Search Results

Search found 13776 results on 552 pages for 'password reset'.

Page 58/552 | < Previous Page | 54 55 56 57 58 59 60 61 62 63 64 65  | Next Page >

  • jquery reset conditional filter

    - by VictorS
    I have 2 dropdown lists on my webform and using jquery trying to filter/reset filter 2nd dropdown elements based on 1st dropdown selection. $(document).ready(function() { $('#dropdown1').change(function(e) { switch ($(this).val()) { case "4": //this removal works $('#dropdown2').filter(function() { return ($(this).val() == 16); }).remove(); break; ................. default: //how would I restore filter here? } } }); Removing part works, so it filters item with no problem, but I have difficulty restoring the filter on dropdown 2 if something else is chosen in dropdown 1. I was trying to use .hide() and .show() instead of .remove() but it doesn't seem to work on IE6 at least.

    Read the article

  • WPF: Reset the positions of scatterviewitems?

    - by sofri
    Hi, I have a scatterview with some items in it which I place with Orientation und Center. Now I want to have the possibility to reset the positions of the scatterviewitems after scaling, rotating and moving them, while the program is running. At the moment I do it this way: private void Reset_ContactTapGesture(object sender, Microsoft.Surface.Presentation.ContactEventArgs e) { item1.Center = new Point(150,150); item1.Orientation = 15; item1.Width = 100; item1.Height = 150; } Is there a better way to do it?

    Read the article

  • After .load() Reset Textbox with User Entered Value using JavaScript and jQuery

    - by Aaron Salazar
    My function below calls a partial view after a user enters a filter-by string into the text box '#DocId'. When the user is done typing, the partial view is displayed with filtered data. Since my textbox needs to be in the partial view, when user is done entering a filter-by string and is shown the filtered data, the textbox is reset and the user entered data is lost. How can I set the value of the textbox back to the user entered string after the partial view is displayed? I'm pretty sure I need to use .val() but I can't seem to get this to work. $(function() { $('#DocId').live('keyup', function() { clearTimeout($.data(this, 'timer')); var val = $(this).val(); var wait = setTimeout(function() { $('#tableContent').load('/CurReport/TableResults', { filter: val }, 500) }, 500); $(this).data('timer', wait); }); }); Thank you, Aaron

    Read the article

  • jQuery > reset popup window 's x and y relative to icon that launched the window

    - by Scott B
    I have a jQuery jPicker colorpicker that is bound to an input text field and opens onclick of a small picker.gif icon. The problem I'm having is that (1) The colorpicker that opens does not appear to be positioned according to the x/y position of the picker.gif (it opens far away from the click point) and (2) The colorpicker does not seem to be aware of the viewport's scroll position (the top of the colorpicker is partially hidden at the top of the window). I'd like to use jQuery to reposition the colorpalette (1) Based on the x/y position of the input that its bound to and (2) reset it's top position based on the viewport's visible Y position. Here is the script where I am creating a new jPicker and binding it to my input text fields for header and sidebar... $('#myHeaderColor').jPicker ( {}, function(color) { $(this).val(color.get_Hex()); }, function(color) { $(this).val(color.get_Hex()); } ); $('#mySidebarColor').jPicker ( {}, function(color) { $(this).val(color.get_Hex()); }, function(color) { $(this).val(color.get_Hex()); } );

    Read the article

  • Reset application and settings on user change

    - by Don
    Currently working on a project where a login will be required to use the application. I'm trying to figure out a smarter way to reset the application if someone is somehow logged out and the next one to login is not the same user. The option I have come up with at the moment is storing all user specific data/information in a DTO but this leaves me with cleaning up some parts of the work area. Is a ResetControls my only option here? I'm afraid that when updating the application someone might forget to update that part, most likely myself now that wrote it out. Anybody with experience in this that could provide some ideas to a simple yet fairly automagic solution?

