
    I have a website running on IIS and using Windows Authentication. All users that are configured to get access to the site are form a AD domain (not local users). In the properties of a Website, I have set to use the AD domain as the realm. Now, when using Firefox, Safari or Chrome - Everything is fine. When the user tries to open the site, he get's the login box. he enters simply "username" and "password" (let's pretend that it's an actual login and password :P) and he get's into the site. When using IE, however, things get nasty. When the user tries to open the site - he get's the login box. User enters the "username" and "password" again, but those get rejected! And when the second time login box pops up - it has the username filled in as "web-server-domain-name\username" which is wrong, because web-server-domain-name is not the domain where all users reside (it's "ad-domain"). I've spent days trying to figure out what's going on... Note, that if I manually enter "ad-domain\username" - I get accepted into the site without problems. So, my guess is that IE sends wrong username if domain is not specified. Anyway, IE is the only browser that triggers this behavior! Is it possible to do a server-side fix? Maybe it's possible to somehow auto-map the users to AD users? If it's not solvable server-side - is there a client-side fix for this? Thank you. PS: I'm more of a programmer than a sys-admin, so configuring servers isn't the strong side of mine... :P UPDATE: @Evan: Yes, "Digest authentication for Windows domain servers" is also enabled. @Eric: IIS version is 6.0. The authentication methods enabled are: Integrated and digest - all other methods are disabled. As for the security log. I looked at it, when doing "username" and "password" login in Chrome/Firefox and when doing "ad-domain\username" and "password" login from IE - the generated log messages are the same (I see no difference, anyway). When entering "username" and "password" I don't see any errors in the security (or any other) log, so can't tell what method it's trying to use. UPDATE 2: As suggested by Eric in the comments - I played around with Fiddler... While playing with it, I noticed, that when "username" and "password" is entered in FF and IE - the "Authorization" header value (encrypted) sent by IE is longer (almost two times) than one sent by FF. I tried to disable Windows Integrated authentication and only leave the Digest enabled - that fixed the problem (meaning, IE used the right realm just like other browsers), but that caused bazillion other problems with my site, because with Digest - user impersonation on the server doesn't work (that causes problems, when connecting to database etc). Any ideas?

    Read the article

  • Postfix + SASLAUTHD + MySQL authentication problems

    - by Or W
    I've been trying to sort this out for the past 6 hours or so, this is the error message I'm facing (Running CentOS x64): /var/log/maillog: Jun 22 20:42:49 ptroa postfix/smtpd[10130]: warning: SASL authentication failure: Password verification failed Jun 22 20:42:49 ptroa postfix/smtpd[10130]: warning: bzq-79-177-192-133.red.bezeqint.net[79.177.192.133]: SASL PLAIN authentication failed: authentication failure Jun 22 20:42:49 ptroa postfix/smtpd[10130]: warning: bzq-79-177-192-133.red.bezeqint.net[79.177.192.133]: SASL LOGIN authentication failed: authentication failure /var/log/messages: Jun 22 20:15:38 ptroa saslauthd[9401]: do_auth : auth failure: [user=myuser] [service=smtp] [realm=domain.com] [mech=pam] [reason=PAM auth error] I have dovecot installed as well and I'm able to receive emails via the MySQL authentication. The problem is when I'm trying to use SMTP to send out emails. Some config files: /etc/postfix/main.cf: # See /usr/share/postfix/main.cf.dist for a commented, more complete version # Debian specific: Specifying a file name will cause the first # line of that file to be used as the name. The Debian default # is /etc/mailname. myorigin = /etc/mailname smtpd_banner = Server Message biff = no # appending .domain is the MUA's job. append_dot_mydomain = no # Uncomment the next line to generate "delayed mail" warnings #delay_warning_time = 4h readme_directory = /usr/share/doc/postfix # TLS parameters smtpd_tls_cert_file = /etc/postfix/smtpd.cert smtpd_tls_key_file = /etc/postfix/smtpd.key smtpd_use_tls = yes smtpd_tls_session_cache_database = btree:${data_directory}/smtpd_scache smtp_tls_session_cache_database = btree:${data_directory}/smtp_scache # See /usr/share/doc/postfix/TLS_README.gz in the postfix-doc package for # information on enabling SSL in the smtp client. myhostname = domain.com alias_maps = hash:/etc/aliases alias_database = hash:/etc/aliases myorigin = /etc/mailname mydestination = relayhost = mynetworks = 127.0.0.0/8 [::ffff:127.0.0.0]/104 [::1]/128 mailbox_size_limit = 0 recipient_delimiter = + inet_interfaces = all html_directory = /usr/share/doc/postfix/html message_size_limit = 30720000 virtual_alias_domains = virtual_alias_maps = proxy:mysql:/etc/postfix/mysql-virtual_forwardings.cf, mysql:/etc/postfix/mysql-virtual_email2email.cf virtual_mailbox_domains = proxy:mysql:/etc/postfix/mysql-virtual_domains.cf virtual_mailbox_maps = proxy:mysql:/etc/postfix/mysql-virtual_mailboxes.cf virtual_mailbox_base = /home/vmail virtual_uid_maps = static:5000 virtual_gid_maps = static:5000 smtpd_sasl_auth_enable = yes broken_sasl_auth_clients = yes smtpd_sasl_authenticated_header = yes smtpd_recipient_restrictions = permit_mynetworks, permit_sasl_authenticated, reject_unauth_destination virtual_create_maildirsize = yes virtual_maildir_extended = yes proxy_read_maps = $local_recipient_maps $mydestination $virtual_alias_maps $virtual_alias_domains $virtual_mailbox_maps $virtual_mailbox_domains $relay_recipient_maps $relay_domains $canonical_maps $sender_canonical_maps $recipient_cano$ virtual_transport = dovecot dovecot_destination_recipient_limit = 1 /etc/default/saslauthd: START=yes DESC="SASL Authentication Daemon" NAME="saslauthd" MECHANISMS="pam" MECH_OPTIONS="" THREADS=5 OPTIONS="-c -m /var/spool/postfix/var/run/saslauthd -r" /etc/pam.d/smtp: #%PAM-1.0 #auth include password-auth #account include password-auth auth required pam_mysql.so user=mail_admin passwd=password host=127.0.0.1 db=mail table=users usercolumn=email passwdcolumn=password crypt=1 verbose=1 account sufficient pam_mysql.so user=mail_admin passwd=password host=127.0.0.1 db=mail table=users usercolumn=email passwdcolumn=password crypt=1 verbose=1

    Read the article

  • IE7 textbox onfocus problem

    - by Craig
    Because IE won't do document.getElementById(ID).setAttribute('type','password') I've re-engineered the way the password field woirks on this site: http://devdae.dialanexchange.com/Default.aspx so it works in accordance with this idea: http://www.folksonomy.org/2009/01/12/changing-input-type-from-text-to-password-in-internet-explorer-hack/ It works fine in IE8 and FF3. It breaks in IE7 just as you click into the password field. I'm now tearing my hair out. Can anyone give me a clue what's wrong as IE7's diagnosis is just "Object expected, code 0"?

    Read the article

  • How do I run a sudo command in Emacs?

    - by Inaimathi
    I tried using shell-command, but it just does something like this: Sudo password: incorrect, try again. Sudo password: incorrect, try again. Sudo password: incorrect, try again. Failed three attempts. without actually asking for a password. I don't want to have to start up Emacs using sudo emacs, but I guess that's an option if nothing else will work.

    Read the article

  • 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

  • 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

< Previous Page | 85 86 87 88 89 90 91 92 93 94 95 96  | Next Page >