Search Results

Search found 40429 results on 1618 pages for 'change password'.

Page 219/1618 | < Previous Page | 215 216 217 218 219 220 221 222 223 224 225 226  | Next Page >

  • Can't access windows 7 shared files on Ubuntu 11.10

    - by Corey
    I just set up ubuntu 11.10 and Samba. I got it to access shares on a Vista machine, but when I try to access the shares on a windows 7 machine it asks for a Username, Domain, and Password. I have no password set up on the windows 7 machine so I put in the username, and domain try to connect and the password prompt keeps appearing...also tried guest and admin with no luck...I've tried many different fixes(modifying registry entries & advanced securities on the win 7 machine) with no luck. Thanks

    Read the article

  • No access to Samba shares

    - by koanhead
    I have three shared folders in my local home directory- that is to say, on my Ubuntu desktop's /home/me/. All were set up using "Sharing Options" in Nautilus' right-click menu. The standard "Music" and "Videos" folders are configured identically: the "Guest Access" box is checked, but the "Allow others to create and delete" is not. The third folder, called "shared", is configured to not allow Guest access but to allow others to modify files. I have not altered /etc/samba/smb.conf by hand, I have only used Sharing Options to create and modify these so-called "shares". My roommates have two Windows 7 computers and one Ubuntu Netbook Remix netbook. I have the aforementioned desktop machine and laptop running 10.04. None of these machines can access any of the shares. Attempts to access the Guest shares result in the message \\machine\directory is not accessible. The network name could not be found. This is the error message generated by a VM running Windows 2000. The other Windows machines generate a similar error. The Ubuntu laptop gives the error Unable to mount location: Failed to mount Windows share. Hurrah, once again, for informative error messages. That really helps a lot. When attempting to browse the folder called "shared" from the laptop, I'm confronted with a password dialog. This behavior is the same will all machines I've tried in the situation. On entering my username and password for the account to which the shares belong, the password dialog briefly disappears and is replaced with an identical dialog. No error message, useful or not, appears. When attempting to browse this folder with the VM, the outcome is the same except that the password dialog helpfully states "incorrect username or password". My assumption is that the username and password in question is that of the user which owns the shares. I have tried all other username and password combinations available in this context and the outcome is the same. I would like to be able to share files. Sharing them with Windows machines is a nice feature, or would be if it was available. Really I consider sharing files between two machines with the same version of the same operating system kind of a minimum condition for network usability. Samba last functioned reliably for me more than ten years ago. I have attempted to use it on and off since then with only intermittent success. Oh, and "Personal File Sharing" from the Preferences menu does not result in an entry in Places → Network → my-server. In fact, the old entry "MY-SERVER" goes away and is replaced by "koanhead's public files on my-server", which when I attempt to open it from the laptop gives a "DBus.Error.NoReply: Message did not receive a reply." I know I come here and gripe about Ubuntu a lot, but on the other hand I spend literally hours every day trying to fix things in Ubuntu. It's a good system which aspires to greatness, which is why things like this either Need to work; or Be adequately documented. Ideally both would be the case. Anyway, rant over. Hopefully someone will have some insight on this issue. Thanks all who bother to read this wall o'text for your time.

    Read the article

  • Login keyring box frozen

    - by Maddie
    I'm new to ubuntu and I'm afraid I've done something really stupid. After booting, When prompted to enter a password to unlock my login keyring, I tried to enter in my password, but it won't let me. I can move my cursor but everything else is frozen. I tried to access the terminal by using Ctrl + Alt + F2, I entered my login and password from there and it worked, I just don't know what to do afterwards. Don't know what other information I can provide other than the fact that I am using Ubuntu 10.04

    Read the article

  • Ubuntu One Client 3.02 for windows - Authentication Failure signing in with existing account

    - by steigerj
    I have logged into ubuntu one using a browser successfully. But using the very same login email address and password in the windows client 3.02 gets me an 'authentication failed' message. I am 100% sure I used the same email address and password. My password does not contain any non-ascii characters. Someone suggested to use an older version of the client. Where could i possibly download a complete installer?

    Read the article

  • Custom Rule Sets in JohnTheRipper

    - by user854619
    I'm trying to create a custom rule set to do hash cracking. I have a SHA1 hash and a rule set that was enforced to create the password. The password must be of the form, 6-8 characters Every other letter changes case Password "shifts" characters at least one degree and at most three One odd number and one even number are at the beginning of the password One special character and one punctuation character are appended to the end of the password How can I defined a brute force attack in JohnTheRipper or similar hash cracking program? I've also attempted to write code to generate a wordlist of possible passwords, with no success. Thanks!

    Read the article

  • How to access UbuntOne when it asks for default keyring which has never been set?

