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  • how to add a sidebar to a .net page based on a master page that doesnt have a sidebar.

    - by UXdesigner
    Hello, I have been told that I should add a sidebar to one page of this .net project, but the master page don't include a sidebar. How can I add a sidebar to one page only ? This is the code for the Master Template, can anyone suggest or help me out here? I'd buy a book and read more, but I have to do this for the next 12 hours. <%@ Master Language="C#" AutoEventWireup="true" CodeFile="Public.master.cs" Inherits="Public" %> <%@ Register Assembly="AjaxControlToolkit" Namespace="AjaxControlToolkit" TagPrefix="cc2" %> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <%--<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> --%><html xmlns="http://www.w3.org/1999/xhtml"> <head runat="server"> <title></title> <%--<link href="favicon.ico" rel="Shortcut Icon" type="image/x-icon" />--%> <link href="<%= Server.MapPath("~/css/main2.css") %>" rel="stylesheet" type="text/css" media="all" /> <link href="<%= Server.MapPath("~/css/dropdown.css") %>" media="screen" rel="stylesheet" type="text/css" /> <link href="<%= Server.MapPath("~/css/default.advanced.css") %>" media="screen" rel="stylesheet" type="text/css" /> <link href="<%= Server.MapPath("~/css/vlightbox.css") %>" rel="stylesheet" type="text/css" /> <link href="<%= Server.MapPath("~/css/visuallightbox.css") %>" rel="stylesheet" type="text/css" media="screen" /> <link href="<%= Server.MapPath("~/boxes.css") %>"rel="stylesheet" type="text/css" media="screen" /> <script src="<%= Server.MapPath("~/engine/js/jquery.min.js") %>" ype="text/javascript"></script> <script src="<%= Server.MapPath("~/js/cufon-yui.js") %>" type="text/javascript"></script> <script src="<%= Server.MapPath("~/js/AFB_400.font.js") %>" type="text/javascript"></script> <style type="text/css"> #vlightbox a#vlb { display:none } </style> <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.3.2/jquery.min.js" ></script> <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jqueryui/1.5.3/jquery-ui.min.js" ></script> <script type="text/javascript"> Cufon.replace('h2'); </script> <script type="text/javascript"> Cufon.replace('h3'); </script> <script type="text/javascript"> Cufon.replace('h5'); </script> <!--[if IE 8]> <style type="text/css"> #footer {display:table;} </style> <![endif]--> <style> ul#nav { width:100%; height:36px; display:block; background-color:#000; background-repeat:repeat-x; } #wrapthatbanner {display:block; float:left; width:100%; height:529px; margin-left:-20px; margin-bottom:0px; } </style> <script type='text/javascript' src='http://ajax.googleapis.com/ajax/libs/jquery/1.3.2/jquery.min.js?ver=1.3.2'></script> <script type="text/javascript"> $(document).ready(function() { $("#footer").stickyFooter(); }); // sticky footer plugin (function($) { var footer; $.fn.extend({ stickyFooter: function(options) { footer = this; positionFooter(); $(window) .scroll(positionFooter) .resize(positionFooter); function positionFooter() { var docHeight = $(document.body).height() - $("#sticky-footer-push").height(); if (docHeight < $(window).height()) { var diff = $(window).height() - docHeight; if (!$("#sticky-footer-push").length > 0) { $(footer).before('<div id="sticky-footer-push"></div>'); } $("#sticky-footer-push").height(diff); } } } }); })(jQuery); </script> </head> <body id="@@(categoria)@@"> <form id="form1" runat="server"> <asp:ScriptManager ID="ScriptManager1" runat="server" EnablePageMethods="true" AsyncPostBackTimeout="900"></asp:ScriptManager> <div id="container"> <asp:UpdatePanel ID="UpdatePanel1" runat="server"> <ContentTemplate> <div id="header"> <div id="headerlink"> <table width="100%" border="0"> <tr> <td height="77px;" width="67%"> <asp:ImageButton PostBackUrl="~/index.aspx" ImageUrl="~/images/Titulos/5.png" runat="server" alt="" name="screen_logo" width="257" hspace="10" vspace="10" border="0" id="screen_logo" title="" /> </td> <td valign="top" align="right" width="33%"> <table> <tr> <td> <asp:Label ID="lblFullMessage" Visible="false" runat="server" Font-Size="X-Small" ForeColor="White" Text="Please enter the {0}, {1} and {2} characters from your password."></asp:Label> </td> </tr> <tr valign="middle"> <td> <img src="../images/login.jpg"</td> <td valign="top"> <asp:TextBox runat="server" Height="16px" Font-Size="Small" ID="txtLogin" Width="100px"></asp:TextBox> <asp:Button ID="btnLogin" Height="20px" Font-Size="X-Small" runat="server" Text="Go" OnClick="btnLogin_Click" /> </td> </tr> <tr> <td> <asp:Label ID="lblError" Visible="false" runat="server" Font-Size="X-Small" ForeColor="Red" Text="Error"></asp:Label> </td> </tr> </table> </td> </tr> </table> </div> </div> </ContentTemplate> </asp:UpdatePanel> <ul id="nav" class="dropdown dropdown-horizontal"> <li><asp:HyperLink NavigateUrl="~/index.aspx" CssClass="dir" runat="server" ID="lnk1">Home</asp:HyperLink></li> <li><asp:HyperLink NavigateUrl="~/PublicSide/link.aspx" CssClass="dir" runat="server" ID="lnk3">link</asp:HyperLink></li> <li><asp:HyperLink NavigateUrl="~/PublicSide/link.aspx" CssClass="dir" runat="server" ID="lnk4">link</asp:HyperLink></li> <li><asp:HyperLink NavigateUrl="~/PublicSide/link.aspx" CssClass="dir" runat="server" ID="lnk7">link</asp:HyperLink></li> <li><asp:HyperLink NavigateUrl="~/PublicSide/link.aspx" CssClass="dir" runat="server" ID="lnk5">link</asp:HyperLink></li> <li><asp:HyperLink NavigateUrl="~/PublicSide/link.aspx" CssClass="dir" runat="server" ID="lnk2">link</asp:HyperLink></li> <li><asp:HyperLink NavigateUrl="~/PublicSide/link.aspx" CssClass="dir" runat="server" ID="lnk6">link</asp:HyperLink></li> </ul> <div id="wmfg"> </div> <div id="content"><asp:ContentPlaceHolder ID="Content1" runat="server"> </asp:ContentPlaceHolder></div> <div id="footer">Footer</div> </div> </form> </body> </html>

