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  • Migrating Identity Providers - specifying a new users password hash.

    - by Stephen Denne
    We'd like to switch Identity Provider (and Web Access Manager), and also the user directory we use, but would like to do so without users needing to change their password. We currently have the SSHA of the passwords. I'm expecting to write code to perform the migration. I don't mind how complex the code has to be, rather my concern is whether such a migration is possible at all. MS Active Directory would be our preferred user store, but I believe that it can not have new users set up in it with a particular password hash. Is that correct? What user directory stores can be populated with users already set up with a SSHA password? What Identity Provider and Access Management products work with those stores?

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  • Find out what resource is triggering bad password attempt?

    - by Craig Tataryn
    Background: Have a problem at work where I am constantly being locked out of my computer. We are in an environment that has a Domain Controller and we use Active Directory for authentication. By going through my normal workflow while on the phone with Desktop Support we were able to track the bad password attempts that were causing the lockouts to an application: "Eclipse". This is the application I use to do software development. I immediately thought it was a cached password for our SVN server that's the culprit, however the desktop support person couldn't tell me which resource the password attempt was being made against (i.e. which URL for instance). Question: Is there a way that I can monitor bad authentication requests made by an application on my desktop and find out what resource they are attempting it against?

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  • In windows 7, how do I disable "Ease of Access" pop ups on the logon screen as I am typing my password?

    - by Silfheed
    So the question is pretty much in the title. Things to note: 'Use On-Screen Keyboard' is un-checked I have disabled Ease of Access by replacing utilman.exe But my problem now is that when I'm typing my password, It'll stop all together on the 'u' or the 'p' character and I have to click on the ease of access button then move focus back to the password to continue. Yes, one obvious solution is to not use a password that has 'u' or 'p' in it (which I've done), but that's obviously not a completely satisfying answer. Thanks! I feel a tumbleweed badge...

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  • How can I prevent Apache from exposing a user's password?

    - by Marius Marais
    When using basic authentication (specifically via LDAP, but htpasswd also) with Apache, it makes the REMOTE_USER variable available to the PHP / Ruby / Python code underneath -- this is very useful for offloading authentication to the webserver. In our office environment we have lots of internal applications working like this over SSL, all quite secure. BUT: Apache exposes the PHP_AUTH_USER (=REMOTE_USER) and PHP_AUTH_PW variables to any application inside PHP. (PHP_AUTH_PW contains the plaintext password the user entered.) This means it's possible for the app to harvest usernames and passwords. Presumably the same information is available to Python and Ruby (all three are currently in use; PHP is being phased out). So how can I prevent Apache from doing this? One idea is to use Kerberos Negotiate authentication (which does not expose the password and has the benefit of being SSO), but that automatically falls back to Basic for some browsers (Chrome and in some cases Firefox), causing the password to be exposed again.

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  • How to disable the Windows 8 lock screen, without disabling the password?

    - by zeel
    Windows 8 now has a slide-away lock screen like so many other OSes designed for a touch interface. However on a non-touch PC/laptop this is just an extra annoying step in logging in. There is an option to disable it, but this also turns off the password requirement. Is there a way to disable the lock screen, and have Windows wake up directly to the password entry screen? Often when pressing a key to hide it there will be a noticeable delay before password entry is accepted, and no indicator, so one can easily lose the first character or so that they attempt to type, thus failing the login attempt.

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  • non-interactive ssh sudo... prompts for the password in plain text

    - by Iain
    I'm running some non-interactive ssh commands. The ssh authentication is taken care of fine through the ssh agent, but if I run a command that requires sudo then the password prompt in my terminal is plain text. For example: ssh remotemachine "sudo -u www mkdir -p /path/to/new/folder" will prompt me for the password in plain text. Does anyone know how I can get it to use the normal secure prompt or that I can pass the password via a switch? (as then I can set up a secure prompt on this side before I send the command) Any help is much appreciated.

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  • How to enable/disable authentication without password when executing commands as superuser?

    - by 44taka
    On a Fedora 19 system which I set up for somebody a while ago I noticed that no authentication is required when commands are executed as the superuser. So, for example, when running Yum Extender, configuring the firewall or running some command with sudo in the terminal, I am not asked to provide a password. (With graphical applications the authentication dialog pops up for a few milliseconds.) For better security I would like to disable this automatic (authentication-less) assumption of superuser privileges. I do not remember if or how I enabled this authentication without a password. I might have enabled it for convenience for the non-pro user of this machine, but did not do any "fancy" things (like editing config files) to do so. I did not edit the sudoer file. I just checked that. I might have checked a "Do not ask for password again" checkbox or something similar. Whatever I did, I would like to undo it and enforce authentication for superuser tasks again.

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  • Using the login Details via Application