    - by obu-tim
    I am trying to set up UbuntuOne on a new computer and after I enter the email and password, it asks for the keyring 'default'. I don't know what it is and I never set it. Makes it difficult to enter so it seems to be a counterproductive security default. I understand that if autologin is set the keyring is called. I tried setting the main user to need a password but if I reboot it doesn't ask for the password so it sort of autoboots still. So How do I set the keyring default password? If I can't set it I can't install UbuntOne

    Read the article

  • No access to Samba shares

    - by koanhead
    I have three shared folders in my local home directory- that is to say, on my Ubuntu desktop's /home/me/. All were set up using "Sharing Options" in Nautilus' right-click menu. The standard "Music" and "Videos" folders are configured identically: the "Guest Access" box is checked, but the "Allow others to create and delete" is not. The third folder, called "shared", is configured to not allow Guest access but to allow others to modify files. I have not altered /etc/samba/smb.conf by hand, I have only used Sharing Options to create and modify these so-called "shares". My roommates have two Windows 7 computers and one Ubuntu Netbook Remix netbook. I have the aforementioned desktop machine and laptop running 10.04. None of these machines can access any of the shares. Attempts to access the Guest shares result in the message \\machine\directory is not accessible. The network name could not be found. This is the error message generated by a VM running Windows 2000. The other Windows machines generate a similar error. The Ubuntu laptop gives the error Unable to mount location: Failed to mount Windows share. Hurrah, once again, for informative error messages. That really helps a lot. When attempting to browse the folder called "shared" from the laptop, I'm confronted with a password dialog. This behavior is the same will all machines I've tried in the situation. On entering my username and password for the account to which the shares belong, the password dialog briefly disappears and is replaced with an identical dialog. No error message, useful or not, appears. When attempting to browse this folder with the VM, the outcome is the same except that the password dialog helpfully states "incorrect username or password". My assumption is that the username and password in question is that of the user which owns the shares. I have tried all other username and password combinations available in this context and the outcome is the same. I would like to be able to share files. Sharing them with Windows machines is a nice feature, or would be if it was available. Really I consider sharing files between two machines with the same version of the same operating system kind of a minimum condition for network usability. Samba last functioned reliably for me more than ten years ago. I have attempted to use it on and off since then with only intermittent success. Oh, and "Personal File Sharing" from the Preferences menu does not result in an entry in Places → Network → my-server. In fact, the old entry "MY-SERVER" goes away and is replaced by "koanhead's public files on my-server", which when I attempt to open it from the laptop gives a "DBus.Error.NoReply: Message did not receive a reply." I know I come here and gripe about Ubuntu a lot, but on the other hand I spend literally hours every day trying to fix things in Ubuntu. It's a good system which aspires to greatness, which is why things like this either Need to work; or Be adequately documented. Ideally both would be the case. Anyway, rant over. Hopefully someone will have some insight on this issue. Thanks all who bother to read this wall o'text for your time.

    Read the article

  • Why doesn't phpMyAdmin connect to MySQL server?

    - by Grafica
    I'm running xampp on Windows 7, and when I type localhost/phpmyadmin, there is an error: phpMyAdmin tried to connect to the MySQL server, and the server rejected the connection. You should check the host, username and password in your configuration and make sure that they correspond to the information given by the administrator of the MySQL server. Here's what I did, but I'm still not able to connect: In config.inc.php, changed from true to false: $cfg['Servers'[$i]['AllowNoPassword'] = false; Changed password here: localhost/security/xamppsecurity.php In resetroot.bat, typed new password where it says 'password': echo REPLACE INTO user VALUES ('localhost', 'pma', 'password', 'N', 'N', 'N', 'N', 'N', 'N', 'N', 'N', 'N', 'N', 'N', 'N', 'N', 'N', 'N', 'N', 'N', 'N', 'N', 'N', 'N', 'N', 'N', 'N', 'N', 'N', 'N', 'N', 'N', '', '', '', '', 0, 0, 0, 0, '', ''); >>resetroot.sql Restarted apache and mySQL I still get the same error message. Thanks in advance!

    Read the article

  • Dual Monitor Lock Screen Problem

    - by Justin Carver
    Ubuntu 12.04 Nvidia GTX 550-Ti x-swat Nvidia driver Problem: Using 2 monitors. When screen locks there is a blue box on top of the wallpaper and password dialog box that hides the field for entering your password or switching users. Problem is on 2 systems with similar hardware (Nvidia card, x-swat driver, dual monitors) You can still type your password in blindly and hit enter to login but it's irritating to not be able to see the dialog box.

    Read the article

  • Remote connect to mysql server?

