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  • C# Active Directory - Check username / password

    - by Michael G
    I'm using the following code on Windows Vista Ultimate SP1 to query our active directory server to check the user name and password of a user on a domain. public Object IsAuthenticated() { String domainAndUsername = strDomain + @"\" + strUser; DirectoryEntry entry = new DirectoryEntry(_path, domainAndUsername, strPass); SearchResult result; try { //Bind to the native AdsObject to force authentication. DirectorySearcher search = new DirectorySearcher(entry) { Filter = ("(SAMAccountName=" + strUser + ")") }; search.PropertiesToLoad.Add("givenName"); // First Name search.PropertiesToLoad.Add("sn"); // Last Name search.PropertiesToLoad.Add("cn"); // Last Name result = search.FindOne(); if (null == result) { return null; } //Update the new path to the user in the directory. _path = result.Path; _filterAttribute = (String)result.Properties["cn"][0]; } catch (Exception ex) { return new Exception("Error authenticating user. " + ex.Message); } return user; } the target is using .NET 3.5, and compiled with VS 2008 standard I'm logged in under a domain account that is a domain admin where the application is running. The code works perfectly on windows XP; but i get the following exception when running it on Vista: System.DirectoryServices.DirectoryServicesCOMException (0x8007052E): Logon failure: unknown user name or bad password. at System.DirectoryServices.DirectoryEntry.Bind(Boolean throwIfFail) at System.DirectoryServices.DirectoryEntry.Bind() at System.DirectoryServices.DirectoryEntry.get_AdsObject() at System.DirectoryServices.DirectorySearcher.FindAll(Boolean findMoreThanOne) at System.DirectoryServices.DirectorySearcher.FindOne() at Chain_Of_Custody.Classes.Authentication.LdapAuthentication.IsAuthenticated() at System.DirectoryServices.DirectoryEntry.Bind(Boolean throwIfFail) at System.DirectoryServices.DirectoryEntry.Bind() at System.DirectoryServices.DirectoryEntry.get_AdsObject() at System.DirectoryServices.DirectorySearcher.FindAll(Boolean findMoreThanOne) at System.DirectoryServices.DirectorySearcher.FindOne() at Chain_Of_Custody.Classes.Authentication.LdapAuthentication.IsAuthenticated() I've tried changing the authentication types, I'm not sure what's going on. See also: http://stackoverflow.com/questions/290548/c-validate-a-username-and-password-against-active-directory

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  • django: cannot import settings, cannot login to admin, cannot change admin password

    - by xpanta
    Hi, It seems that I am completely lost here. Yesterday I noticed that I cannot login to the admin panel (don't use it much, so it's been some weeks since last login). I thought that I might have changed the admin password and now I can't remember it (though I doubt it). I tried django-admin.py changepassword (using django 1.2.1) but it said that 'changepassword' is unknown command (I have all the necessary imports in my settings.py. Admin interface used to work ok). Then I gave a django-admin.py validate. Then the hell begun. django-admin.py validate gave me this error: Error: Settings cannot be imported, because environment variable DJANGO_SETTINGS_MODULE is undefined. I then gave a set DJANGO_SETTINGS_MODULE=myproject.settings and then again a django-admin.py validate This is what I get now: Error: Could not import settings 'myproject.settings' (Is it on sys.path? Does it have syntax errors?): No module named myproject.settings and now I am lost. I tried django console and sys.path.append('c:\workspace') or sys.append('c:\workspace\myproject') but still get the same errors. I use windows 7 and my project dir is c:\workspace. I don't use a PYTHONPATH variable (although I tried setting it temporarily to C:\workspace but I still get the same error). I don't use Apache, just the django development server. What am I doing wrong? My web page works fine. I think that the fact that I can't login as admin is related to the previous import error, no? PS: I also tried this: http://coderseye.com/2007/howto-reset-the-admin-password-in-django.html but still I couldn't change admin password for some reason. Although I could create another admin user (with which I couldn't login).