    Read the article

  • ASP.NET: How to "reset" validation after calling Page_ClientValidate

    - by Josh Young
    I have an ASP.NET page with a jQuery dialog that is displayed to change some data. I am setting up the jQuery dialog so that when the user clicks the OK button it calls ASP.NET's Page_ClientValidate('validationGroup') via javascript, finds all the invalid controls and changes their CSS class. So here's the scenario: the user opens the dialog, keys in some invalid data, clicks OK (receiving the validation messages), and then clicks Cancel. Now the dialog is closed, but the validation messages are still there, so that when they open the dialog again, the data goes back to the way it was initially, but the form is still in the invalid state (the validation messages are still displaying). What I need is a "reset" function of sorts to call after calling Page_ClientValidate('validationGroup'). Does this exist?

    Read the article

  • How to reset buttons when using touchesBegan

    - by FireStorm
    When I use touchesBegan on my sprite kit game the buttons press down fine the first time but are unresponsive the second time, how do I "reset" these buttons if you will. Buttons are as follows: - (void)touchesBegan:(NSSet *)touches withEvent:(UIEvent *)event { UITouch *touch = [touches anyObject]; CGPoint location = [touch locationInNode:self]; SKNode *node = [self nodeAtPoint:location]; if ([node.name isEqualToString:@"Button"]) { [self runAction:[SKAction playSoundFileNamed:@"sound" waitForCompletion: NO]]; .... Thanks for your help!

    Read the article

  • 40k Event Log Errors an hour Unknown Username or bad password

    - by ErocM
    I am getting about 200k of these an hour: An account failed to log on. Subject: Security ID: SYSTEM Account Name: TGSERVER$ Account Domain: WORKGROUP Logon ID: 0x3e7 Logon Type: 4 Account For Which Logon Failed: Security ID: NULL SID Account Name: administrator Account Domain: TGSERVER Failure Information: Failure Reason: Unknown user name or bad password. Status: 0xc000006d Sub Status: 0xc0000064 Process Information: Caller Process ID: 0x334 Caller Process Name: C:\Windows\System32\svchost.exe Network Information: Workstation Name: TGSERVER Source Network Address: - Source Port: - Detailed Authentication Information: Logon Process: Advapi Authentication Package: Negotiate Transited Services: - Package Name (NTLM only): - Key Length: 0 This event is generated when a logon request fails. It is generated on the computer where access was attempted. The Subject fields indicate the account on the local system which requested the logon. This is most commonly a service such as the Server service, or a local process such as Winlogon.exe or Services.exe. The Logon Type field indicates the kind of logon that was requested. The most common types are 2 (interactive) and 3 (network). The Process Information fields indicate which account and process on the system requested the logon. The Network Information fields indicate where a remote logon request originated. Workstation name is not always available and may be left blank in some cases. The authentication information fields provide detailed information about this specific logon request. - Transited services indicate which intermediate services have participated in this logon request. - Package name indicates which sub-protocol was used among the NTLM protocols. - Key length indicates the length of the generated session key. This will be 0 if no session key was requested. On my server... I changed my adminstrative username to something else and since then I've been inidated with these messages. I found on http://technet.microsoft.com/en-us/library/cc787567(v=WS.10).aspx that the 4 means "Batch logon type is used by batch servers, where processes may be executing on behalf of a user without their direct intervention." which really doesn't shed any light on it for me. I checked the services and they are all logging in as local system or network service. Nothing for administrator. Anyone have any idea how I tell where these are coming from? I would assume this is a program that is crapping out... Thanks in advance!

    Read the article

  • Why enumerator structs are a really bad idea (redux)