    - by LF4
    I've been trying to figure out why I keep getting this error when I try to connect to the MySQL server with the following commands. $~ mysql -u username -h SQLserver -p Enter password: ERROR 1045 (28000): Access denied for user 'username'@'myIP' (using password: YES) I've done the following: Port is open in the firewall other wise I wouldn't get the error it'd just timeout. MySQL server not running with skip-networking or bind-address username has host as '%' and I can connect locally so the password is correct. GRANT USAGE ON *.* TO username@% IDENTIFIED BY 'password'; FLUSH PRIVILEGES; I wanted to know if anyone had ideas or ran into this issue before and solved it? mysql> select user, host from mysql.user where user='username'; +----------+------+ | user | host | +----------+------+ | username | % | +----------+------+ 1 row in set (0.00 sec) mysql> show grants for 'username'; +----------------------------------------------------------------------------------------------------------------+ | Grants for username@% | +----------------------------------------------------------------------------------------------------------------+ | GRANT ALL PRIVILEGES ON *.* TO 'username'@'%' IDENTIFIED BY PASSWORD '*F42AD03PASSWORDHASHADF4021C86B' | | GRANT ALL PRIVILEGES ON `DB2`.* TO 'username'@'%' | +----------------------------------------------------------------------------------------------------------------+ 2 rows in set (0.00 sec)

    Read the article

  • Having problem to login after upgrading from 11.10 to 12.04

    - by LinuxIsMyFriend
    I just update to 12.04 from 11.10 and I get to the login screen w/o problems. When I enter my password the screen turns black and returns to the login screen half a second later. There is a related question out there which was solved by creating more space on the disk but my disks (or rather partitions) are all below 30%. I can log in as guest. I can also login at the cmd prompt (going to tty with Alt+Ctrl+F1) with my normal user credentials. When logged in as guest I can also install programs using my normal account password. There is the normal authentication error when I mistype my password so I'm also sure the password works. Any suggestions?

    Read the article

  • Why my register form don't accept spry validation in dw? [migrated]

    - by shaghayeh
    Why my register form don't accept spry validation in dw? Here is my code: <?php if(isset($_POST['go_register'])) { $last_login=jdate('Y/n/j ,H:i:s'); $firstname=trim($_POST['firstname']); $lastname=trim($_POST['lastname']); $email=trim($_POST['username']); $password=sha1($_POST['password']); $register_date=$_POST['register_date']; if(!empty($firstname)&& !empty($lastname)&& !empty($email)&& !empty($password)&& !empty($register_date)){ $reg_query="insert into users(firstname,lastname,email,password,register_date,last_login)values('$firstname','$lastname','$email','$password','$register_date','$last_login')"; $reg_result=mysqli_query($connection,$reg_query); if($reg_result){ echo "ok"; } }//end of not empty }//end of isset ?> <div class="sectionclass"> <script src="../SpryAssets/SpryValidationTextField.js" type="text/javascript"></script> <link href="../SpryAssets/SpryValidationTextField.css" rel="stylesheet" type="text/css"> <h2>??? ??? ???</h2> <form method="post" action="<?php echo $_SERVER['PHP_SELF']."?catid=2"; ?>"> <div> <span style="color:#F00">*</span> <label>??? </label> <span id="sprytextfield1"> <input type="text" name="firstname" /> <span class="textfieldRequiredMsg">A value is required.</span></span></div> <div> <span style="color:#F00">*</span> <label>??? ????????</label><input type="text" name="lastname" /> </div> <div> <span style="color:#F00">*</span> <label>??? ??????(?????)</label><input dir="ltr" type="text" name="username" /> </div> <div> <span style="color:#F00">*</span> <label>??? ????</label><input type="password" name="password" /> </div> <input type="hidden" name="register_date" value="<?php echo jdate('Y/n/j'); ?>"> <div> <input type="submit" name="go_register" value="??? ???"/> </div> </form> </div> <script type="text/javascript"> var sprytextfield1 = new Spry.Widget.ValidationTextField("sprytextfield1"); </script>

    Read the article

  • How to change TXSDateTime SOAP serialization in Delphi 7?

    - by LukLed
    I am trying to use Java based webservice and have soap request: <?xml version="1.0"?> <SOAP-ENV:Envelope xmlns:SOAP-ENV="http://schemas.xmlsoap.org/soap/envelope/" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:SOAP-ENC="http://schemas.xmlsoap.org/soap/encoding/"> <SOAP-ENV:Body xmlns:NS1="http://something/"> <NS1:getRequest id="1"> <sessionId xsi:type="xsd:string"></sessionId> <reportType xsi:type="NS1:reportType">ALL</reportType> <xsd:dateFrom xsi:type="xsd:dateTime">2010-05-30T23:29:43.088+02:00</xsd:dateFrom> <xsd:dateTo xsi:type="xsd:dateTime">2010-05-31T23:29:43.728+02:00</xsd:dateTo> </NS1:getRequest> <parameters href="#1" /> </SOAP-ENV:Body> </SOAP-ENV:Envelope> It doesn't work, because webservice doesn't recognize dates as parameters. When I change <xsd:dateFrom xsi:type="xsd:dateTime">2010-05-30T23:29:43.088+02:00</xsd:dateFrom> <xsd:dateTo xsi:type="xsd:dateTime">2010-05-31T23:29:43.728+02:00</xsd:dateTo> to <dateFrom xsi:type="xsd:dateTime">2010-05-30T23:29:43.088+02:00</xsd:dateFrom> <dateTo xsi:type="xsd:dateTime">2010-05-31T23:29:43.728+02:00</xsd:dateTo> everything works ok, but Delphi (without Delphi source code changes) doesn't allow to change generated XML, it has only some options. Is it possible to set conversion options, so TSXDateTime is converted to <dateFrom, not <xsd:dateFrom tag? Did you meet that problem?