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  • Generate an ID via COM interop

    - by Erik van Brakel
    At the moment, we've got an unmaintanable ball of code which offers an interface to a third party application. The third party application has a COM assembly which MUST be used to create new entries. This process involves two steps: generate a new object (basically an ID), and update that object with new field values. Because COM interop is so slow, we only use that to generate the ID (and related objects) in the database. The actual update is done using a regular SQL query. What I am trying to figure out if it's possible to use NHibernate to do some of the heavy lifting for us, without bypassing the COM assembly. Here's the code for saving something to the database as I envision it: using(var s = sessionFactory.OpenSession()) using(var t = s.BeginTransaction()) { MyEntity entity = new MyEntity(); s.Save(entity); t.Commit(); } Regular NH code I'd say. Now, this is where it gets tricky. I think I have to supply my own implementation of NHibernate.Id.IIdentifierGenerator which calls the COM assembly in the Generate method. That's not a problem. What IS a problem is that the COM assembly requires initialisation, which does take a bit of time. It also doesn't like multiple instances in the same process, for some reason. What I would like to know is if there's a way to properly access an external service in the generator code. I'm free to use any technique I want, so if it involves something like an IoC container that's no problem. The thing I am looking for is where exactly to hook-up my code so I can access the things I need in my generator, without having to resort to using singletons or other nasty stuff.

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  • When to use certain optimizations such as -fwhole-program and -fprofile-generate with several shared libraries

    - by James
    Probably a simple answer; I get quite confused with the language used in the GCC documentation for some of these flags! Anyway, I have three libraries and a programme which uses all these three. I compile each of my libraries seperately with individual (potentially) different sets of warning flags. However, I compile all three libraries with the same set of optimisation flags. I then compile my main programme linking in these three libraries with its own set of warning flags and the same optimisation flags used during the libraries' compilation. 1) Do I have to compile the libraries with optimisation flags present or can I just use these flags when compiling the final programme and linking to the libraries? If the latter, will it then optimise all or just some (presumably that which is called) of the code in these libraries? 2) I would like to use -fwhole-program -flto -fuse-linker-plugin and the linker plugin gold. At which stage do I compile with these on ... just the final compilation or do these flags need to be present during the compilation of the libraries? 3) Pretty much the same as 2) however with, -fprofile-generate -fprofile-arcs and -fprofile-use. I understand one first runs a programme with generate, and then with use. However, do I have to compile each of the libraries with generate/use etc. or just the final programme? And if it is just the last programme, when I then compeil with -fprofile-use will it also optimise the libraries functionality? Many thanks, James

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  • How do I generate a connection reset programatically?

    - by Brock Adams
    Hi, I'm sure you've seen the "the connection was reset" message displayed when trying to browse web pages. (The text is from Firefox, other browsers differ.) I need to generate that message/error/condition on demand, to test workarounds. So, how do I generate that condition programmatically? (How to generate a TCP RST from PHP -- or one of the other web-app languages?) Caveats and Conditions: It cannot be a general IP block. The test client must still be able to see the test server when not triggering the condition. Ideally, it would be done at the web-application level (Python, PHP, Coldfusion, Javascript, etc.). Access to routers is problematic. Access to Apache config is a pain. Ideally, it would be triggered by fetching a specific web-page. Bonus if it works on a standard, commercial web host. Update: Sending RST is not enough to cause this condition. See my partial answer, below. I've a solution that works on a local machine, Now need to get it working on a remote host.