    - by ramin ss
    I have a CURL(in C++) to send my user and pass to remauth.php file so i think i do something wrong on remuth.php ( because i am basic in php and my program can not run because the auth not passed.) I use login via Application. my CURL: bool Auth_PerformSessionLogin(const char* username, const char* password) { curl_global_init(CURL_GLOBAL_ALL); CURL* curl = curl_easy_init(); if (curl) { char url[255]; _snprintf(url, sizeof(url), "http://%s/remauth.php", "SITEADDRESS.com"); char buf[8192] = {0}; char postBuf[8192]; _snprintf(postBuf, sizeof(postBuf), "%s&&%s", username, password); curl_easy_setopt(curl, CURLOPT_URL, url); curl_easy_setopt(curl, CURLOPT_WRITEFUNCTION, AuthDataReceived); curl_easy_setopt(curl, CURLOPT_WRITEDATA, (void*)&buf); curl_easy_setopt(curl, CURLOPT_USERAGENT, "IW4M"); curl_easy_setopt(curl, CURLOPT_FAILONERROR, true); curl_easy_setopt(curl, CURLOPT_POST, 1); curl_easy_setopt(curl, CURLOPT_POSTFIELDS, postBuf); curl_easy_setopt(curl, CURLOPT_POSTFIELDSIZE, -1); curl_easy_setopt(curl, CURLOPT_SSL_VERIFYPEER, false); CURLcode code = curl_easy_perform(curl); curl_easy_cleanup(curl); curl_global_cleanup(); if (code == CURLE_OK) { return Auth_ParseResultBuffer(buf); } else { Auth_Error(va("Could not reach the SITEADDRESS.comt server. Error code from CURL: %x.", code)); } return false; } curl_global_cleanup(); return false; } and my remauth.php: <?php ob_start(); $host=""; // Host name $dbusername=""; // Mysql username $dbpassword=""; // Mysql password $db_name=""; // Database name $tbl_name=""; // Table name // Connect to server and select databse. mysql_connect("$host", "$dbusername", "$dbpassword") or die(mysql_error()); mysql_select_db("$db_name") or die(mysql_error()); // Define $username and $password //$username=$username; //$password=md5($_POST['password']); //$password=$password; $username=$_POST['username']; $password=$_POST['password']; //$post_item[]='action='.$_POST['submit']; // To protect MySQL injection (more detail about MySQL injection) $username = stripslashes($username); $password = stripslashes($password); $username = mysql_real_escape_string($username); $password = mysql_real_escape_string($password); $sql="SELECT * FROM $tbl_name WHERE username='$username'"; $result=mysql_query($sql); // Mysql_num_row is counting table row $count=mysql_num_rows($result); // If result matched $username and $password, table row must be 1 row if($count==1){ $row = mysql_fetch_assoc($result); if (md5(md5($row['salt']).md5($password)) == $row['password']){ session_register("username"); session_register("password"); echo "#"; return true; } else { echo "o"; return false; } } else{ echo "o"; return false; } ob_end_flush(); ?> ///////////////////////////////////

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  • PHP Multiple User Login Form - Navigation to Different Pages Based on Login Credentials

    - by Zulu Irminger
    I am trying to create a login page that will send the user to a different index.php page based on their login credentials. For example, should a user with the "IT Technician" role log in, they will be sent to "index.php", and if a user with the "Student" role log in, they will be sent to the "student/index.php" page. I can't see what's wrong with my code, but it's not working... I'm getting the "wrong login credentials" message every time I press the login button. My code for the user login page is here: <?php session_start(); if (isset($_SESSION["manager"])) { header("location: http://www.zuluirminger.com/SchoolAdmin/index.php"); exit(); } ?> <?php if (isset($_POST["username"]) && isset($_POST["password"]) && isset($_POST["role"])) { $manager = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["username"]); $password = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["password"]); $role = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["role"]); include "adminscripts/connect_to_mysql.php"; $sql = mysql_query("SELECT id FROM Users WHERE username='$manager' AND password='$password' AND role='$role' LIMIT 1"); $existCount = mysql_num_rows($sql); if (($existCount == 1) && ($role == 'IT Technician')) { while ($row = mysql_fetch_array($sql)) { $id = $row["id"]; } $_SESSION["id"] = $id; $_SESSION["manager"] = $manager; $_SESSION["password"] = $password; $_SESSION["role"] = $role; header("location: http://www.zuluirminger.com/SchoolAdmin/index.php"); } else { echo 'Your login details were incorrect. Please try again <a href="http://www.zuluirminger.com/SchoolAdmin/index.php">here</a>'; exit(); } } ?> <?php if (isset($_POST["username"]) && isset($_POST["password"]) && isset($_POST["role"])) { $manager = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["username"]); $password = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["password"]); $role = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["role"]); include "adminscripts/connect_to_mysql.php"; $sql = mysql_query("SELECT id FROM Users WHERE username='$manager' AND password='$password' AND role='$role' LIMIT 1"); $existCount = mysql_num_rows($sql); if (($existCount == 1) && ($role == 'Student')) { while ($row = mysql_fetch_array($sql)) { $id = $row["id"]; } $_SESSION["id"] = $id; $_SESSION["manager"] = $manager; $_SESSION["password"] = $password; $_SESSION["role"] = $role; header("location: http://www.zuluirminger.com/SchoolAdmin/student/index.php"); } else { echo 'Your login details were incorrect. Please try again <a href="http://www.zuluirminger.com/SchoolAdmin/index.php">here</a>'; exit(); } } ?> And the form that the data is pulled from is shown here: <form id="LoginForm" name="LoginForm" method="post" action="http://www.zuluirminger.com/SchoolAdmin/user_login.php"> User Name:<br /> <input type="text" name="username" id="username" size="50" /><br /> <br /> Password:<br /> <input type="password" name="password" id="password" size="50" /><br /> <br /> Log in as: <select name="role" id="role"> <option value="">...</option> <option value="Head">Head</option> <option value="Deputy Head">Deputy Head</option> <option value="IT Technician">IT Technician</option> <option value="Pastoral Care">Pastoral Care</option> <option value="Bursar">Bursar</option> <option value="Secretary">Secretary</option> <option value="Housemaster">Housemaster</option> <option value="Teacher">Teacher</option> <option value="Tutor">Tutor</option> <option value="Sanatorium Staff">Sanatorium Staff</option> <option value="Kitchen Staff">Kitchen Staff</option> <option value="Parent">Parent</option> <option value="Student">Student</option> </select><br /> <br /> <input type="submit" name = "button" id="button" value="Log In" onclick="javascript:return validateLoginForm();" /> </h3> </form> Once logged in (and should the correct page be loaded, the validation code I have at the top of the script looks like this: <?php session_start(); if (!isset($_SESSION["manager"])) { header("location: http://www.zuluirminger.com/SchoolAdmin/user_login.php"); exit(); } $managerID = preg_replace('#[^0-9]#i', '', $_SESSION["id"]); $manager = preg_replace('#[^A-Za-z0-9]#i', '', $_SESSION["manager"]); $password = preg_replace('#[^A-Za-z0-9]#i', '', $_SESSION["password"]); $role = preg_replace('#[^A-Za-z0-9]#i', '', $_SESSION["role"]); include "adminscripts/connect_to_mysql.php"; $sql = mysql_query("SELECT id FROM Users WHERE username='$manager' AND password='$password' AND role='$role' LIMIT 1"); $existCount = mysql_num_rows($sql); if ($existCount == 0) { header("location: http://www.zuluirminger.com/SchoolAdmin/index.php"); exit(); } ?> Just so you're aware, the database table has the following fields: id, username, password and role. Any help would be greatly appreciated! Many thanks, Zulu