    Read the article

  • Securing Web Service communication with SSL using CXF

    - by reef
    Hi all, I am trying to secure communications via SSL/TLS for one of our Web Service using CXF 2.2.5. I am wondering how to update client and server Spring configuration file to activate this feature. I found some information on CXF's website (CXF Wiki) for the client configuration, here is the given example: <http:conduit name="{http://apache.org/hello_world}HelloWorld.http-conduit"> <http:tlsClientParameters> <sec:keyManagers keyPassword="password"> <sec:keyStore type="JKS" password="password" file="src/test/java/org/apache/cxf/systest/http/resources/Morpit.jks"/> </sec:keyManagers> <sec:trustManagers> <sec:keyStore type="JKS" password="password" file="src/test/java/org/apache/cxf/systest/http/resources/Truststore.jks"/> </sec:trustManagers> <sec:cipherSuitesFilter> <!-- these filters ensure that a ciphersuite with export-suitable or null encryption is used, but exclude anonymous Diffie-Hellman key change as this is vulnerable to man-in-the-middle attacks --> <sec:include>.*_EXPORT_.*</sec:include> <sec:include>.*_EXPORT1024_.*</sec:include> <sec:include>.*_WITH_DES_.*</sec:include> <sec:include>.*_WITH_NULL_.*</sec:include> <sec:exclude>.*_DH_anon_.*</sec:exclude> </sec:cipherSuitesFilter> </http:tlsClientParameters> <http:authorization> <sec:UserName>Betty</sec:UserName> <sec:Password>password</sec:Password> </http:authorization> <http:client AutoRedirect="true" Connection="Keep-Alive"/> </http:conduit> Concerning this configuration, the Concerning the server side configuration I am unable to launch the server properly, here is the configuration I have: <http:destination name="{urn:ihe:iti:xds-b:2007}DocumentRepository_Port_Soap12.http-destination"> </http:destination> <httpj:engine-factory> <httpj:engine port="9043"> <httpj:tlsServerParameters> <sec:keyManagers keyPassword="changeit"> <sec:keyStore type="JKS" password="changeit" file="security/keystore.jks" /> </sec:keyManagers> <sec:trustManagers> <sec:keyStore type="JKS" password="changeit" file="security/cacerts.jks" /> </sec:trustManagers> <sec:cipherSuitesFilter> <!-- these filters ensure that a ciphersuite with export-suitable or null encryption is used, but exclude anonymous Diffie-Hellman key change as this is vulnerable to man-in-the-middle attacks --> <sec:include>.*_EXPORT_.*</sec:include> <sec:include>.*_EXPORT1024_.*</sec:include> <sec:include>.*_WITH_DES_.*</sec:include> <sec:include>.*_WITH_NULL_.*</sec:include> <sec:exclude>.*_DH_anon_.*</sec:exclude> </sec:cipherSuitesFilter> <sec:clientAuthentication want="true" required="true" /> </httpj:tlsServerParameters> </httpj:engine> </httpj:engine-factory> But when I run my application server (JOnas) with this configuration I have the following error message: Line 20 in XML document from ServletContext resource [/WEB-INF/beans.xml] is invalid; nested exception is org.xml.sax.SAXParseException: cvc-complex-type.2.4.c: The matching wildcard is strict, but no declaration can be found for element 'httpj:engine-factory'. Do you guys know how to solve this issue? Thanks in advance,

    Read the article

  • iphone app scroll view is horizontal but i want it vertical - how do i change this?

    - by Billy Jones
    Hi Folks, Got this code for a viewscroller from the apple developers site. @synthesize scrollView1, scrollView2; const CGFloat kScrollObjHeight = 467.0; const CGFloat kScrollObjWidth = 320.0; const NSUInteger kNumImages = 6; (void)layoutScrollImages { UIImageView *view = nil; NSArray *subviews = [scrollView1 subviews]; // reposition all image subviews in a horizontal serial fashion CGFloat curXLoc = 0; for (view in subviews) { if ([view isKindOfClass:[UIImageView class]] && view.tag 0) { CGRect frame = view.frame; frame.origin = CGPointMake(curXLoc, 0); view.frame = frame; curXLoc += (kScrollObjWidth); } } // set the content size so it can be scrollable [scrollView1 setContentSize:CGSizeMake((kNumImages * kScrollObjWidth), [scrollView1 bounds].size.height)]; } it works brilliantly the images are rotated by pulling them left or right. but I want to pull the images up and down to change them. Can anyone help me out with this? this part in the code looks as if it controls the direction but i'm not sure how to change it. // reposition all image subviews in a horizontal serial fashion CGFloat curXLoc = 0; for (view in subviews) { if ([view isKindOfClass:[UIImageView class]] && view.tag 0) { CGRect frame = view.frame; frame.origin = CGPointMake(curXLoc, 0); view.frame = frame; curXLoc += (kScrollObjWidth); } } Any help most appreciated, thanks! Billy