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  • Outlook Interop: Password protected PST file headache

    - by Ed Manet
    Okay, I have no problem identifying the .PST file using the Outlook Interop assemblies in a C# app. But as soon as I hit a password protected file, I am prompted for a password. We are in the process of disabling the use of PSTs in our organization and one of the steps is to unload the PST files from the users' Outlook profile. I need to have this app run silently and not prompt the user. Any ideas? Is there a way to create the Outlook.Application object with no UI and then just try to catch an Exception on password protected files? // create the app and namespace Application olApp = new Application(); NameSpace olMAPI = olApp.GetNamespace("MAPI"); // get the storeID of the default inbox string rootStoreID = olMAPI.GetDefaultFolder(OlDefaultFolders.olFolderInbox).StoreID; // loop thru each of the folders foreach (MAPIFolder fo in olMAPI.Folders) { // compare the first 75 chars of the storeid // to prevent removing the Inbox folder. string s1 = rootStoreID.Substring(1, 75); string s2 = fo.StoreID.Substring(1, 75); if (s1 != s2) { // unload the folder olMAPI.RemoveStore(fo); } } olApp.Quit();

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  • Python-based password tracker (or dictionary)

    - by Arrieta
    Hello: Where we work we need to remember about 10 long passwords which need to change every so often. I would like to create a utility which can potentially save these passwords in an encrypted file so that we can keep track of them. I can think of some sort of dictionary passwd = {'host1':'pass1', 'host2':'pass2'}, etc, but I don't know what to do about encryption (absolutely zero experience in the topic). So, my question is really two questions: Is there a Linux-based utility which lets you do that? If you were to program it in Python, how would you go about it? A perk of approach two, would be for the software to update the ssh public keys after the password has been changed (you know the pain of updating ~15 tokens once you change your password). As it can be expected, I have zero control over the actual network configuration and the management of scp keys. I can only hope to provide a simple utility to me an my very few coworkers so that, if we need to, we can retrieve a password on demand. Cheers.

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  • Saving Email/Password to Keychain in iOS

    - by Jason
    I'm very new to iOS development so forgive me if this is a newbie question. I have a simple authentication mechanism for my app that takes a user's email address and password. I also have a switch that says 'Remember me'. If the user toggles that switch on, I'd like to preserve their email/password so those fields can be auto-populated in the future. I've gotten this to work with saving to a plist file but I know that's not the best idea since the password is unencrypted. I found some sample code for saving to the keychain, but to be honest, I'm a little lost. For the function below, I'm not sure how to call it and how to modify it to save the email address as well. I'm guessing to call it would be: saveString(@"passwordgoeshere"); Thank you for any help!!! + (void)saveString:(NSString *)inputString forKey:(NSString *)account { NSAssert(account != nil, @"Invalid account"); NSAssert(inputString != nil, @"Invalid string"); NSMutableDictionary *query = [NSMutableDictionary dictionary]; [query setObject:(id)kSecClassGenericPassword forKey:(id)kSecClass]; [query setObject:account forKey:(id)kSecAttrAccount]; [query setObject:(id)kSecAttrAccessibleWhenUnlocked forKey:(id)kSecAttrAccessible]; OSStatus error = SecItemCopyMatching((CFDictionaryRef)query, NULL); if (error == errSecSuccess) { // do update NSDictionary *attributesToUpdate = [NSDictionary dictionaryWithObject:[inputString dataUsingEncoding:NSUTF8StringEncoding] forKey:(id)kSecValueData]; error = SecItemUpdate((CFDictionaryRef)query, (CFDictionaryRef)attributesToUpdate); NSAssert1(error == errSecSuccess, @"SecItemUpdate failed: %d", error); } else if (error == errSecItemNotFound) { // do add [query setObject:[inputString dataUsingEncoding:NSUTF8StringEncoding] forKey:(id)kSecValueData]; error = SecItemAdd((CFDictionaryRef)query, NULL); NSAssert1(error == errSecSuccess, @"SecItemAdd failed: %d", error); } else { NSAssert1(NO, @"SecItemCopyMatching failed: %d", error); } }

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  • generate k distinct number less then n