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  • Where am I going wrong in my Xml Schema?

    - by chobo2
    Hi I am trying to make a XML Schema but everytime I use it and try to validate my data I get an error. I get this error: Validation of the XML Document failed! Error message(s): Could not find schema information for the element 'Email'. Line: 1 Column:1213 http://www.xmlforasp.net/SchemaValidator.aspx My Xml file I am trying to validate. <?xml version="1.0" encoding="utf-8" ?> <School> <SchoolPrefix>BCIT</SchoolPrefix> <TeacherAccounts> <Account> <StudentNumber>A00140000</StudentNumber> <Password>123456</Password> <Email>[email protected]</Email> </Account> <Account> <StudentNumber>A00000041</StudentNumber> <Password>1234567</Password> <Email>[email protected]</Email> </Account> <Account> <StudentNumber>A0400100</StudentNumber> <Password>1234567</Password> <Email>[email protected]</Email> </Account> </TeacherAccounts> <FullTimeAccounts> <Account> <StudentNumber>A00000000</StudentNumber> <Password>1234567</Password> <Email>[email protected]</Email> </Account> <Account> <StudentNumber>A00141000</StudentNumber> <Password>1234567</Password> <Email>[email protected]</Email> </Account> </FullTimeAccounts> <PartTimeAccounts> <Account> <StudentNumber>A81020409</StudentNumber> <Password>1234567</Password> <Email>[email protected]</Email> </Account> <Account> <StudentNumber>A040014000</StudentNumber> <Password>1234567</Password> <Email>[email protected]</Email> </Account> <Account> <StudentNumber>A00024040</StudentNumber> <Password>1234567</Password> <Email>[email protected]</Email> </Account> <Account> <StudentNumber>A00004101</StudentNumber> <Password>1234567</Password> <Email>[email protected]</Email> </Account> </PartTimeAccounts> </School> XMl Schema <?xml version="1.0" encoding="utf-8"?> <xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema" targetNamespace="http://www.nothing.com" xmlns="http://www.nothing.com" elementFormDefault="qualified"> <xs:element name="School"> <xs:complexType> <xs:sequence> <xs:element name="SchoolPrefix" minOccurs="1" maxOccurs="1"> <xs:simpleType> <xs:restriction base="xs:string"> <xs:minLength value="2" /> <xs:maxLength value="8" /> </xs:restriction> </xs:simpleType> </xs:element> <xs:element name="TeacherAccounts" minOccurs="1" maxOccurs="1"> <xs:complexType> <xs:sequence> <xs:element name="Account" type="UserInfo" minOccurs="0" maxOccurs="unbounded" /> </xs:sequence> </xs:complexType> </xs:element> <xs:element name="FullTimeAccounts"> <xs:complexType> <xs:sequence> <xs:element name="Account" type="UserInfo" minOccurs="0" maxOccurs="unbounded"/> </xs:sequence> </xs:complexType> </xs:element> <xs:element name="PartTimeAccounts"> <xs:complexType> <xs:sequence> <xs:element name="Account" type="UserInfo" minOccurs="0" maxOccurs="unbounded" /> </xs:sequence> </xs:complexType> </xs:element> </xs:sequence> </xs:complexType> </xs:element> <xs:complexType name="UserInfo"> <xs:sequence> <xs:element name="StudentNumber"> <xs:simpleType> <xs:restriction base="xs:string"> <xs:minLength value="1"/> <xs:maxLength value="50"/> </xs:restriction> </xs:simpleType> </xs:element> <xs:element name="Password"> <xs:simpleType> <xs:restriction base="xs:string"> <xs:minLength value="6"/> <xs:maxLength value="50"/> </xs:restriction> </xs:simpleType> </xs:element> <xs:element name="Email"> <xs:simpleType> <xs:restriction base="xs:string"> <xs:pattern value="\w+([-+.']\w+)*@\w+([-.]\w+)*\.\w+([-.]\w+)*"/> </xs:restriction> </xs:simpleType> </xs:element> </xs:sequence> </xs:complexType> </xs:schema>