    Read the article

  • ASP.NET Membership API not working on Win2008 server/IIS7

    - by Program.X
    I have a very odd problem. I have a web app that uses the .NET Membership API to provide login functionality. This works fine on my local dev machine, using WebDev 4.0 server. I'm using .NET 4.0 with some URL Rewriting, but not on the pages where login is required. I have a Windows Server 2008 with IIS7 However, the Membership API seemingly does not work on the server. I have set up remote debugging and the LoginUser.LoggedIn event of the LoginUser control gets fired okay, but the MembershipUser is null. I get no answer about the username/password being invalid so it seems to be recognising it. If I enter an invalid username/password, I get an invalid username/password response. Some code, if it helps: <asp:ValidationSummary ID="LoginUserValidationSummary" runat="server" CssClass="validation-error-list" ValidationGroup="LoginUserValidationGroup"/> <div class="accountInfo"> <fieldset class="login"> <legend>Account Information</legend> <p> <asp:Label ID="UserNameLabel" runat="server" AssociatedControlID="UserName">Username:</asp:Label> <asp:TextBox ID="UserName" runat="server" CssClass="textEntry"></asp:TextBox> <asp:RequiredFieldValidator ID="UserNameRequired" runat="server" ControlToValidate="UserName" CssClass="validation-error" Display="Dynamic" ErrorMessage="User Name is required." ToolTip="User Name is required." ValidationGroup="LoginUserValidationGroup">*</asp:RequiredFieldValidator> </p> <p> <asp:Label ID="PasswordLabel" runat="server" AssociatedControlID="Password">Password:</asp:Label> <asp:TextBox ID="Password" runat="server" CssClass="passwordEntry" TextMode="Password"></asp:TextBox> <asp:RequiredFieldValidator ID="PasswordRequired" runat="server" ControlToValidate="Password" CssClass="validation-error" Display="Dynamic" ErrorMessage="Password is required." ToolTip="Password is required." ValidationGroup="LoginUserValidationGroup">*</asp:RequiredFieldValidator> </p> <p> <asp:CheckBox ID="RememberMe" runat="server"/> <asp:Label ID="RememberMeLabel" runat="server" AssociatedControlID="RememberMe" CssClass="inline">Keep me logged in</asp:Label> </p> </fieldset> <p class="login-action"> <asp:Button ID="LoginButton" runat="server" CommandName="Login" CssClass="submitButton" Text="Log In" ValidationGroup="LoginUserValidationGroup"/> </p> and the code behind: protected void Page_Load(object sender, EventArgs e) { LoginUser.LoginError += new EventHandler(LoginUser_LoginError); LoginUser.LoggedIn += new EventHandler(LoginUser_LoggedIn); } void LoginUser_LoggedIn(object sender, EventArgs e) { // this code gets run so it appears logins work Roles.DeleteCookie(); // this behaviour has been removed for testing - no difference } void LoginUser_LoginError(object sender, EventArgs e) { HtmlGenericControl htmlGenericControl = LoginUser.FindControl("errorMessageSpan") as HtmlGenericControl; if (htmlGenericControl != null) htmlGenericControl.Visible = true; } I have "Fiddled" with the Login form reponse and I get the following Cookie-Set headers: Set-Cookie: ASP.NET_SessionId=lpyyiyjw45jjtuav1gdu4jmg; path=/; HttpOnly Set-Cookie: .ASPXAUTH=A7AE08E071DD20872D6BBBAD9167A709DEE55B352283A7F91E1066FFB1529E5C61FCEDC86E558CEA1A837E79640BE88D1F65F14FA8434AA86407DA3AEED575E0649A1AC319752FBCD39B2A4669B0F869; path=/; HttpOnly Set-Cookie: .ASPXROLES=; expires=Mon, 11-Oct-1999 23:00:00 GMT; path=/; HttpOnly I don't know what is useful here because it is obviously encrypted but I find the .APXROLES cookie having no value interesting. It seems to fail to register the cookie, but passes authentication

    Read the article

  • How to change disclosure style when user enters in edit mode of a UITableView?