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  • How do I securely delete cookies and saved passwords?

    - by Matt Solnit
    Is there a way, from inside the browser, to delete cookies and saved passwords in such a way that the data is unrecoverable (similar to Secure Empty Trash)? I know I can probably track down all the files involved and then manually do this, but I'm curious if there's an easier way. Or maybe browsers already do it out of the box? I'm specifically looking for solutions for Firefox and Safari, on Mac OS X.

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  • Can a dropdown box using PHP / MySQL automatically select what was last saved in the database with it?

    - by wertz8090
    I am currently working on trying to get a dropdown box to display the most recently saved field each time I open a page for editing an item in an inventory. For example, let's say that my dropdown box selections are Iron, Copper, Gold, Silver, Titanium. The first time the dropdown box loads, I would like it to load in that order (the default order that I set it to). However, once a user submits a form that specifies a selection, I would like that selection to appear as the new default for the dropdown box (for example, if the user picks Gold, Gold will now display as the default). However - the new default that is selected and submitted can only appear for those fields which it is saved under (for example, since the user selected Gold, their preference is now saved as Gold and will appear as such under the field Necklace - however, Gold should not be the default for the fields Ring, Bracelet, and Earring when I open a page to view them). Basically, I need a dropdown box that can save the option of the latest selection on a page after the form is submitted, while at the same time making sure not to affect the default order of the dropdown box across other pages. Your help is much appreciated.

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  • JavaOne Session Report: “50 Tips in 50 Minutes for GlassFish Fans”

    - by Janice J. Heiss
    At JavaOne 2012 on Monday, Oracle’s Engineer Chris Kasso, and Technology Evangelist Arun Gupta, presented a head-spinning session (CON4701) in which they offered 50 tips for GlassFish fans. Kasso and Gupta alternated back and forth with each presenting 10 tips at a time. An audience of about (appropriately) 50 attentive and appreciative developers was on hand in what has to be one of the most information-packed sessions ever at JavaOne!Aside: I experienced one of the quiet joys of JavaOne when, just before the session began, I spotted Java Champion and JavaOne Rock Star Adam Bien sitting nearby – Adam is someone I have been fortunate to know for many years.GlassFish is a freely available, commercially supported Java EE reference implementation. The session prioritized quantity of tips over depth of information and offered tips that are intended for both seasoned and new users, that are meant to increase the range of functional options available to GlassFish users. The focus was on lesser-known dimensions of GlassFish. Attendees were encouraged to pursue tips that contained new information for them. All 50 tips can be accessed here.Below are several examples of more elaborate tips and a final practical tip on how to get in touch with these folks. Tip #1: Using the login Command * To execute a remote command with asadmin you must provide the admin's user name and password.* The login command allows you to store the login credentials to be reused in subsequent commands.* Can be logged into multiple servers (distinguish by host and port). Example:     % asadmin --host ouch login     Enter admin user name [default: admin]>     Enter admin password>     Login information relevant to admin user name [admin]     for host [ouch] and admin port [4848] stored at     [/Users/ckasso/.asadminpass] successfully.     Make sure that this file remains protected.     Information stored in this file will be used by     asadmin commands to manage the associated domain.     Command login executed successfully.     % asadmin --host ouch list-clusters     c1 not running     Command list-clusters executed successfully.Tip #4: Using the AS_DEBUG Env Variable* Environment variable to control client side debug output* Exposes: command processing info URL used to access the command:                           http://localhost:4848/__asadmin/uptime Raw response from the server Example:   % export AS_DEBUG=true  % asadmin uptime  CLASSPATH= ./../glassfish/modules/admin-cli.jar  Commands: [uptime]  asadmin extension directory: /work/gf-3.1.2/glassfish3/glassfish/lib/asadm      ------- RAW RESPONSE  ---------   Signature-Version: 1.0   message: Up 7 mins 10 secs   milliseconds_value: 430194   keys: milliseconds   milliseconds_name: milliseconds   use-main-children-attribute: false   exit-code: SUCCESS  ------- RAW RESPONSE  ---------Tip #11: Using Password Aliases * Some resources require a password to access (e.g. DB, JMS, etc.).* The resource connector is defined in the domain.xml.Example:Suppose the DB resource you wish to access requires an entry like this in the domain.xml:     <property name="password" value="secretp@ssword"/>But company policies do not allow you to store the password in the clear.* Use password aliases to avoid storing the password in the domain.xml* Create a password alias:     % asadmin create-password-alias DB_pw_alias     Enter the alias password>     Enter the alias password again>     Command create-password-alias executed successfully.* The password is stored in domain's encrypted keystore.* Now update the password value in the domain.xml:     <property name="password" value="${ALIAS=DB_pw_alias}"/>Tip #21: How to Start GlassFish as a Service * Configuring a server to automatically start at boot can be tedious.* Each platform does it differently.* The create-service command makes this easy.   Windows: creates a Windows service Linux: /etc/init.d script Solaris: Service Management Facility (SMF) service * Must execute create-service with admin privileges.* Can be used for the DAS or instances* Try it first with the --dry-run option.* There is a (unsupported) _delete-serverExample:     # asadmin create-service domain1     The Service was created successfully. Here are the details:     Name of the service:application/GlassFish/domain1     Type of the service:Domain     Configuration location of the service:/work/gf-3.1.2.2/glassfish3/glassfish/domains     Manifest file location on the system:/var/svc/manifest/application/GlassFish/domain1_work_gf-3.1.2.2_glassfish3_glassfish_domains/Domain-service-smf.xml.     You have created the service but you need to start it yourself. Here are the most typical Solaris commands of interest:     * /usr/bin/svcs  -a | grep domain1  // status     * /usr/sbin/svcadm enable domain1 // start     * /usr/sbin/svcadm disable domain1 // stop     * /usr/sbin/svccfg delete domain1 // uninstallTip #34: Posting a Command via REST* Use wget/curl to execute commands on the DAS.Example:  Deploying an application   % curl -s -S \       -H 'Accept: application/json' -X POST \       -H 'X-Requested-By: anyvalue' \       -F id=@/path/to/application.war \       -F force=true http://localhost:4848/management/domain/applications/application* Use @ before a file name to tell curl to send the file's contents.* The force option tells GlassFish to force the deployment in case the application is already deployed.* Use wget/curl to execute commands on the DAS.Example:  Deploying an application   % curl -s -S \       -H 'Accept: application/json' -X POST \       -H 'X-Requested-By: anyvalue' \       -F id=@/path/to/application.war \       -F force=true http://localhost:4848/management/domain/applications/application* Use @ before a file name to tell curl to send the file's contents.* The force option tells GlassFish to force the deployment in case the application is already deployed.Tip #46: Upgrading to a Newer Version * Upgrade applications and configuration from an earlier version* Upgrade Tool: Side-by-side upgrade– GUI: asupgrade– CLI: asupgrade --c– What happens ?* Copies older source domain -> target domain directory* asadmin start-domain --upgrade* Update Tool and pkg: In-place upgrade– GUI: updatetool, install all Available Updates– CLI: pkg image-update– Upgrade the domain* asadmin start-domain --upgradeTip #50: How to reach us?* GlassFish Forum: http://www.java.net/forums/glassfish/glassfish* [email protected]* @glassfish* facebook.com/glassfish* youtube.com/GlassFishVideos* blogs.oracle.com/theaquariumArun Gupta acknowledged that their method of presentation was experimental and actively solicited feedback about the session. The best way to reach them is on the GlassFish user forum.In addition, check out Gupta’s new book Java EE 6 Pocket Guide.