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  • MAMP Pro Uninstaller Throws "The privileged action failed." Error Even After Entering Password

    - by BigM
    I have followed, and successfully completed, every step in this SO article to remove MAMP Pro in favor of installing the free version as I only need it for one site anyway. However, when I run the uninstaller I still get "The privileged action failed." after providing the password. Can anybody shed a little light on this maybe? I'm just trying to get MAMP Pro uninstalled so I can install the free MAMP stack.

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  • How to make Shared Keys .ssh/authorized_keys and sudo work together?

    - by farinspace
    I've setup the .ssh/authorized_keys and am able to login with the new "user" using the pub/private key ... I have also added "user" to the sudoers list ... the problem I have now is when I try to execute a sudo command, something simple like: $ sudo cd /root it will prompt me for my password, which I enter, but it doesn't work (I am using the private key password I set) Also, ive disabled the users password using $ passwd -l user What am I missing? Somewhere my initial remarks are being misunderstood ... I am trying to harden my system ... the ultimate goal is to use pub/private keys to do logins versus simple password authentication. I've figured out how to set all that up via the authorized_keys file. Additionally I will ultimately prevent server logins through the root account. But before I do that I need sudo to work for a second user (the user which I will be login into the system with all the time). For this second user I want to prevent regular password logins and force only pub/private key logins, if I don't lock the user via" passwd -l user ... then if i dont use a key, i can still get into the server with a regular password. But more importantly I need to get sudo to work with a pub/private key setup with a user whos had his/her password disabled. Edit: Ok I think I've got it (the solution): 1) I've adjusted /etc/ssh/sshd_config and set PasswordAuthentication no This will prevent ssh password logins (be sure to have a working public/private key setup prior to doing this 2) I've adjusted the sudoers list visudo and added root ALL=(ALL) ALL dimas ALL=(ALL) NOPASSWD: ALL 3) root is the only user account that will have a password, I am testing with two user accounts "dimas" and "sherry" which do not have a password set (passwords are blank, passwd -d user) The above essentially prevents everyone from logging into the system with passwords (a public/private key must be setup). Additionally users in the sudoers list have admin abilities. They can also su to different accounts. So basically "dimas" can sudo su sherry, however "dimas can NOT do su sherry. Similarly any user NOT in the sudoers list can NOT do su user or sudo su user. NOTE The above works but is considered poor security. Any script that is able to access code as the "dimas" or "sherry" users will be able to execute sudo to gain root access. A bug in ssh that allows remote users to log in despite the settings, a remote code execution in something like firefox, or any other flaw that allows unwanted code to run as the user will now be able to run as root. Sudo should always require a password or you may as well log in as root instead of some other user.

