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  • apache using mod_auth_kerb always asks for the password twice

    - by DrStalker
    (Debian Squeeze) I'm trying to set apache up to use Kerberos authentication to allow AD users to log in. It is working, but prompts the user twice for a username and password, with the first time being ignored (no matter what is put it in.) Only the second prompt includes the AuthName string from the config (i.e.: the first windows is a generic username/password one, the second includes the title "Kerberos Login") I'm not worried about integrated windows authentication working at this stage, I just want users to be able to login with their AD account so we don't need to set up a second repository of user accounts. How do I fix this to eliminate that first useless prompt? The directives in the apache2.conf file: <Directory /var/www/kerberos> AuthType Kerberos AuthName "Kerberos Login" KrbMethodNegotiate On KrbMethodK5Passwd On KrbAuthRealms ONEVUE.COM.AU.LOCAL Krb5KeyTab /etc/krb5.keytab KrbServiceName HTTP/[email protected] require valid-user </Directory> krb5.conf: [libdefaults] default_realm = ONEVUE.COM.AU.LOCAL [realms] ONEVUE.COM.AU.LOCAL = { kdc = SYD01PWDC01.ONEVUE.COM.AU.LOCAL master_kdc = SYD01PWDC01.ONEVUE.COM.AU.LOCAL admin_server = SYD01PWDC01.ONEVUE.COM.AU.LOCAL default_domain = ONEVUE.COM.AU.LOCAL } [login] krb4_convert = true krb4_get_tickets = false The access log when accessing the secured directory (note the two seperate 401's) 192.168.10.115 - - [24/Aug/2012:15:52:01 +1000] "GET /kerberos/ HTTP/1.1" 401 710 "-" "Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.1 (KHTML, like Gecko) Chrome/21.0.1180.83 Safari/537.1" 192.168.10.115 - - [24/Aug/2012:15:52:06 +1000] "GET /kerberos/ HTTP/1.1" 401 680 "-" "Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.1 (KHTML, like Gecko) Chrome/21.0.1180.83 Safari/537.1" 192.168.10.115 - [email protected] [24/Aug/2012:15:52:10 +1000] "GET /kerberos/ HTTP/1.1" 200 375 "-" "Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.1 (KHTML, like Gecko) Chrome/21.0.1180.83 Safari/537.1" And one line in error.log [Fri Aug 24 15:52:06 2012] [error] [client 192.168.0.115] gss_accept_sec_context(2) failed: An unsupported mechanism was requested (, Unknown error)

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  • Shut Out of XP - No Admin Password or CDR

    - by ashes999
    I inherited an old WinXP/Linux dual-boot machine from the stoneage. Because it has Linux, the regular boot process is replaced with the Fedora boot loader; I cannot, therefore, press F8 strategically to tell my PC to boot from CD. Even if I could, it's a moot point; the CDR doesn't seem to recognize any CDs. To make things worse, there's no option to network boot. The original user is probably long gone; I don't know the password for any of the Administrator group users. I can login using my corp account, but that's unprivileged on this machine. Since I'm not an admin, I can't do crazy things, like looking at boot.ini. Or deleting files. I only have 500MB free on my C drive. I'm pretty sure I can't boot from a USB, since I didn't see any settings for this in my BIOS. How can I get admin access for my user? Edit: Things I've tried: Boot from CD (CD not recognized) Launch CD from XP (CD not recognized) Install Daemon Tools Lite so I can install from an ISO -- don't have admin privileges XP password recovery tool -- requires admin privileges Adding an admin user -- no access to Control Panel Users since I'm not an admin Logging in as both the admin users on the system (trying some standard passwords) Using Fedora to chntpw (the Fedora version installed is ancient -- 2.7)

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  • Hadoop initscript askes password

    - by Ramesh
    I have installed hadoop on my ubuntu 12.04 single node .I am trying to execute an init script to make the hadoop run on start up but it asks password every time i execute. #!/bin/sh ### BEGIN INIT INFO # Provides: hadoop services # Required-Start: $network # Required-Stop: $network # Default-Start: 2 3 4 5 # Default-Stop: 0 1 6 # Description: Hadoop services # Short-Description: Enable Hadoop services including hdfs ### END INIT INFO PATH=/usr/local/sbin:/usr/local/bin:/sbin:/bin:/usr/sbin:/usr/bin HADOOP_BIN=/home/naveen/softwares/hadoop-1.0.3/bin NAME=hadoop DESC=hadoop USER=naveen ROTATE_SUFFIX= test -x $HADOOP_BIN || exit 0 RETVAL=0 set -e cd / start_hadoop () { set +e su $USER -s /bin/sh -c $HADOOP_BIN/start-all.sh > /var/log/hadoop/startup_log case "$?" in 0) echo SUCCESS RETVAL=0 ;; 1) echo TIMEOUT - check /var/log/hadoop/startup_log RETVAL=1 ;; *) echo FAILED - check /var/log/hadoop/startup_log RETVAL=1 ;; esac set -e } stop_hadoop () { set +e if [ $RETVAL = 0 ] ; then su $USER -s /bin/sh -c $HADOOP_BIN/stop-all.sh > /var/log/hadoop/shutdown_log RETVAL=$? if [ $RETVAL != 0 ] ; then echo FAILED - check /var/log/hadoop/shutdown_log fi else echo No nodes running RETVAL=0 fi set -e } restart_hadoop() { stop_hadoop start_hadoop } case "$1" in start) echo -n "Starting $DESC: " start_hadoop echo "$NAME." ;; stop) echo -n "Stopping $DESC: " stop_hadoop echo "$NAME." ;; force-reload|restart) echo -n "Restarting $DESC: " restart_hadoop echo "$NAME." ;; *) echo "Usage: $0 {start|stop|restart|force-reload}" >&2 RETVAL=1 ;; esac exit $RETVAL Please tell me how to run hadoop without entering password.