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  • Can't get SSH public key authentication to work

    - by Trey Parkman
    My server is running CentOS 5.3. I'm on a Mac running Leopard. I don't know which is responsible for this: I can log on to my server just fine via password authentication. I've gone through all of the steps for setting up PKA (as described at http://www.centos.org/docs/5/html/Deployment_Guide-en-US/s1-ssh-beyondshell.html), but when I use SSH, it refuses to even attempt publickey verification. Using the command ssh -vvv user@host (where -vvv cranks up verbosity to the maximum level) I get the following relevant output: debug2: key: /Users/me/.ssh/id_dsa (0x123456) debug1: Authentications that can continue: publickey,gssapi-with-mic,password debug3: start over, passed a different list publickey,gssapi-with-mic,password debug3: preferred keyboard-interactive,password debug3: authmethod_lookup password debug3: remaining preferred: ,password debug3: authmethod_is_enabled password debug1: Next authentication method: password followed by a prompt for my password. If I try to force the issue with ssh -vvv -o PreferredAuthentications=publickey user@host I get debug2: key: /Users/me/.ssh/id_dsa (0x123456) debug1: Authentications that can continue: publickey,gssapi-with-mic,password debug3: start over, passed a different list publickey,gssapi-with-mic,password debug3: preferred publickey debug3: authmethod_lookup publickey debug3: No more authentication methods to try. So, even though the server says it accepts the publickey authentication method, and my SSH client insists on it, I'm rebutted. (Note the conspicuous absence of an "Offering public key:" line above.) Any suggestions?

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  • Passwordless SSH not working - keys copied and permissions set

    - by Comcar
    I know this question has been asked, but I'm certain I've done what all the other answers suggest. Machine A: used keygen -t rsa to create id_rsa.pub in ~/.ssh/ copied Machine A's id_rsa.pub to Machine B user's home directory Made the file permissions of id_rsa.pub 600 Machine B added Machine A's pub key to authorised_keys and authorised_keys2: cat ~/id_rsa.pub ~/.ssh/authorised_keys2 made the file permissions of id_rsa.pub 600 I've also ensured both the .ssh directories have the permission 700 on both machine A and B. If I try to login to machine B from machine A, I get asked for the password, not the ssh pass phrase. I've got the root users on both machines to talk to each other using password-less ssh, but I can't get a normal user to do it. Do the user names have to be the same on both sides? Or is there some setting else where I've missed. Machine A is a Ubuntu 10.04 virtual machine running inside VirtualBox on a Windows 7 PC, Machine B is a dedicated Ubuntu 9.10 server UPDATE : I've run ssh with the option -vvv, which provides many many lines of output, but this is the last few commands: debug3: check_host_in_hostfile: filename /home/pete/.ssh/known_hosts debug3: check_host_in_hostfile: match line 1 debug1: Host '192.168.1.19' is known and matches the RSA host key. debug1: Found key in /home/pete/.ssh/known_hosts:1 debug2: bits set: 504/1024 debug1: ssh_rsa_verify: signature correct debug2: kex_derive_keys debug2: set_newkeys: mode 1 debug1: SSH2_MSG_NEWKEYS sent debug1: expecting SSH2_MSG_NEWKEYS debug3: Wrote 16 bytes for a total of 1015 debug2: set_newkeys: mode 0 debug1: SSH2_MSG_NEWKEYS received debug1: SSH2_MSG_SERVICE_REQUEST sent debug3: Wrote 48 bytes for a total of 1063 debug2: service_accept: ssh-userauth debug1: SSH2_MSG_SERVICE_ACCEPT received debug2: key: /home/pete/.ssh/identity ((nil)) debug2: key: /home/pete/.ssh/id_rsa (0x7ffe1baab9d0) debug2: key: /home/pete/.ssh/id_dsa ((nil)) debug3: Wrote 64 bytes for a total of 1127 debug1: Authentications that can continue: publickey,password debug3: start over, passed a different list publickey,password debug3: preferred gssapi-keyex,gssapi-with-mic,gssapi,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: Trying private key: /home/pete/.ssh/identity debug3: no such identity: /home/pete/.ssh/identity debug1: Offering public key: /home/pete/.ssh/id_rsa debug3: send_pubkey_test debug2: we sent a publickey packet, wait for reply debug3: Wrote 368 bytes for a total of 1495 debug1: Authentications that can continue: publickey,password debug1: Trying private key: /home/pete/.ssh/id_dsa debug3: no such identity: /home/pete/.ssh/id_dsa debug2: we did not send a packet, disable method debug3: authmethod_lookup password debug3: remaining preferred: ,password debug3: authmethod_is_enabled password debug1: Next authentication method: password