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  • Can the Firefox password manager store and manage passwords for multiple sub-domains or different URLs in the same domain?

    - by Howiecamp
    Can the Firefox password manager store and manage passwords for multiple sub-domains, or for multiple URLs in the same domain? The default behavior of Firefox is that all requests for *.domain.com are treated as the same. I'd like to have Firefox do the following: Store and manage passwords separately for multiple sub-domains, e.g. mail.google.com and picasa.google.com Store and manage passwords separately for different URLs in the same domain, e.g. http://mail.google.com/a/company1.com and http://mail.google.com/a/company2.com

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  • How to grant su access to wheel without asking for password on FreeBSD?

    - by cstamas
    I would like to grant users of the wheel group (other sysadmins) su access without being asked for password. I know how to do it with pam in linux, but the question now is for FreeBSD. I am not familiar with the syntax for FreeBSD's PAM subsystem. What shall I enter in /etc/pam.d/su instead of the default: auth sufficient pam_rootok.so no_warn auth sufficient pam_self.so no_warn auth requisite pam_group.so no_warn group=wheel root_only fail_safe ruser auth include system # account account include system # session session required pam_permit.so

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  • How can I read password protected Word files on OS X ?

    - by Ohad
    I receive Word documents by mail and read them using the built-in Gmail reader. Sometimes the documents are password protected and I need to obtain access to a Windows machine with Office installed in order to read them. Is there a quicker / less hassle requiring method ? I don't want to have to install Vmware / Parallels nor Office on my fresh and sterile macbook.

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  • On a local network, are you able to password protect certain folders and how (in windows xp)?

    - by Derek
    I have a local network set up for my small office which consists of me, the manager, my wife, the secretary, and a few sales people/others. I would like to share passwords over the network and other such things privately to my wife, the secretary, but would not like the sales people and others to have access to it, yet I need the others to have access to other folders/documents that I'd like to share. How would I go about doing this if not by password? Thanks in advance

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  • Is there a way to password protect a windows application on Windows 7 without changing the exe?

    - by M Anthony
    I want to password-protect a few applications (not just files/folders) on Windows 7, such that, there is no change to the protected application's .exe file itself. Since the .exe files can't be changed, "Empathy" and "Protect Exe" can't be used as they change the .exe file. I tried using WinGuard Pro and LockThis! also but they don't work for Microsoft Outlook and some other apps (that I need). Is it possible at all? Please help.

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  • Ubuntu maverick username and password works with ssh but not with smb when I connect from MacOS 10.6.5

    - by biomed
    I have an Ubuntu Maverick desktop that I can easily connect using ssh but when I want to see the shared directories using "go to server", MacOs connects to the Ubuntu machine and I can see the shared directories but when I enter my username and passord to get access it complains about me entering wrong username and/or password. Any ideas? What more information would you need to give me some advise? Is there a step by step how-to manual to get this done? thanks

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  • How do I completely uninstall mySQL on XP, including the root password?

    - by user341219
    All I need is to be able to log in using root, but have forgotten the password. None of the steps to reset i found online work (i don't even have some of the executables mentioned such as mysql-nt.exe) However I have no problem deleting all databases (i have scripts) and intallations and starting completely from scratch... but uninstalling and deleting directories doesn't work. Thanks.

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  • Can I password protect a location (not directory) in apache using htaccess?