    - by davit-datuashvili
    hi i have following question task is this generate k distinct positive numbers less then n without duplication my method is following first create array size of k where we should write these numbers int a[]=new int[k]; //now i am going to cretae another array where i check if (at given number position is 1 then generate number again else put this number in a array and continue cycle i put here a piece of code and explanations int a[]=new int[k]; int t[]=new int[n+1]; Random r=new Random(); for (int i==0;i<t.length;i++){ t[i]=0;//initialize it to zero } int m=0;//initialize it also for (int i=0;i<a.length;i++){ m=r.nextInt(n);//random element between 0 and n if (t[m]==1){ //i have problem with this i want in case of duplication element occurs repeats this steps afain until there will be different number else{ t[m]=1; x[i]=m; } } so i fill concret my problem if t[m]==1 it means that this element occurs already so i want to generate new number but problem is that number of generated numbers will not be k beacuse if i==0 and occurs duplicate element and we write continue then it will switch at i==1 i need like goto for repeat step or for (int i=0;i<x.length;i++){ loop: m=r.nextInt(n); if ( x[m]==1){ continue loop; } else{ x[m]=1; a[i]=m; continue;//continue next step at i=1 and so on } } i need this code in java please help

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  • get random password with puppet function

    - by ninja-2
    I have a function that allow me to generate random password. My function is working well without a puppetmaster. When i tried with a master an error appear when I called the function : Error 400 on SERVER: bad value for range Here is my function module Puppet::Parser::Functions newfunction(:get_random_password, :type => :rvalue, :doc => <<-EOS Returns a random password. EOS ) do |args| raise(Puppet::ParseError, "get_random_password(): Wrong number of arguments " + "given (#{args.size} for 1)") if args.size != 1 specials = ((33..33).to_a + (35..38).to_a + (40..47).to_a + (58..64).to_a + (91..93).to_a + (95..96).to_a + (123..125).to_a).pack('U*').chars.to_a numbers = (0..9).to_a alphal = ('a'..'z').to_a alphau = ('A'..'Z').to_a length = args[0] CHARS = (alphal + specials + numbers + alphau) pwd = CHARS.sort_by { rand }.join[0...length] return pwd end end The function is called in both case with $pwd = get_random_password(10). When I specified the length directly in the function to 10 for example. the password is well generated in master mode. Have you any idea why i can't specify the lentgth value ? Thanks for any help.

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  • sudo in Debian squeeze inside linux-vserver always wants password

    - by mark
    Every since I upgraded all my linux-vserver Debian guests from Lenny to Squeeze I've the apparent problem that whenever I want to use sudo it asks me for my password. Every time. I've configured sudo to have a timeout of 30 minutes: Defaults timestamp_timeout=30 . This has been configured when it was still Lenny (note: as suggested by EightBitTony I've also tried without this setting - no change). I've a hard time figuring out what the problem here is, since I think my configuration is right. I thought about it being a problem with the file used to record the timestamp, maybe a permission issue, but was unlucky to find any hard evidence. I've compared the contents of /var/lib/sudo/ between a working and a non-working system but couldn't spot any difference. The version of sudo used in both environments is 1.7.4p4-2.squeeze.3. My non-working system(s): find /var/lib/sudo/ -ls 17319289 4 drwx------ 4 root root 4096 Jan 1 1985 /var/lib/sudo/ 17319286 4 drwx------ 2 root mark 4096 Jan 1 1985 /var/lib/sudo/mark 17319312 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/6 17319361 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/9 17319490 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/10 17319326 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/4 17319491 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/2 A working system: find /var/lib/sudo -ls 2598921 4 drwx------ 5 root root 4096 Jan 1 1985 /var/lib/sudo 1999522 4 drwx------ 2 root mark 4096 Jan 1 1985 /var/lib/sudo/mark 2000781 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/8 1998998 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/17 1999459 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/26 1998930 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/24 2000771 4 -rw------- 1 root mark 40 Jun 25 11:39 /var/lib/sudo/mark/4 2000773 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/5 1999223 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/0 1998908 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/14 2000769 4 -rw------- 1 root mark 40 Jul 9 13:30 /var/lib/sudo/mark/2 2000770 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/3 2000782 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/9 2000778 4 -rw------- 1 root mark 40 Jul 8 00:11 /var/lib/sudo/mark/7 1998892 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/19 1999264 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/23 2000789 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/12 1999093 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/25 1998880 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/18 1998853 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/20 2000790 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/15 1998878 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/16 1998874 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/13 2000774 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/6 2000786 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/11 1998893 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/22 2000783 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/10 1998949 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/1 Despite the obvious (some up2date timestamps on the working system) I don't see anything wrong here, so it could be as well be a wrong track. Here's my current /etc/sudoers: # /etc/sudoers # # This file MUST be edited with the 'visudo' command as root. # # See the man page for details on how to write a sudoers file. # Defaults env_reset # Host alias specification # User alias specification User_Alias FULLADMIN = user1, user2, user3 # Cmnd alias specification # User privilege specification root ALL=(ALL) ALL FULLADMIN ALL = (ALL) ALL # Allow members of group sudo to execute any command # (Note that later entries override this, so you might need to move # it further down) %sudo ALL=(ALL) ALL # #includedir /etc/sudoers.d #Defaults always_set_home,timestamp_timeout=30