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  • Odd log entries when starting up PotgreSQL

    - by Shadow
    When restarting pgSQL, I get the following log entries: 2010-02-10 16:08:05 EST LOG: received smart shutdown request 2010-02-10 16:08:05 EST LOG: autovacuum launcher shutting down 2010-02-10 16:08:05 EST LOG: shutting down 2010-02-10 16:08:05 EST LOG: database system is shut down 2010-02-10 16:08:07 EST LOG: database system was shut down at 2010-02-10 16:08:05 EST 2010-02-10 16:08:07 EST LOG: autovacuum launcher started 2010-02-10 16:08:07 EST LOG: database system is ready to accept connections 2010-02-10 16:08:07 EST LOG: connection received: host=[local] 2010-02-10 16:08:07 EST LOG: incomplete startup packet 2010-02-10 16:08:07 EST LOG: connection received: host=[local] 2010-02-10 16:08:07 EST FATAL: password authentication failed for user "postgres" 2010-02-10 16:08:08 EST LOG: connection received: host=[local] 2010-02-10 16:08:08 EST FATAL: password authentication failed for user "postgres" 2010-02-10 16:08:08 EST LOG: connection received: host=[local] 2010-02-10 16:08:08 EST FATAL: password authentication failed for user "postgres" 2010-02-10 16:08:09 EST LOG: connection received: host=[local] 2010-02-10 16:08:09 EST FATAL: password authentication failed for user "postgres" 2010-02-10 16:08:09 EST LOG: connection received: host=[local] 2010-02-10 16:08:09 EST FATAL: password authentication failed for user "postgres" 2010-02-10 16:08:10 EST LOG: connection received: host=[local] 2010-02-10 16:08:10 EST FATAL: password authentication failed for user "postgres" 2010-02-10 16:08:10 EST LOG: connection received: host=[local] 2010-02-10 16:08:10 EST FATAL: password authentication failed for user "postgres" 2010-02-10 16:08:11 EST LOG: connection received: host=[local] 2010-02-10 16:08:11 EST FATAL: password authentication failed for user "postgres" 2010-02-10 16:08:11 EST LOG: connection received: host=[local] 2010-02-10 16:08:11 EST FATAL: password authentication failed for user "postgres" 2010-02-10 16:08:12 EST LOG: connection received: host=[local] 2010-02-10 16:08:12 EST FATAL: password authentication failed for user "postgres" 2010-02-10 16:08:12 EST LOG: connection received: host=[local] 2010-02-10 16:08:12 EST FATAL: password authentication failed for user "postgres" 2010-02-10 16:08:12 EST LOG: connection received: host=[local] 2010-02-10 16:08:12 EST LOG: incomplete startup packet My question regarding a potential consequence of this is posted here: http://stackoverflow.com/questions/2238954/mdb2-says-connection-failed-db-logs-say-otherwise , but I didn't realize this was happening when I asked that question, and I figured this [part of the] problem is for SF. Edit: I can connect to the database and manipulate things normally with the psql CLI and the postgres user.

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  • Recieving a 500 Internal Server error after login success?

    - by Jeremy Quick
    I created my first member login form which takes the typical username and password and then sends it to the code below: checklogin.php: mysql_connect($db_host, $db_username, $db_password) or die(mysql_error()); mysql_select_db($db_database) or die(mysql_error()); $username=$_POST['username']; $password=$_POST['password']; $username = stripslashes($username); $password = stripslashes($password); $username = mysql_real_escape_string($username); $password = mysql_real_escape_string($password); $sql="SELECT * FROM users WHERE username='$username' and password='$password'"; $result=mysql_query($sql); $count=mysql_num_rows($result); if($count == 1) //ERROR APPEARS TO TAKE PLACE HERE { session_start(); $_SESSION['username'] = $username; $_SESSION['password'] = $password; header('login_success.php'); } else { header("location:login_fail.php"); } If I type in the wrong information everything works properly so I know the error appears to be taking effect in the marked if statement. I have been searching the internet now looking for solutions but none seem to match mine or I am overlooking them. I've made a few changes which brought me to this point, before I was receiving deprecation warnings. Also, I have checked the logs and they are empty of errors relating to this.

