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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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  • 48-bit bitwise operations in Javascript?

    - by randomhelp
    I've been given the task of porting Java's Java.util.Random() to JavaScript, and I've run across a huge performance hit/inaccuracy using bitwise operators in Javascript on sufficiently large numbers. Some cursory research states that "bitwise operators in JavaScript are inherently slow," because internally it appears that JavaScript will cast all of its double values into signed 32-bit integers to do the bitwise operations (see https://developer.mozilla.org/En/Core_JavaScript_1.5_Reference/Operators/Bitwise_Operators for more on this.) Because of this, I can't do a direct port of the Java random number generator, and I need to get the same numeric results as Java.util.Random(). Writing something like this.next = function(bits) { if (!bits) { bits = 48; } this.seed = (this.seed * 25214903917 + 11) & ((1 << 48) - 1); return this.seed >>> (48 - bits); }; (which is an almost-direct port of the Java.util.Random()) code won't work properly, since Javascript can't do bitwise operations on an integer that size.) I've figured out that I can just make a seedable random number generator in 32-bit space using the Lehmer algorithm, but the trick is that I need to get the same values as I would with Java.util.Random(). What should I do to make a faster, functional port?

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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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  • Recognizing terminals in a CFG production previously not defined as tokens.

    - by kmels
    I'm making a generator of LL(1) parsers, my input is a CoCo/R language specification. I've already got a Scanner generator for that input. Suppose I've got the following specification: COMPILER 1. CHARACTERS digit="0123456789". TOKENS number = digit{digit}. decnumber = digit{digit}"."digit{digit}. PRODUCTIONS Expression = Term{"+"Term|"-"Term}. Term = Factor{"*"Factor|"/"Factor}. Factor = ["-"](Number|"("Expression")"). Number = (number|decnumber). END 1. So, if the parser generated by this grammar receives a word "1+1", it'd be accepted i.e. a parse tree would be found. My question is, the character "+" was never defined in a token, but it appears in the non-terminal "Expression". How should my generated Scanner recognize it? It would not recognize it as a token. Is this a valid input then? Should I add this terminal in TOKENS and then consider an error routine for a Scanner for it to skip it? How does usual language specifications handle this?

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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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  • 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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  • 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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  • 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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  • Login loop in Snow Leopard