    - by R31n4ld0
    Hello, guys. I have a UITableView that in 'normal' mode, show a UITableViewCellAccessoryDisclosureIndicator meaning if the user taps the row, another list is showed, like HIG says: "Disclosure indicator. When this element is present, users know they can tap anywhere in the row to see the next level in the hierarchy or the choices associated with the list item. Use a disclosure indicator in a row when selecting the row results in the display of another list. Don’t use a disclosure indicator to reveal detailed information about the list item; instead, use a detail disclosure button for this purpose." When the user tap the edit button in the top bar of the UITableView, I think I have to change the disclosure because if the user tap it, a view for changing the information of the current row is showed (see the bold line above), again, like HIG says: "Detail disclosure button. Users tap this element to see detailed information about the list item. (Note that you can use this element in views other than table views, to reveal additional details about something; see “Detail Disclosure Buttons” for more information.) In a table view, use a detail disclosure button in a row to display details about the list item. Note that the detail disclosure button, unlike the disclosure indicator, can perform an action that is separate from the selection of the row. For example, in Phone Favorites, tapping the row initiates a call to the contact; tapping the detail disclosure button in the row reveals more information about the contact." Have I miss understood the HIG, or I really do have to change the disclosure style in edit mode of UITableView? If yes, how I can intercept the edit mode when the user taps the Edit button? Thanks in advance.

    Read the article

  • Eclipse - How can I change a 'Project Facet' from Tomcat 6 to Tomcat 5.5?

    - by pcimring
    (Eclipse 3.4, Ganymede) I have an existing Dynamic Web Application project in Eclipse. When I created the project, I specified 'Default configuration for Apache Tomcat v6' under the 'Configuration' drop down. It's a month or 2 down the line, and I would now like to change the configuration to Tomcat 'v5.5'. (This will be the version of Tomcat on the production server.) I have tried the following steps (without success): I selected Targeted Runtimes under the Project Properties The Tomcat v5.5 option was disabled and The UI displayed this message: If the runtime you want to select is not displayed or is disabled you may need to uninstall one or more of the currently installed project facets. I then clicked on the Uninstall Facets... link. Under the Runtimes tab, only Tomcat 6 displayed. For Dynamic Web Module, I selected version 2.4 in place of 2.5. Under the Runtimes tab, Tomcat 5.5 now displayed. However, the UI now displayed this message: Cannot change version of project facet Dynamic Web Module to 2.4. The Finish button was disabled - so I reached a dead-end. I CAN successfully create a NEW Project with a Tomcat v5.5 configuration. For some reason, though, it will not let me downgrade' an existing Project. As a work-around, I created a new Project and copied the source files from the old Project. Nonetheless, the work-around was fairly painful and somewhat clumsy. Can anyone explain how I can 'downgrade' the Project configuration from 'Tomcat 6' to 'Tomcat 5'? Or perhaps shed some light on why this happened? Thanks Pete

    Read the article

  • C#, AES encryption check!

    - by Data-Base
    I have this code for AES encryption, can some one verify that this code is good and not wrong? it works fine, but I'm more concern about the implementation of the algorithm // Plaintext value to be encrypted. //Passphrase from which a pseudo-random password will be derived. //The derived password will be used to generate the encryption key. //Password can be any string. In this example we assume that this passphrase is an ASCII string. //Salt value used along with passphrase to generate password. //Salt can be any string. In this example we assume that salt is an ASCII string. //HashAlgorithm used to generate password. Allowed values are: "MD5" and "SHA1". //SHA1 hashes are a bit slower, but more secure than MD5 hashes. //PasswordIterations used to generate password. One or two iterations should be enough. //InitialVector (or IV). This value is required to encrypt the first block of plaintext data. //For RijndaelManaged class IV must be exactly 16 ASCII characters long. //KeySize. Allowed values are: 128, 192, and 256. //Longer keys are more secure than shorter keys. //Encrypted value formatted as a base64-encoded string. public static string Encrypt(string PlainText, string Password, string Salt, string HashAlgorithm, int PasswordIterations, string InitialVector, int KeySize) { byte[] InitialVectorBytes = Encoding.ASCII.GetBytes(InitialVector); byte[] SaltValueBytes = Encoding.ASCII.GetBytes(Salt); byte[] PlainTextBytes = Encoding.UTF8.GetBytes(PlainText); PasswordDeriveBytes DerivedPassword = new PasswordDeriveBytes(Password, SaltValueBytes, HashAlgorithm, PasswordIterations); byte[] KeyBytes = DerivedPassword.GetBytes(KeySize / 8); RijndaelManaged SymmetricKey = new RijndaelManaged(); SymmetricKey.Mode = CipherMode.CBC; ICryptoTransform Encryptor = SymmetricKey.CreateEncryptor(KeyBytes, InitialVectorBytes); MemoryStream MemStream = new MemoryStream(); CryptoStream CryptoStream = new CryptoStream(MemStream, Encryptor, CryptoStreamMode.Write); CryptoStream.Write(PlainTextBytes, 0, PlainTextBytes.Length); CryptoStream.FlushFinalBlock(); byte[] CipherTextBytes = MemStream.ToArray(); MemStream.Close(); CryptoStream.Close(); return Convert.ToBase64String(CipherTextBytes); } public static string Decrypt(string CipherText, string Password, string Salt, string HashAlgorithm, int PasswordIterations, string InitialVector, int KeySize) { byte[] InitialVectorBytes = Encoding.ASCII.GetBytes(InitialVector); byte[] SaltValueBytes = Encoding.ASCII.GetBytes(Salt); byte[] CipherTextBytes = Convert.FromBase64String(CipherText); PasswordDeriveBytes DerivedPassword = new PasswordDeriveBytes(Password, SaltValueBytes, HashAlgorithm, PasswordIterations); byte[] KeyBytes = DerivedPassword.GetBytes(KeySize / 8); RijndaelManaged SymmetricKey = new RijndaelManaged(); SymmetricKey.Mode = CipherMode.CBC; ICryptoTransform Decryptor = SymmetricKey.CreateDecryptor(KeyBytes, InitialVectorBytes); MemoryStream MemStream = new MemoryStream(CipherTextBytes); CryptoStream cryptoStream = new CryptoStream(MemStream, Decryptor, CryptoStreamMode.Read); byte[] PlainTextBytes = new byte[CipherTextBytes.Length]; int ByteCount = cryptoStream.Read(PlainTextBytes, 0, PlainTextBytes.Length); MemStream.Close(); cryptoStream.Close(); return Encoding.UTF8.GetString(PlainTextBytes, 0, ByteCount); } Thank you