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  • Simple way to create possible case

    - by bugbug
    I have lists of data such as a = [1,2,3,4] b = ["a","b","c","d","e"] c = ["001","002","003"] And I want to create new another list that was mixed from all possible case of a,b,c like this d = ["1a001","1a002","1a003",...,"4e003"] Is there any module or method to generate d without write many for loop?

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  • How can I login to lightdm with password for fingerprint-enabled user after 12.10 upgrade?

    - by jxn
    Sorry for the long question. I have a laptop with ubuntu quantal 12.10, a fingerprint scanner, and a few active user accounts. When the machine boots up to lightdm, I get a prompt toenter my password or scan my finger print. Every now and then, fingerprint scanning just doesn't seem to work. Before the 12.10 upgrade, I was always able to enter my password for this user when fingerprint failed. Now, no matter what, I have to scan my prints to login as this user. If I try to login as a different user (fingerprint is not enabled for any others), I can see the password is typed out -- asterisks show in the password input box as I type them -- and get in. Not so for the fingerprint user. Any clues on how to figure out what's gone wrong?

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  • Disable Password Complexity/Expiration etc. Policy on Windows Server 2008

    - by Sahil Malik
    Ad:: SharePoint 2007 Training in .NET 3.5 technologies (more information). One of the things I like to do, for development environments only is to get rid of that excessively bothersome password policies. I like to have my password as something like p@ssword1, so they are easy to remember etc. etc. Obviously never do this in production. However, Windows Server 2008 comes with a password policy that expires my passwords every 90 days, and requires me to pick complex passwords, can’t reuse passwords etc. etc. Well here is how you disable password policy on a Windows Server 2008 machine - Run Group Policy Management (gpmc.msc) Expand to your domain, look for Forest\Domains\yourdomain\default domain policy. Go to the settings tab, right click on the tab, and choose “Edit”. This will open the Group Policy Management Editor, in which - Go to Computer Configuration\Policies\Windows Settings\Security Settings\Account Policies\Password Policy, and change the policy to whatever that suits you. Close everything, and run command prompt as administrator, and issue a “gpupdate /force” command to force the group policy update on the machine. Restart, and you’re done! :) Comment on the article ....

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  • Should maven generate jaxb java code or just use java code from source control?

    - by Peter Turner
    We're trying to plan how to mash together a build server for our shiny new java backend. We use a lot of jaxb XSD code generation and I was getting into a heated argument with whoever cared that the build server should delete jaxb created structures that were checked in generate the code from XSD's use code generated from those XSD's Everyone else thought that it made more sense to just use the code they checked in (we check in the code generated from the XSD because Eclipse pretty much forces you to do this as far as I can tell). My only stale argument is in my reading of the Joel test is that making the build in one step means generating from the source code and the source code is not the java source, but the XSD's because if you're messing around with the generated code you're gonna get pinched eventually. So, given that we all agree (you may not agree) we should probably be checking in our generate java files, should we use them to generate our code or should we generate it using the XSD's?