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  • Exception:Cannot Start your application.The Workgroup information file is missing or opened exclusiv

    - by Jeev
    We were getting this error when trying   to connect  to a password protected access file. This is what the connection string looked likestring conString =@"Provider=Microsoft.Jet.OLEDB.4.0;Data Source="Path to your access file";User Id=;Password=password";To fix the issue this is what we didstring conString =@"Provider=Microsoft.Jet.OLEDB.4.0;Data Source="Path to your access file";Jet OLEDB:Database Password=password";  We removed the User id and changed the password to Jet OLEDB:Database Password Hope this helps someone   

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  • pass username and password to get-credential or run sql query without using invoke-sqlcmd in Powersh

    - by Emo
    I am trying to connect to a remote sql database and simply run the "select @@servername" query in Powershell. I'm trying to do this without using integrated security. I've been struggling with "get-credential" and "invoke-sqlcmd", only to find (I think), that you can't pass the password from "get-credential" to another Powershell cmdlets. Here's the code I'm using: add-pssnapin sqlserverprovidersnapin100 add-pssnapin sqlservercmdletsnapin100 load assemblies [Reflection.Assembly]::Load("Microsoft.SqlServer.Smo, Version=9.0.242.0, Culture=neutral, PublicKeyToken=89845dcd8080cc91") [Reflection.Assembly]::Load("Microsoft.SqlServer.SqlEnum, Version=9.0.242.0, Culture=neutral, PublicKeyToken=89845dcd8080cc91") [Reflection.Assembly]::Load("Microsoft.SqlServer.SmoEnum, Version=9.0.242.0, Culture=neutral, PublicKeyToken=89845dcd8080cc91") [Reflection.Assembly]::Load("Microsoft.SqlServer.ConnectionInfo, Version=9.0.242.0, Culture=neutral,PublicKeyToken=89845dcd8080cc91") connect to SQL Server $serverName = "HLSQLSRV03" $server = New-Object -typeName Microsoft.SqlServer.Management.Smo.Server -argumentList $serverName login using SQL authentication $server.ConnectionContext.LoginSecure=$false; $credential = Get-Credential $userName = $credential.UserName -replace("\","") $pass = $credential.Password $server.ConnectionContext.set_Login($userName) $server.ConnectionContext.set_SecurePassword($credential.Password) $DB = "Master" invoke-sqlcmd -query "select @@Servername" -database $DB -serverinstance $servername -username $username -password $pass If if just hardcode the password in at the end of the "invoke-sqlcmd" line, it works. Is this because you can't use "get-credential" with "invoke-sqlcmd"? If so...what are my alternatives? Thanks so much for you help Emo

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  • Create a view in database to fetch username and password of all users having a particular roie (Form

    - by naveen
    Hi Guys, Background ========== I am using ASP.NET Forms Authentication for my application. I have made my Password Format "Clear", so no problems with the password encryption. Requirement =========== I need to create a view(in SQL server) to display all Administrators in my System. ie aspnet_Roles.LoweredRoleName='administrator' The fields needed are UserName and Password Can you guys give point out the most optimal way to do so? Regards, Naveen Jose

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  • java script is not working

    - by Piyush
    i am comparing password and confirm password through java script.my code- function validate_form(thisform) { with (thisform) { if (validate_required(password,"<b>Error: </b>Password must be filled out!")==false) {password.focus();return false;} else if (validate_required(cnfpassword,"<b>Error: </b>Confirm Password must be filled out!")==false) {cnfpassword.focus();return false;} else if (document.getElementById('password').value != document.getElementById('cnfpassword').value) {password.focus();Sexy.error("<b>Error: </b>Passwords entered are not same!"); password.value="";cnfpassword.value="";return false;} } validate_required() function is working fine, it is showing alert msg but password compare is not working. But the same code is working fine in some other page.I have written some php code to avoid page caching- <?php session_start(); session_cache_limiter('nocache'); header('Pragma: no-cache'); ?> what's the problem???

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  • How to update the user profile through tokenid when a particular user logs in using his username and password