    Read the article

  • How do I change the spacing between fields in a DataForm?

    - by Simon_Weaver
    How do I change the spacing between fields in a DataForm in Silverlight? I've tried editing the template but cannot find what I need. I thought all I needed to do was change the MinHeight and Margin of the DataField style, but that doesn't seem to do it. <Style TargetType="dataFormToolkit:DataField"> <Setter Property="IsTabStop" Value="False"/> <Setter Property="Margin" Value="2"/> <Setter Property="MinHeight" Value="5"/> <Setter Property="Template"> <Setter.Value> <ControlTemplate TargetType="dataFormToolkit:DataField"> <ContentControl x:Name="ContentControl" Foreground="{TemplateBinding Foreground}" HorizontalContentAlignment="Stretch" IsTabStop="False" VerticalAlignment="Center"/> </ControlTemplate> </Setter.Value> </Setter> </Style> I've found a number of articles about styling DataForm but many of them seem to be out of date. I don't see anything in the complete extracted template in Blend that corresponds to spacing.

    Read the article

  • How can I change the text in a <span></span> element using jQuery?

    - by Eric Reynolds
    I have a span element as follows: <span id="download">Download</span>. This element is controlled by a few radio buttons. Basically, what I want to do is have 1 button to download the item selected by the radio buttons, but am looking to make it a little more "flashy" by changing the text inside the <span> to say more specifically what they are downloading. The span is the downloading button, and I have it animated so that the span calls slideUp(), then should change the text, then return by slideDown().Here is the code I am using that does not want to work. $("input[name=method]").change(function() { if($("input[name=method]").val() == 'installer') { $('#download').slideUp(500); $('#download').removeClass("downloadRequest").removeClass("styling").css({"cursor":"default"}); $('#download').text("Download"); $('#download').addClass("downloadRequest").addClass("styling").css({"cursor":"pointer"}); $('#download').slideDown(500); } else if($("input[name=method]").val() == 'url') { $('#download').slideUp(500); $('#download').removeClass("downloadRequest").removeClass("styling").css({"cursor":"default"}); $('#download').text("Download From Vendor Website"); $('#download').addClass("styling").addClass("downloadRequest").css({"cursor":"pointer"}); $('#download').slideDown(500); } }); I changed the code a bit to be more readable so I know that it doesn't have the short code that jQuery so eloquently allows. Everything in the code works, with the exception of the changing of the text inside the span. I'm sure its a simple solution that I am just overlooking. Any help is appreciated, Eric R.

    Read the article

  • Unable to change the system zone setting on Windows Server 2008 R2.

    - by Ganesh
    Hi All, I have an MFC application that tries to change the system zone setting on the Windows Server 2008 R2. I am using the SetTimeZoneInformation() API which fails with the error code 1314 .i.e. “A required privilege is not held by the client.”. Please refer the sample code below: TIME_ZONE_INFORMATION l_TimeZoneInfo; DWORD l_dwRetVal = 0; ZeroMemory(&l_TimeZoneInfo, sizeof(TIME_ZONE_INFORMATION)); l_TimeZoneInfo.Bias = -330; l_TimeZoneInfo.StandardBias = 0; l_TimeZoneInfo.StandardDate.wDay = 0; l_TimeZoneInfo.StandardDate.wDayOfWeek = 0; l_TimeZoneInfo.StandardDate.wHour = 0; l_TimeZoneInfo.StandardDate.wMilliseconds = 0; l_TimeZoneInfo.StandardDate.wMinute = 0; l_TimeZoneInfo.StandardDate.wMonth = 0; l_TimeZoneInfo.StandardDate.wSecond = 0; l_TimeZoneInfo.StandardDate.wYear = 0; CString l_csDaylightName = _T("India Daylight Time"); CString l_csStdName = _T("India Standard Time"); wcscpy(l_TimeZoneInfo.DaylightName,l_csDaylightName.GetBuffer(l_csDaylightName.GetLength())); wcscpy(l_TimeZoneInfo.StandardName,l_csStdName.GetBuffer(l_csStdName.GetLength())); ::SetLastError(0); if(0 == ::SetTimeZoneInformation(&l_TimeZoneInfo)) { l_dwRetVal = ::GetLastError(); CString l_csErr = _T(""); l_csErr.Format(_T("%d"),l_dwRetVal); } The MFC application has been developed using Visual Studio 2008 and is UAC aware i.e. the application has UAC enabled in its manifest file with the UAC execution level set to “HighestAvailable”. I have administrator privileges and when I run the application it still fails to change the system zone setting. Thanks in Advance, Ganesh

    Read the article

  • How can we change views in a UISplitViewController other than using the popover and selecting?