    - by hgpc
    I can't get out of a login loop of a particular admin user. After entering the password the login screen is shown again after about a minute. Other users work fine. It started happening after a simple reboot. Can you please help me? Thank you! Tried to no avail: Change the password Remove the password Repair disk (no errors) Boot in safe mode Reinstall Snow Leopard and updating to 10.6.6 Remove content of ~/Library/Caches Removed content of ~/Library/Preferences Replaced /etc/authorization with Install DVD copy The system.log mentions a crash report. I'm including both below. system.log Jan 8 02:43:30 loginwindow218: Login Window - Returned from Security Agent Jan 8 02:43:30 loginwindow218: USER_PROCESS: 218 console Jan 8 02:44:42 kernel[0]: Jan 8 02:44:43: --- last message repeated 1 time --- Jan 8 02:44:43 com.apple.launchd[1] (com.apple.loginwindow218): Job appears to have crashed: Bus error Jan 8 02:44:43 com.apple.UserEventAgent-LoginWindow223: ALF error: cannot find useragent 1102 Jan 8 02:44:43 com.apple.UserEventAgent-LoginWindow223: plugin.UserEventAgentFactory: called with typeID=FC86416D-6164-2070-726F-70735C216EC0 Jan 8 02:44:43 /System/Library/CoreServices/loginwindow.app/Contents/MacOS/loginwindow233: Login Window Application Started Jan 8 02:44:43 SecurityAgent228: CGSShutdownServerConnections: Detaching application from window server Jan 8 02:44:43 com.apple.ReportCrash.Root232: 2011-01-08 02:44:43.936 ReportCrash232:2903 Saved crash report for loginwindow218 version ??? (???) to /Library/Logs/DiagnosticReports/loginwindow_2011-01-08-024443_localhost.crash Jan 8 02:44:44 SecurityAgent228: MIG: server died: CGSReleaseShmem : Cannot release shared memory Jan 8 02:44:44 SecurityAgent228: kCGErrorFailure: Set a breakpoint @ CGErrorBreakpoint() to catch errors as they are logged. Jan 8 02:44:44 SecurityAgent228: CGSDisplayServerShutdown: Detaching display subsystem from window server Jan 8 02:44:44 SecurityAgent228: HIToolbox: received notification of WindowServer event port death. Jan 8 02:44:44 SecurityAgent228: port matched the WindowServer port created in BindCGSToRunLoop Jan 8 02:44:44 loginwindow233: Login Window Started Security Agent Jan 8 02:44:44 WindowServer234: kCGErrorFailure: Set a breakpoint @ CGErrorBreakpoint() to catch errors as they are logged. Jan 8 02:44:44 com.apple.WindowServer234: Sat Jan 8 02:44:44 .local WindowServer234 <Error>: kCGErrorFailure: Set a breakpoint @ CGErrorBreakpoint() to catch errors as they are logged. Jan 8 02:44:54 SecurityAgent243: NSSecureTextFieldCell detected a field editor ((null)) that is not a NSTextView subclass designed to work with the cell. Ignoring... Crash report Process: loginwindow 218 Path: /System/Library/CoreServices/loginwindow.app/Contents/MacOS/loginwindow Identifier: loginwindow Version: ??? (???) Code Type: X86-64 (Native) Parent Process: launchd [1] Date/Time: 2011-01-08 02:44:42.748 +0100 OS Version: Mac OS X 10.6.6 (10J567) Report Version: 6 Exception Type: EXC_BAD_ACCESS (SIGBUS) Exception Codes: 0x000000000000000a, 0x000000010075b000 Crashed Thread: 0 Dispatch queue: com.apple.main-thread Thread 0 Crashed: Dispatch queue: com.apple.main-thread 0 com.apple.security 0x00007fff801c6e8b Security::ReadSection::at(unsigned int) const + 25 1 com.apple.security 0x00007fff801c632f Security::DbVersion::open() + 123 2 com.apple.security 0x00007fff801c5e41 Security::DbVersion::DbVersion(Security::AppleDatabase const&, Security::RefPointer<Security::AtomicBufferedFile> const&) + 179 3 com.apple.security 0x00007fff801c594e Security::DbModifier::getDbVersion(bool) + 330 4 com.apple.security 0x00007fff801c57f5 Security::DbModifier::openDatabase() + 33 5 com.apple.security 0x00007fff801c5439 Security::Database::_dbOpen(Security::DatabaseSession&, unsigned int, Security::AccessCredentials const*, void const*) + 221 6 com.apple.security 0x00007fff801c4841 Security::DatabaseManager::dbOpen(Security::DatabaseSession&, Security::DbName const&, unsigned int, Security::AccessCredentials const*, void const*) + 77 7 com.apple.security 0x00007fff801c4723 Security::DatabaseSession::DbOpen(char const*, cssm_net_address const*, unsigned int, Security::AccessCredentials const*, void const*, long&) + 285 8 com.apple.security 0x00007fff801d8414 cssm_DbOpen(long, char const*, cssm_net_address const*, unsigned int, cssm_access_credentials const*, void const*, long*) + 108 9 com.apple.security 0x00007fff801d7fba CSSM_DL_DbOpen + 106 10 com.apple.security 0x00007fff801d62f6 Security::CssmClient::DbImpl::open() + 162 11 com.apple.security 0x00007fff801d8977 SSDatabaseImpl::open(Security::DLDbIdentifier const&) + 53 12 com.apple.security 0x00007fff801d8715 SSDLSession::DbOpen(char const*, cssm_net_address const*, unsigned int, Security::AccessCredentials const*, void const*, long&) + 263 13 com.apple.security 0x00007fff801d8414 cssm_DbOpen(long, char const*, cssm_net_address const*, unsigned int, cssm_access_credentials const*, void const*, long*) + 108 14 com.apple.security 0x00007fff801d7fba CSSM_DL_DbOpen + 106 15 com.apple.security 0x00007fff801d62f6 Security::CssmClient::DbImpl::open() + 162 16 com.apple.security 0x00007fff802fa786 Security::CssmClient::DbImpl::unlock(cssm_data const&) + 28 17 com.apple.security 0x00007fff80275b5d Security::KeychainCore::KeychainImpl::unlock(Security::CssmData const&) + 89 18 com.apple.security 0x00007fff80291a06 Security::KeychainCore::StorageManager::login(unsigned int, void const*, unsigned int, void const*) + 3336 19 com.apple.security 0x00007fff802854d3 SecKeychainLogin + 91 20 com.apple.loginwindow 0x000000010000dfc5 0x100000000 + 57285 21 com.apple.loginwindow 0x000000010000cfb4 0x100000000 + 53172 22 com.apple.Foundation 0x00007fff8721e44f __NSThreadPerformPerform + 219 23 com.apple.CoreFoundation 0x00007fff82627401 __CFRunLoopDoSources0 + 1361 24 com.apple.CoreFoundation 0x00007fff826255f9 __CFRunLoopRun + 873 25 com.apple.CoreFoundation 0x00007fff82624dbf CFRunLoopRunSpecific + 575 26 com.apple.HIToolbox 0x00007fff8444493a RunCurrentEventLoopInMode + 333 27 com.apple.HIToolbox 0x00007fff8444473f ReceiveNextEventCommon + 310 28 com.apple.HIToolbox 0x00007fff844445f8 BlockUntilNextEventMatchingListInMode + 59 29 com.apple.AppKit 0x00007fff80b01e64 _DPSNextEvent + 718 30 com.apple.AppKit 0x00007fff80b017a9 -NSApplication nextEventMatchingMask:untilDate:inMode:dequeue: + 155 31 com.apple.AppKit 0x00007fff80ac748b -NSApplication run + 395 32 com.apple.loginwindow 0x0000000100004b16 0x100000000 + 19222 33 com.apple.loginwindow 0x0000000100004580 0x100000000 + 17792 Thread 1: Dispatch queue: com.apple.libdispatch-manager 0 libSystem.B.dylib 0x00007fff8755216a kevent + 10 1 libSystem.B.dylib 0x00007fff8755403d _dispatch_mgr_invoke + 154 2 libSystem.B.dylib 0x00007fff87553d14 _dispatch_queue_invoke + 185 3 libSystem.B.dylib 0x00007fff8755383e _dispatch_worker_thread2 + 252 4 libSystem.B.dylib 0x00007fff87553168 _pthread_wqthread + 353 5 libSystem.B.dylib 0x00007fff87553005 start_wqthread + 13 Thread 0 crashed with X86 Thread State (64-bit): rax: 0x000000010075b000 rbx: 0x00007fff5fbfd990 rcx: 0x00007fff875439da rdx: 0x0000000000000000 rdi: 0x00007fff5fbfd990 rsi: 0x0000000000000000 rbp: 0x00007fff5fbfd5d0 rsp: 0x00007fff5fbfd5d0 r8: 0x0000000000000007 r9: 0x0000000000000000 r10: 0x00007fff8753beda r11: 0x0000000000000202 r12: 0x0000000100133e78 r13: 0x00007fff5fbfda50 r14: 0x00007fff5fbfda50 r15: 0x00007fff5fbfdaa0 rip: 0x00007fff801c6e8b rfl: 0x0000000000010287 cr2: 0x000000010075b000