    - by Rani
    i have an application in which when the user enters his username and password in the login form , his username and password are submitted to the server database for validation and if he is a valid user a tokenid is generated for him through JSON response and he is allowed to log in.this is the JSON response that i get in the console when a valid user log in through the log in form : {"TokenID":"kaenn43otO","isError":false,"ErrorMessage":"","Result":[{"UserId":"164","FirstName":"Indu","LastName":"nair","Email":"[email protected]","ProfileImage":null,"ThumbnailImage":null,"DeviceInfoId":"22"}],"ErrorCode":900} 2011-06-24 11:56:57.639 Journey[5526:207] this is token id:kaenn43otO so when the same user wants to update his profile by clicking on the update tab using the tokenid So that if he want he can his firstname,lastname,password,email.etc.i i have done the following code to update the user profile using tokenid but it does not work.i get tokenid when the user log in and this tokenid i have to pass to other class: this is my code to update user profile: -(void)sendRequest { //this is the class in which i get the tokenid and i am passing it to this class apicontroller *apitest = [[apicontroller alloc]init]; NSString *tokentest = apitest.tokenapi; NSString *post = [NSString stringWithFormat:@"Token=%@&firstname=%@&lastname=%@&Username=%@&Password=%@&Email=%@",tokentest,txtfirstName.text,txtlast.text,txtUserName.text,txtPassword.text,txtEmail.text]; NSData *postData = [post dataUsingEncoding:NSASCIIStringEncoding allowLossyConversion:YES]; NSString *postLength = [NSString stringWithFormat:@"%d", [postData length]]; NSLog(@"%@",postLength); NSMutableURLRequest *request = [[[NSMutableURLRequest alloc] init] autorelease]; [request setURL:[NSURL URLWithString:@"http://192.168.0.1:96/JourneyMapperAPI?RequestType=Register&Command=SET"]]; [request setHTTPMethod:@"POST"]; [request setValue:postLength forHTTPHeaderField:@"Content-Length"]; [request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Content-Type"]; [request setHTTPBody:postData]; NSURLConnection *theConnection = [[NSURLConnection alloc] initWithRequest:request delegate:self]; if (theConnection) { webData = [[NSMutableData data] retain]; NSLog(@"%@",webData); [theConnection start]; } else { } } -(void)connection:(NSURLConnection *)connection didReceiveResponse:(NSURLResponse *)response { [webData setLength: 0]; } -(void)connection:(NSURLConnection *)connection didReceiveData:(NSData *)data { [webData appendData:data]; } -(void)connection:(NSURLConnection *)connection didFailWithError:(NSError *)error { [connection release]; [webData release]; } -(void)connectionDidFinishLoading:(NSURLConnection *)connection { NSString *loginStatus = [[NSString alloc] initWithBytes: [webData mutableBytes] length:[webData length] encoding:NSUTF8StringEncoding]; NSLog(@"%@",loginStatus); [loginStatus release]; [connection release]; [webData release]; } -(IBAction)click:(id)sender { [self sendRequest]; } this is my apicontroller.m file in which i am creating login form so that user login and in this page tokenid is generated -(void)sendRequest { UIDevice *device = [UIDevice currentDevice]; NSString *udid = [device uniqueIdentifier]; //NSString *sysname = [device systemName]; NSString *sysver = [device systemVersion]; NSString *model = [device model]; NSLog(@"idis:%@",[device uniqueIdentifier]); NSLog(@"system nameis :%@",[device systemName]); NSLog(@"System version is:%@",[device systemVersion]); NSLog(@"System model is:%@",[device model]); NSLog(@"device orientation is:%d",[device orientation]); NSString *post = [NSString stringWithFormat:@"Loginkey=%@&Password=%@&DeviceCode=%@&Firmware=%@&IMEI=%@",txtUserName.text,txtPassword.text,model,sysver,udid]; NSData *postData = [post dataUsingEncoding:NSASCIIStringEncoding allowLossyConversion:YES]; NSString *postLength = [NSString stringWithFormat:@"%d", [postData length]]; NSLog(@"%@",postLength); NSMutableURLRequest *request = [[[NSMutableURLRequest alloc] init] autorelease]; [request setURL:[NSURL URLWithString:@"http://192.168.0.1:96/JourneyMapperAPI?RequestType=Login"]]; [request setHTTPMethod:@"POST"]; [request setValue:postLength forHTTPHeaderField:@"Content-Length"]; [request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Content-Type"]; [request setHTTPBody:postData]; NSURLConnection *theConnection = [[NSURLConnection alloc] initWithRequest:request delegate:self]; if (theConnection) { webData = [[NSMutableData data] retain]; NSLog(@"%@",webData); } else { } } //to select username and password from database. -(void)check { //app.journeyList = [[NSMutableArray alloc]init]; [self createEditableCopyOfDatabaseIfNeeded]; NSString *filePath = [self getWritableDBPath]; sqlite3 *database; if(sqlite3_open([filePath UTF8String], &database) == SQLITE_OK) { const char *sqlStatement = "SELECT Username,Password FROM UserInformation where Username=? and Password=?"; sqlite3_stmt *compiledStatement; if(sqlite3_prepare_v2(database, sqlStatement, -1, &compiledStatement, NULL) == SQLITE_OK) { sqlite3_bind_text(compiledStatement, 1, [txtUserName.text UTF8String], -1, SQLITE_TRANSIENT); sqlite3_bind_text(compiledStatement, 2, [txtPassword.text UTF8String], -1, SQLITE_TRANSIENT); //NSString *loginname= [NSString stringWithUTF8String:(char *)sqlite3_column_text(compiledStatement, 1)]; // NSString *loginPassword = [NSString stringWithUTF8String:(char *)sqlite3_column_text(compiledStatement, 2)]; } if(sqlite3_step(compiledStatement) != SQLITE_ROW ) { NSLog( @"Save Error: %s", sqlite3_errmsg(database) ); UIAlertView *alert = [[UIAlertView alloc]initWithTitle:@"UIAlertView" message:@"User is not valid" delegate:self cancelButtonTitle:@"OK" otherButtonTitles:nil]; [alert show]; [alert release]; alert = nil; } else { isUserValid = YES; if (isUserValid) { UIAlertView *alert = [[UIAlertView alloc]initWithTitle:@"UIAlertView" message:@"Valid User" delegate:self cancelButtonTitle:@"OK" otherButtonTitles:nil]; [alert show]; [alert release]; } } sqlite3_finalize(compiledStatement); } sqlite3_close(database); } -(void)connection:(NSURLConnection *)connection didReceiveResponse:(NSURLResponse *)response { [webData setLength: 0]; } -(void)connection:(NSURLConnection *)connection didReceiveData:(NSData *)data { [webData appendData:data]; } -(void)connection:(NSURLConnection *)connection didFailWithError:(NSError *)error { [connection release]; [webData release]; } -(void)connectionDidFinishLoading:(NSURLConnection *)connection { NSString *loginStatus = [[NSString alloc] initWithBytes: [webData mutableBytes] length:[webData length] encoding:NSUTF8StringEncoding]; NSLog(@"%@",loginStatus); //this is to perfrom insert opertion on the userinformation table NSString *json_string = [[NSString alloc] initWithData:webData encoding:NSUTF8StringEncoding]; NSDictionary *result = [json_string JSONValue]; //here i am storing the tokenid tokenapi = [result objectForKey:@"TokenID"]; NSLog(@"this is token id:%@",tokenapi); // BOOL errortest = [[result objectForKey:@"isError"] boolValue]; if(errortest == FALSE) { values = [result objectForKey:@"Result"]; NSLog(@"Valid User"); } else { NSLog(@"Invalid User"); } NSMutableArray *results = [[NSMutableArray alloc] init]; for (int index = 0; index<[values count]; index++) { NSMutableDictionary * value = [values objectAtIndex:index]; Result * result = [[Result alloc] init]; //through this i get the userid result.UserID = [value objectForKey:@"UserId"]; result.FirstName = [value objectForKey:@"FirstName"]; result.Username =[value objectForKey:@"Username"]; result.Email =[value objectForKey:@"Email"]; result.ProfileImage =[value objectForKey:@"ProfileImage"]; result.ThumbnailImage =[value objectForKey:@"ThumbnailImage"]; result.DeviceInfoId =[value objectForKey:@"DeviceInfoId"]; NSLog(@"%@",result.UserID); [results addObject:result]; [result release]; } for (int index = 0; index<[results count]; index++) { Result * result = [results objectAtIndex:index]; //save the object variables to database here [self createEditableCopyOfDatabaseIfNeeded]; NSString *filePath = [self getWritableDBPath]; sqlite3 *database; if(sqlite3_open([filePath UTF8String], &database) == SQLITE_OK) { const char *sqlStatement = "insert into UserInformation(UserID,DeviceId,Username,Password,FirstName,Email) VALUES (?,?,?,?,?,?)"; sqlite3_stmt *compiledStatement; if(sqlite3_prepare_v2(database, sqlStatement, -1, &compiledStatement, NULL) == SQLITE_OK) { //sqlite3_bind_text( compiledStatement, 1, [journeyid UTF8String],-1,SQLITE_TRANSIENT); sqlite3_bind_text( compiledStatement, 1, [result.UserID UTF8String],-1,SQLITE_TRANSIENT); sqlite3_bind_text(compiledStatement, 2, [result.DeviceInfoId UTF8String],-1,SQLITE_TRANSIENT); sqlite3_bind_text(compiledStatement, 3, [txtUserName.text UTF8String],-1,SQLITE_TRANSIENT); sqlite3_bind_text(compiledStatement, 4, [txtPassword.text UTF8String],-1,SQLITE_TRANSIENT); sqlite3_bind_text (compiledStatement, 5, [result.FirstName UTF8String],-1,SQLITE_TRANSIENT); sqlite3_bind_text (compiledStatement, 6, [result.Email UTF8String],-1,SQLITE_TRANSIENT); } if(sqlite3_step(compiledStatement) != SQLITE_DONE ) { NSLog( @"Save Error: %s", sqlite3_errmsg(database) ); } else { UIAlertView *alert = [[UIAlertView alloc]initWithTitle:@"UIAlertView" message:@"Record added" delegate:self cancelButtonTitle:@"OK" otherButtonTitles:nil]; [alert show]; [alert release]; alert = nil; } sqlite3_finalize(compiledStatement); } sqlite3_close(database); } [loginStatus release]; [connection release]; [webData release]; } -(IBAction)click:(id)sender { [self sendRequest]; //this is to select username and password from database. [self check]; } for a particular user who logs in only his profile must be update by using tokenid and userid.Please help me in solving this problem.Thanks