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  • PhP Login/Register system [migrated]

    - by Marian
    I found this good tutorial on creating a login/register system using PhP and MySQL. The forum is around 5 years old (edited last year) but it can still be usefull. Beginner Simple Register-Login system There seems to be an issue with both login and register pages. <?php function register_form(){ $date = date('D, M, Y'); echo "<form action='?act=register' method='post'>" ."Username: <input type='text' name='username' size='30'><br>" ."Password: <input type='password' name='password' size='30'><br>" ."Confirm your password: <input type='password' name='password_conf' size='30'><br>" ."Email: <input type='text' name='email' size='30'><br>" ."<input type='hidden' name='date' value='$date'>" ."<input type='submit' value='Register'>" ."</form>"; } function register(){ $connect = mysql_connect("host", "username", "password"); if(!$connect){ die(mysql_error()); } $select_db = mysql_select_db("database", $connect); if(!$select_db){ die(mysql_error()); } $username = $_REQUEST['username']; $password = $_REQUEST['password']; $pass_conf = $_REQUEST['password_conf']; $email = $_REQUEST['email']; $date = $_REQUEST['date']; if(empty($username)){ die("Please enter your username!<br>"); } if(empty($password)){ die("Please enter your password!<br>"); } if(empty($pass_conf)){ die("Please confirm your password!<br>"); } if(empty($email)){ die("Please enter your email!"); } $user_check = mysql_query("SELECT username FROM users WHERE username='$username'"); $do_user_check = mysql_num_rows($user_check); $email_check = mysql_query("SELECT email FROM users WHERE email='$email'"); $do_email_check = mysql_num_rows($email_check); if($do_user_check > 0){ die("Username is already in use!<br>"); } if($do_email_check > 0){ die("Email is already in use!"); } if($password != $pass_conf){ die("Passwords don't match!"); } $insert = mysql_query("INSERT INTO users (username, password, email) VALUES ('$username', '$password', '$email')"); if(!$insert){ die("There's little problem: ".mysql_error()); } echo $username.", you are now registered. Thank you!<br><a href=login.php>Login</a> | <a href=index.php>Index</a>"; } switch($act){ default; register_form(); break; case "register"; register(); break; } ?> Once pressed the register button the page does nothing, fields are erased and no data is added inside the database or error given. I tought that the problem might be the switch($act){ part so I removed it and changed the page using a require require('connect.php'); where connect.php is <?php mysql_connect("localhost","host","password"); mysql_select_db("database"); ?> Removed the function register_form(){ and echo part turning it into an HTML code: <form action='register' method='post'> Username: <input type='text' name='username' size='30'><br> Password: <input type='password' name='password' size='30'><br> Confirm your password: <input type='password' name='password_conf' size='30'><br> Email: <input type='text' name='email' size='30'><br> <input type='hidden' name='date' value='$date'> <input type='submit' name="register" value='Register'> </form> And instead of having a function register(){ I replaced it with a if($register){ So when the Register button is pressed it runs the php code, but this edit doesn't seem to work either. So what can the problem be? If needed I can re-add this code on my Domain The login page has the same issue, nothing happens when the button is pressed beside emptying the fields.

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  • How do I make the PolicyKit authentication agent window not dissapear when I enter faulty password in Ubuntu 12.04?

    - by Petar
    As far as I remember in previous versions of Ubuntu, whenever authentication was required and when the PolicyKit authentication agent window was presented, it stayed there even after I would enter a faulty password. But now, whenever I make a mistake, the window is closed immediately. I find this behaviour irritating. For instance I use Synaptic rather frequently, and I prefer to start it using Synapse. I press Ctrl+Space to invoke Synapse, then I enter "syn" (s-shows SMplayer, sy- shows System Monitor) and than I press Enter so that Synaptic is invoked. Then I'm presented with the PolicyKit authentication agent window. As my password is rather complicated - using special characters and big letters, it's easy to make a mistake. If I do make a mistake while typing my password, I'm forced to redo all the previous steps. It's annoying as hell, knowing that this is not the way the PolicyKit authentication agent window behaved before. It used to warn me that the password was not correct and than wait for the correct input. I'm not sure if it allowed trying for the correct password indefinitely, or it was limited to 3 retries which is a much saner behaviour than the current one. I'm using Gnome 3, but the same thing happens in Unity too, although the window looks different.