    - by Simon Cooper
    My previous blog post went into some detail as to why calling MoveNext on a BCL generic collection enumerator didn't quite do what you thought it would. This post covers the Reset method. To recap, here's the simple wrapper around a linked list enumerator struct from my previous post (minus the readonly on the enumerator variable): sealed class EnumeratorWrapper : IEnumerator<int> { private LinkedList<int>.Enumerator m_Enumerator; public EnumeratorWrapper(LinkedList<int> linkedList) { m_Enumerator = linkedList.GetEnumerator(); } public int Current { get { return m_Enumerator.Current; } } object System.Collections.IEnumerator.Current { get { return Current; } } public bool MoveNext() { return m_Enumerator.MoveNext(); } public void Reset() { ((System.Collections.IEnumerator)m_Enumerator).Reset(); } public void Dispose() { m_Enumerator.Dispose(); } } If you have a look at the Reset method, you'll notice I'm having to cast to IEnumerator to be able to call Reset on m_Enumerator. This is because the implementation of LinkedList<int>.Enumerator.Reset, and indeed of all the other Reset methods on the BCL generic collection enumerators, is an explicit interface implementation. However, IEnumerator is a reference type. LinkedList<int>.Enumerator is a value type. That means, in order to call the reset method at all, the enumerator has to be boxed. And the IL confirms this: .method public hidebysig newslot virtual final instance void Reset() cil managed { .maxstack 8 L_0000: nop L_0001: ldarg.0 L_0002: ldfld valuetype [System]System.Collections.Generic.LinkedList`1/Enumerator<int32> EnumeratorWrapper::m_Enumerator L_0007: box [System]System.Collections.Generic.LinkedList`1/Enumerator<int32> L_000c: callvirt instance void [mscorlib]System.Collections.IEnumerator::Reset() L_0011: nop L_0012: ret } On line 0007, we're doing a box operation, which copies the enumerator to a reference object on the heap, then on line 000c calling Reset on this boxed object. So m_Enumerator in the wrapper class is not modified by the call the Reset. And this is the only way to call the Reset method on this variable (without using reflection). Therefore, the only way that the collection enumerator struct can be used safely is to store them as a boxed IEnumerator<T>, and not use them as value types at all.

    Read the article

  • X509Certificate.CreateFromCertFile - the specified network password is not correct

    - by pcampbell
    I have a .NET application that I want to use as a client to call an SSL SOAP web service. I have been supplied with a valid client certificate called foo.pfx. There is a password on the certificate itself. I've located the certificate at the following location: C:\certs\foo.pfx To call the web service, I need to attach the client certificate. Here's the code: public X509Certificate GetCertificateFromDisk(){ try{ string certPath = ConfigurationManager.AppSettings["MyCertPath"].ToString(); //this evaluates to "c:\\certs\\foo.pfx". So far so good. X509Certificate myCert = X509Certificate.CreateFromCertFile(certPath); // exception is raised here! "The specified network password is not correct" return cert; } catch (Exception ex){ throw; } } It sounds like the exception is around the .NET application trying to read the disk. The method CreateFromCertFile is a static method that should create a new instance of X509Certificate. The method isn't overridden, and has only one argument: the path. When I inspect the Exception, I find this: _COMPlusExceptionCode = -532459699 Source=mscorlib Question: does anyone know what the cause of the exception "The specified network password is not correct" ?

    Read the article

  • Windows server 2003 default administrator password

    - by Jason Baker
    Sorry if this is an overly simplistic question, but I'm a bit stuck here. :) I need a windows machine for me to do some programming for class. Since I have my Macbook with me everywhere I go, I figured that it would be easiest to install a vm. And since I can get a copy of Windows server 2k3 for free via dreamspark, I thought I'd try to do that. Here's what happened though: I installed windows server (disk one). When the system booted up, vmware automatically installed VMWare tools and prompted me to restart. There was also a prompt to start the installation of disc 2, but I figured it would be better to restart before doing that. When the machine came back up, I was prompted to log in as the administrator. The problem is that I wasn't prompted to make an administrator account or password. Is there a default password I can use? I've tried all the obvious ones (blank, password, etc) and googling, but I didn't come up with anything.