    - by wolverine
    I have done a sample app with UISplitViewController studying the example they have provided. I have created three detailviews and have configured them to change by the default means. Either using the left/master view in landscape AND using the popover in the portrait orientation. Now I am trying to move to another view(previous/next) from the currentView by using left/right swipe in each view. For that, what I did was just created a function in the RootViewController. I copy-pasted the same code as that of the tablerow selection used by the popover from the RootViewController. I am calling this function from my current view's controller and is passing the respective index of the view(to be displayed next) from the current view. Function is being called but nothing is happening. Plz help me OR is anyother way to do it other than this complex step? I am giving the function that I used to change the view. - (void) rearrangeViews:(int)viewRow { UIViewController <SubstitutableDetailViewController> *detailViewController = nil; if (viewRow == 0) { DetailViewController *newDetailViewController = [[DetailViewController alloc] initWithNibName:@"DetailView" bundle:nil]; detailViewController = newDetailViewController; } if (viewRow == 1) { SecondDetailViewController *newDetailViewController = [[SecondDetailViewController alloc] initWithNibName:@"SecondDetailView" bundle:nil]; detailViewController = newDetailViewController; } if (viewRow == 2) { ThirdDetailViewController *newDetailViewController = [[ThirdDetailViewController alloc] initWithNibName:@"ThirdDetailView" bundle:nil]; detailViewController = newDetailViewController; } // Update the split view controller's view controllers array. NSArray *viewControllers = [[NSArray alloc] initWithObjects:self.navigationController, detailViewController, nil]; splitViewController.viewControllers = viewControllers; [viewControllers release]; if (rootPopoverButtonItem != nil) { [detailViewController showRootPopoverButtonItem:self.rootPopoverButtonItem]; } [detailViewController release]; }

    Read the article

  • JavaFX 2.0 - How to change legend color of a LineChart dynamically?

    - by marie
    I am trying to style my JavaFX linechart but I have some trouble with the legend. I know how to change the legend color of a line chart in the css file: .default-color0.chart-series-line { -fx-stroke: #FF0000, white; } .default-color1.chart-series-line { -fx-stroke: #00FF00, white; } .default-color2.chart-series-line { -fx-stroke: #0000FF, white; } .default-color0.chart-line-symbol { -fx-background-color: #FF0000, white; } .default-color1.chart-line-symbol { -fx-background-color: #00FF00, white; } .default-color2.chart-line-symbol { -fx-background-color: #0000FF, white; } But this is not enough for my purposes. I have three or more colored toggle buttons and a series of data for every button. The data should be displayed in the same color the button has after I have selected the button. This should be possible with a multiselection of the buttons, so that more than one series of data can be displayed simultaneously. For the chart lines I have managed it by changing the style after I clicked the button: .. dataList.add(series); .. series.getNode().setStyle("-fx-stroke: rgba(" + rgba + ")"); If I deselect the button I remove the data from the list. dataList.remove(series); That is working fine for the strokes, but how can I do the same for the legend? You can see an example below. First I clicked the red button, thus the stroke and the legend is red (default-color0). After that I clicked the blue button. Here you can see the problem. The stroke is blue but the legend is green, because default color1 is used and I do not know how to change the legend color.

    Read the article

  • Number of simple mutations to change one string to another?

    - by mstksg
    Hi; I'm sure you've all heard of the "Word game", where you try to change one word to another by changing one letter at a time, and only going through valid English words. I'm trying to implement an A* Algorithm to solve it (just to flesh out my understanding of A*) and one of the things that is needed is a minimum-distance heuristic. That is, the minimum number of one of these three mutations that can turn an arbitrary string a into another string b: 1) Change one letter for another 2) Add one letter at a spot before or after any letter 3) Remove any letter Examples aabca => abaca: aabca abca abaca = 2 abcdebf => bgabf: abcdebf bcdebf bcdbf bgdbf bgabf = 4 I've tried many algorithms out; I can't seem to find one that gives the actual answer every time. In fact, sometimes I'm not sure if even my human reasoning is finding the best answer. Does anyone know any algorithm for such purpose? Or maybe can help me find one? Thanks.

    Read the article

< Previous Page | 215 216 217 218 219 220 221 222 223 224 225 226  | Next Page >