    - by dimvic
    I have used code like this in apache configuration to protect locations with password <Location ~ "/admin.*"> AuthType Basic AuthName "Protected Area" AuthUserFile /home/user/public_html/.htpasswd Require valid-user </Location> is there a way to do the same thing using an htaccess file? the locations I want to protect don't really exist on the filesystem, it's locations available thanks to mod_rewrite

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  • KnpLabs / DoctrineBehaviors Translatable - currentLocale = null

    - by Ruben
    Using the trait \Knp\DoctrineBehaviors\Model\Translatable\Translation inside an Entity, I see that the property currentLocale is never setted , so we always obtain the default locale ('en') in all the calls to $this->translate(). If I change this getDefaultLocale, all the translations are made correctly, so I think that is no problem with the fallback. I tried debug the currentLocaleCallable. I see that if I put a "var_dump ($this-container-get('request'));" in the contructor of currentLocaleCallable, the request have a locale to null. And outside the request has the correct locale.It seems that container is not in the scope: request , i don't know how can I get it to work I post an issue in github https://github.com/KnpLabs/DoctrineBehaviors/issues/71 EDITED This service is defined in vendor/knplabs/doctrine-behaviors/config/orm-services.yml and is: knp.doctrine_behaviors.reflection.class_analyzer: class: "%knp.doctrine_behaviors.reflection.class_analyzer.class%" public: false knp.doctrine_behaviors.translatable_listener: class: "%knp.doctrine_behaviors.translatable_listener.class%" public: false arguments: - "@knp.doctrine_behaviors.reflection.class_analyzer" - "%knp.doctrine_behaviors.reflection.is_recursive%" - "@knp.doctrine_behaviors.translatable_listener.current_locale_callable" tags: - { name: doctrine.event_subscriber } knp.doctrine_behaviors.translatable_listener.current_locale_callable: class: "%knp.doctrine_behaviors.translatable_listener.current_locale_callable.class%" arguments: - "@service_container" # lazy request resolution public: false EDIT 2: My composer.json "php": ">=5.3.3", "symfony/symfony": "2.3.*", "doctrine/orm": ">=2.2.3,<2.4-dev", "doctrine/doctrine-bundle": "1.2.*", "twig/extensions": "1.0.*", "symfony/assetic-bundle": "2.3.*", "symfony/swiftmailer-bundle": "2.3.*", "symfony/monolog-bundle": "2.3.*", "sensio/distribution-bundle": "2.3.*", "sensio/framework-extra-bundle": "2.3.*", "sensio/generator-bundle": "2.3.*", "incenteev/composer-parameter-handler": "~2.0", "friendsofsymfony/user-bundle": "1.3.*", "avalanche123/imagine-bundle": "v2.1", "raulfraile/ladybug-bundle": "~1.0", "genemu/form-bundle": "2.2.*", "friendsofsymfony/rest-bundle": "0.12.0", "stof/doctrine-extensions-bundle": "dev-master", "sonata-project/admin-bundle": "dev-master", "a2lix/translation-form-bundle": "1.*@dev", "sonata-project/user-bundle": "dev-master", "psliwa/pdf-bundle": "dev-master", "jms/serializer-bundle": "dev-master", "jms/di-extra-bundle": "dev-master", "knplabs/doctrine-behaviors": "dev-master", "sonata-project/doctrine-orm-admin-bundle": "dev-master", "knplabs/knp-paginator-bundle": "dev-master", "friendsofsymfony/jsrouting-bundle": "~1.1", "zendframework/zend-validator": ">=2.0.0-rc2", "zendframework/zend-barcode": ">=2.0.0-rc2"

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  • Mapping issue with multi-field primary keys using hibernate/JPA annotations

    - by Derek Clarkson
    Hi all, I'm stuck with a database which is using multi-field primary keys. I have a situation where I have a master and details table, where the details table's primary key contains fields which are also the foreign key's the the master table. Like this: Master primary key fields: master_pk_1 Details primary key fields: master_pk_1 details_pk_2 details_pk_3 In the Master class we define the hibernate/JPA annotations like this: @Id @GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "idGenerator") @Column(name = "master_pk_1") private long masterPk1; @OneToMany(cascade=CascadeType.ALL) @JoinColumn(name = "master_pk_1", referencedColumnName = "master_pk_1") private List<Details> details = new ArrayList<Details>(); And in the details class I have defined the id and back reference like this: @EmbeddedId @AttributeOverrides( { @AttributeOverride( name = "masterPk1", column = @Column(name = "master_pk_1")), @AttributeOverride(name = "detailsPk2", column = @Column(name = "details_pk_2")), @AttributeOverride(name = "detailsPk2", column = @Column(name = "details_pk_2")) }) private DetailsPrimaryKey detailsPrimaryKey = new DetailsPrimaryKey(); @ManyToOne @JoinColumn(name = "master_pk_1", referencedColumnName = "master_pk_1", insertable=false) private Master master; The goal of all of this was that I could create a new master, add some details to it, and when saved, JPA/Hibernate would generate the new id for master in the masterPk1 field, and automatically pass it down to the details records, storing it in the matching masterPk1 field in the DetailsPrimaryKey class. At least that's what the documentation I've been looking at implies. What actually happens is that hibernate appears to corectly create and update the records in the database, but not pass the key to the details classes in memory. Instead I have to manually set it myself. I also found that without the insertable=true added to the back reference to master, that hibernate would create sql that had the master_pk_1 field listed twice in the insert statement, resulting in the database throwing an exception. My question is simply is this arrangement of annotations correct? or is there a better way of doing it?