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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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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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  • Source Control - XCode - Visual Studio 2005/2008 / 2010

    - by Mick Walker
    My apologies if this has been asked before, I wasnt quite sure if this question should be asked on a programming forum, as it more relates to programming environment than a particular technology, so please accept my (double) appologies if I am posting this in the wrong place, my logic in this case was if it effects the code I write, then this is the place for it. At home, I do a lot of my development on a Mac Pro, I do development for the Mac, iPhone and Windows on this machine (Xcode & Visual Studio - (multiple versions installed in bootcamp, but generally I run it via Parallels)). When visiting a client, I have a similar setup, but on my MacBook Pro. What I want is a source control solution to install on the Mac Pro, that will support both XCode and multiple versions of visual studio, so that when I visit a client, I can simply grab the latest copy from source control via the MacBook Pro. Whilst visiting the client, he / she may suggest changes, and minor ones I would tend to make on site, so I need the ability to merge any modified code back into the trunk of the project / solution when I return home. At the moment, I am using no source control at all, and rely on simply coping folders and overwriting them when I return from a client- thats my 'merge'!!! I was wondering if anyone had any ideas of a source provider I could use, which would support both Windows and Mac development environments, and is cheap (free would be better).

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  • Security benefits from a second opinion, are there flaws in my plan to hash & salt user passwords vi

    - by Tchalvak
    Here is my plan, and goals: Overall Goals: Security with a certain amount of simplicity & database-to-database transferrability, 'cause I'm no expert and could mess it up and I don't want to have to ask a lot of users to reset their passwords. Easy to wipe the passwords for publishing a "wiped" databased of test data. (e.g. I'd like to be able to use a postgresql statement to simply reset all passwords to something simple so that testers can use that testing data for themselves). Plan: Hashing the passwords Account creation records the original email that an account is created with, forever. A global salt is used, e.g. "90fb16b6901dfceb73781ba4d8585f0503ac9391". An account specific salt, the original email the account was created with, is used, e.g. "[email protected]". The users's password is used, e.g. "password123" (I'll be warning against weak passwords in the signup form) The combination of the global salt, account specific salt, and password is hashed via some hashing method in postgresql (haven't been able to find documentation for hashing functions in postgresql, but being able to use sha-2 or something like that would be nice if I could find it). The hash gets stored in the database. Recovering an account To change their password, they have to go through standard password reset (and that reset email gets sent to the original email as well as the most recent account email that they have set). Flaws? Are there any flaws with this that I need to address? And are there best practices to doing hashing fully within postgresql?

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  • Rosetta Stone data on Dropbox

    - by jnman
    Hi, I'm currently trying to learn a new language using Rosetta Stone. Given that I have 2 computers (mac desktop and laptop), I was wondering if anyone knew if it was possible to have the data files in Dropbox so that I can synchronize my learning across computers.

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  • admin can't view non admin user's folder in osx

    - by adolf garlic
    I'm trying to add a new keyboard layout for a non admin user on my mac. I had thought that the keyboard layout would be applied for all users when I added it to mine but alas no. I cannot get into the Users\\library\keyboard layouts folder, as it won't let me (but I'm an admin FFS!) I even went into 'get info' and set it to 'everyone read and write' but it still tells me that I don't have permission How on earth can I update the other user's keyboard layout folder?

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