    Read the article

  • ASP.NET Membership ChangePassword control - Need to check for previous password

    - by Steve
    Hi guys, I have a new table that hold old passwords, I need to check if there is a match. If there is a match I need the ChangePassword contol to NOT change the password. I need to tell the user that this password was used and pic a new one. I can't seem to be able to interrupt the control from changing the password. Maybe I am using the wrong event. Here is a piece of my code, or how I wish it would work. I appreciate all your help. protected void ChangePassword1_ChangedPassword(object sender, EventArgs e) { MembershipUser user = Membership.GetUser(); string usrName = ""; if (user != null) { string connStr = ConfigurationManager.ConnectionStrings["LocalSqlServer"].ConnectionString; SqlConnection mySqlConnection = new SqlConnection(connStr); SqlCommand mySqlCommand = mySqlConnection.CreateCommand(); mySqlCommand.CommandText = "Select UserName from OldPasswords where UserName = 'test'"; mySqlConnection.Open(); SqlDataReader mySqlDataReader = mySqlCommand.ExecuteReader(CommandBehavior.Default); while (mySqlDataReader.Read()) { usrName = mySqlDataReader["UserName"].ToString(); if (usrName == user.ToString()) { Label1.Text = "Match"; } else { Label1.Text = "NO Match!"; } }

    Read the article

  • 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

    Read the article

  • 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).

    Read the article

  • 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();

    Read the article

  • 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.

    Read the article

  • 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); } }

    Read the article

  • 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.

    Read the article

  • 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

    Read the article

  • Very different font sizes across browsers

    - by Yang
    Chrome/WebKit and Firefox have different rendering engines which render fonts differently, in particular with differing dimensions. This isn't too surprising, but what's surprising is the magnitude of some of the differences. I can always tweak individual elements on a page to be more similar, but that's tedious, to say the least. I've been searching for more systematic solutions, but many resources (e.g. SO answers) simply say "use a reset package." While I'm sure this fixes a bunch of other things like padding and spacing, it doesn't seem to make any difference for font dimensions. For instance, if I take the reset package from http://html5reset.org/, I can show pretty big differences (note the layout dimensions shown in the inspectors). [The images below are actually higher res than shown/resized in this answer.] <h1 style="font-size:64px; background-color: #eee;">Article Header</h1> With Helvetica, Chrome is has the shorter height instead. <h1 style="font-size:64px; background-color: #eee; font-family: Helvetica">Article Header</h1> Using a different font, Chrome again renders a much taller font, but additionally the letter spacing goes haywire (probably due to the boldification of the font): <style> @font-face { font-family: "MyriadProRegular"; src: url("fonts/myriadpro-regular-webfont.eot"); src: local("?"), url("fonts/myriadpro-regular-webfont.woff") format("woff"), url("fonts/myriadpro-regular-webfont.ttf") format("truetype"), url("fonts/myriadpro-regular-webfont.svg#webfonteknRmz0m") format("svg"); font-weight: normal; font-style: normal; } @font-face { font-family: "MyriadProLight"; src: url("fonts/myriadpro-light-webfont.eot"); src: local("?"), url("fonts/myriadpro-light-webfont.woff") format("woff"), url("fonts/myriadpro-light-webfont.ttf") format("truetype"), url("fonts/myriadpro-light-webfont.svg#webfont2SBUkD9p") format("svg"); font-weight: normal; font-style: normal; } @font-face { font-family: "MyriadProSemibold"; src: url("fonts/myriadpro-semibold-webfont.eot"); src: local("?"), url("fonts/myriadpro-semibold-webfont.woff") format("woff"), url("fonts/myriadpro-semibold-webfont.ttf") format("truetype"), url("fonts/myriadpro-semibold-webfont.svg#webfontM3ufnW4Z") format("svg"); font-weight: normal; font-style: normal; } </style> ... <h1 style="font-size:64px; background-color: #eee; font-family: Helvetica">Article Header</h1> I've tried a few resets/normalize packages to no avail. I just wanted to confirm here that this is indeed a fact of life (even omitting the more glaring offenders like IE and mobile) and I'm not missing some super-awesome solution to this mess.

    Read the article

  • 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?

    Read the article

  • 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 ....

    Read the article

  • 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.

    Read the article

  • 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.

    Read the article

  • 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

    Read the article

< Previous Page | 54 55 56 57 58 59 60 61 62 63 64 65  | Next Page >