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  • Git push current branch to a remote with Heroku

    - by cmaughan
    I'm trying to create a staging branch on Heroku, but there's something I don't quite get. Assuming I've already created a heroku app and setup the remote to point to staging-remote, If I do: git checkout -b staging staging-remote/master I get a local branch called 'staging' which tracks staging-remote/master - or that's what I thought.... But: git remote show staging-remote Gives me this: remote staging Fetch URL: [email protected]:myappname.git Push URL: [email protected]:myappname.git HEAD branch: master Remote branch: master tracked Local branch configured for 'git pull': staging-remote merges with remote master Local ref configured for 'git push': master pushes to master (up to date) As you can see, the pull looks reasonable, but the default push does not. It implies that if I do: git push staging-remote I'm going to push my local master branch up to the staging branch. But that's not what I want.... Basically, I want to merge updates into my staging branch, then easily push it to heroku without having to specify the branch like so: git push staging-remote mybranch:master The above isn't hard to do, but I want to avoid accidentally doing the previous push and pushing the wrong branch... This is doubly important for the production branch I'd like to create! I've tried messing with git config, but haven't figured out how to get this right yet...

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  • How can you connect to a password protected MS Access Database from a Spring JdbcTemplate?

    - by Tim Visher
    I need to connect to a password protected MS Access 2003 DB using the JDBC-ODBC bridge. I can't find out how to specify the password in the connect string, or even if that is the correct method of connecting. It would probably be relevant to mention that this is a Spring App which is accessing the database through a JdbcTemplate configured as a datasource bean in our application context file. Some relevant snippets: from application-context.xml <bean id="jdbcTemplate" class="org.springframework.jdbc.core.JdbcTemplate"> <property name="dataSource" ref="legacyDataSource" /> </bean> <bean id="jobsheetLocation" class="java.lang.String"> <constructor-arg value="${jobsheet.location}"/> </bean> <bean id="legacyDataSource" class="org.springframework.jdbc.datasource.DriverManagerDataSource"> <property name="driverClassName" value="${jdbc.legacy.driverClassName}" /> <property name="url" value="${jdbc.legacy.url}"/> <property name="password" value="-------------" /> </bean> from our build properties jdbc.legacy.driverClassName=sun.jdbc.odbc.JdbcOdbcDriver jdbc.legacy.url=jdbc:odbc:Driver\={Microsoft Access Driver (*.mdb)};Dbq\=@LegacyDbPath@;DriverID\=22;READONLY\=true Any thoughts?

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  • Forms authentication: how do you store username password in web.config?

    - by Nick G
    I'm used to using Forms Authentication with a database, but I'm writing a little internal utility and the app doesn't have a database so I want to store the username and password in web.config. However for some reason, forms authentication is still trying to access SQL Server and I can't see how to stop it doing this and pick up the credentials from web.config. What am I doing wrong? I just get the error "Failed to generate a user instance of SQL Server due to a failure in impersonating the client. The connection will be closed." Here are the relevant sections of my web.config: <configuration> <system.web> <authentication mode="Forms"> <forms loginUrl="~/Login.aspx" timeout="60" name=".LoginCookie" path="/" > <credentials passwordFormat="Clear"> <user name="user1" password="[pass]" /> <user name="user2" password="[pass]" /> </credentials> </forms> </authentication> <authorization> <deny users="?" /> </authorization> </system.web> </configuration>

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