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  • After locking the screen in Ubuntu 14.04, password is not accepted, How can it be fixed?

    - by Itai Ganot
    I'm running Ubuntu 14.04 fully updated on my laptop. Since the last update every time I lock the screen (when leaving my room for example) - when I get back and input my password, it is not accepted even though it's the correct password, the error I get is: Password incorrect, please try again I found that clicking the "Switch Account" fixes the issue but it is very annoying, if you know any way to fix it, it would be nice. Thanks in advance

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  • How do I include a password with SSH command? (want to make shell script)

    - by Evan
    I'm trying to SSH to a server on startup with a .sh script, but that will require me to enter the password for the account on the server that I'm SSHing to. I did some RTFMing, and I see in "-o" that it has "PasswordAuthentication" but I'm not sure how or if I could use that option. As this will be in a shell script, obviously I'd like to have the password in that file, or in any case not have to enter in the password manually every time the script runs.

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  • How to enter the Default Keyring password via the command line?

    - by Jerkofalltrades
    Is there a way to enter the default keyring password using the command line? For instance: You have a remote setup of Ubuntu 10.10 thats set to auto login. You don't want to remove the keyring password. All right the system boots up and logs in automatically, then asks for the keyring password now at this point you can create ssh connections but you can't remote desktop. What can you do to enter the keyring password at this point? Also, to better clarify, this is from a remote connection using the command line.

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  • is it possible to ssh into ubuntu server with keys and from another computer using password?

    - by Sandro Dzneladze
    I have ubuntu server at home, and I use SSH keys to log in via terminal from my laptop. Rarely, but mostly from work, I need to access this server, but I dont want to copy keys on my work laptop - Id rather just use plain password for that. So if I enable password log in from ssh config file, it will ask me password on my laptop too even though it contains keys, right? Is there a way to stop this behavior? Thanks.

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  • Using PhysX, how can I predict where I will need to generate procedural terrain collision shapes?

    - by Sion Sheevok
    In this situation, I have terrain height values I generate procedurally. For rendering, I use the camera's position to generate an appropriate sized height map. For collision, however, I need to have height fields generated in areas where objects may intersect. My current potential solution, which may be naive, is to iterate over all "awake" physics actors, use their bounds/extents and velocities to generate spheres in which they may reside after a physics update, then generate height values for ranges encompassing clustered groups of actors. Much of that data is likely already calculated by PhysX already, however. Is there some API, maybe a set of queries, even callbacks from the spatial system, that I could use to predict where terrain height values will be needed?

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  • ADODB Connection String: Workgroup Information file is Missing?

    - by Mohgeroth
    I have a few data sources in access that I need to connect to programatically to do things with behind the scenes and keep visibility away from users. Said datasource has a password 'pass' as I'm going to call it here. Using this connection method I get an error attempting to use the open method Dim conn as ADODB.Connection Set ROBBERS.conn = New ADODB.Connection conn.open "Provider=Microsoft.Jet.OLEDB.4.0;" _ & "Data Source=\\pep-home\projects\billing\autobilling\DPBilling2.mdb;" _ & "Jet OLEDB:Database Password=pass;", "admin", "pass" "Cannot start your application. The workgroup information file is missing or opened exclusively by another user." Due to planning to move into 2007, we are not using nor have ever used a workgroup identification file through access. The database password on the data source was set through the Set Databa Password which had to be done on an exclusive open. Ive spent a good while changing around my connection options, where to put the passwords etc and either cannot find the right format, or (why I'm asking here) I think there may be some other unknown that I must setup to do this. Anyone out there got some useful information?

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