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  • Custom login in Django

    - by alpgs
    Django newbie here. I wrote simplified login form which takes email and password. It works great if both email and password are supplied, but if either is missing i get KeyError exception. According to django documentation this should never happen: By default, each Field class assumes the value is required, so if you pass an empty value -- either None or the empty string ("") -- then clean() will raise a ValidationError exception I tried to write my own validators for fields (clean_email and clean_password), but it doesn't work (ie I get KeyError exception). What am I doing wrong? class LoginForm(forms.Form): email = forms.EmailField(label=_(u'Your email')) password = forms.CharField(widget=forms.PasswordInput, label=_(u'Password')) def clean_email(self): data = self.cleaned_data['email'] if not data: raise forms.ValidationError(_("Please enter email")) return data def clean_password(self): data = self.cleaned_data['password'] if not data: raise forms.ValidationError(_("Please enter your password")) return data def clean(self): try: username = User.objects.get(email__iexact=self.cleaned_data['email']).username except User.DoesNotExist: raise forms.ValidationError(_("No such email registered")) password = self.cleaned_data['password'] self.user = auth.authenticate(username=username, password=password) if self.user is None or not self.user.is_active: raise forms.ValidationError(_("Email or password is incorrect")) return self.cleaned_data

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  • Why does tomcat-admin install require adding admin and manager to tomcat-users.xml manually?

    - by J G
    I installed tomcat6 on lucid using apt-get. All working. I installed tomcat-admin. Not working. I amended the /etc/tomcat6/tomcat-users.xml file to uncomment the users and roles (from the default) to be like the following: <role rolename="tomcat"/> <role rolename="role1"/> <user username="tomcat" password="password" roles="tomcat"/> <user username="both" password="password" roles="tomcat,role1"/> <user username="role1" password="password" roles="role1"/> This still didn't work. Then from the following page I added. <role rolename="manager"/> <user username="admin" password="secret" roles="manager"/> then it worked. Why doesn't this occur as part of the install? (Why isn't this in the Ubuntu Manual on Tomcat ?)

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  • Why acegi (Spring Security) converts password to uppercase before comparing ?

    - by Tony
    One of my colleague in QA team reported a bug to me, the bug said that can't change password to lowercase, otherwise login is rejected,using number or uppercase is all fine. The login system was implemented using acegi 1.0 (now called Spring Security). This was a very strange bug,changing password is done by encrypting the user input string into MD5 string, I implemented this without using anything related acegi, I don't if the is the origin cause of the problem. When the login is rejected, through debugging, I find that, the user input is converted into uppercase by acegi when passing to the acegi comparing logic. At first, I didn't believe this, when I checkout the acegi source and debugging with it, I find it does convert both username and password to uppercase (source code line 121), Can you tell me why it does this? This can cause password encoding mismatch!

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