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  • Windows 7 Professional Cannot Connect to Share - Wrong password

    - by henryford
    I know that this question has actually been asked a few times before, but every solution I found didn't yield any results on my end, I can't get my head around it: When I am trying to connect to a share on the network, I always get the response "The specified network password is incorrect". However, the password is definetly correct and it works if I connect from another machine. I changed the LAN Manager authentication level to "Send LM & NTLM - use NTLMv2 session security if negiotated", I configured Kerberos encryption types to include all suites, rebooted (several times), but still - no luck. I can connect if I use my regular account with which I am logged in, but I need to connect with a different user since my log-in user has not enough privileges on the share. When I do that, the error above comes up. I'm really frustrated at the moment, this problem is driving me crazy. I'd be gladful for any possible solution to this. At the moment I'm using a workaround: I connect to a different machine via RDP, login with the user I have to use for the network-share connection and then I can map the drive and copy/paste from the RDP session to my local workstation. This is also working when I am connecting via RDP with my current login user and map the drive with the other user who has sufficent privileges. Tanks in advance, Thomas

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  • SSH Keys Authentication keeps asking for password

    - by Rhyuk
    Im trying to set access from ServerA(SunOS) to ServerB(Some custom Linux with Keyboard Interactive login) with SSH Keys. As a proof of concept I was able to do it between 2 virtual machines. Now in my real life scenario it isnt working. I created the keys in ServerA, copied them to ServerB, chmod'd .ssh folders to 700 on both ServerA,B. Here is the log of what I get. debug1: SSH2_MSG_KEXINIT sent debug1: SSH2_MSG_KEXINIT received debug1: kex: server->client aes128-ctr hmac-md5 none debug1: kex: client->server aes128-ctr hmac-md5 none debug1: Peer sent proposed langtags, ctos: debug1: Peer sent proposed langtags, stoc: debug1: We proposed langtags, ctos: en-US debug1: We proposed langtags, stoc: en-US debug1: SSH2_MSG_KEX_DH_GEX_REQUEST sent debug1: expecting SSH2_MSG_KEX_DH_GEX_GROUP debug1: dh_gen_key: priv key bits set: 125/256 debug1: bits set: 1039/2048 debug1: SSH2_MSG_KEX_DH_GEX_INIT sent debug1: expecting SSH2_MSG_KEX_DH_GEX_REPLY debug1: Host 'XXX.XXX.XXX.XXX' is known and matches the RSA host key. debug1: Found key in /XXX/.ssh/known_hosts:1 debug1: bits set: 1061/2048 debug1: ssh_rsa_verify: signature correct debug1: newkeys: mode 1 debug1: set_newkeys: setting new keys for 'out' mode debug1: SSH2_MSG_NEWKEYS sent debug1: expecting SSH2_MSG_NEWKEYS debug1: newkeys: mode 0 debug1: set_newkeys: setting new keys for 'in' mode debug1: SSH2_MSG_NEWKEYS received debug1: done: ssh_kex2. debug1: send SSH2_MSG_SERVICE_REQUEST debug1: got SSH2_MSG_SERVICE_ACCEPT debug1: Authentications that can continue: publickey,keyboard-interactive debug1: Next authentication method: publickey debug1: Trying private key: /XXXX/.ssh/identity debug1: Trying public key: /xxx/.ssh/id_rsa debug1: Authentications that can continue: publickey,keyboard-interactive debug1: Trying private key: /xxx/.ssh/id_dsa debug1: Next authentication method: keyboard-interactive Password: Password: ServerB has pretty limited actions since its a custom propietary linux. What could be happening? EDIT WITH ANSWER: Problem was that I didnt have those settings enabled in the sshd_config (Refer to accepted answer) AND that while pasting the key from ServerA to ServerB it would interpret the key as 3 separate lines. What I did was, in case you cant use ssh-copy-id like I couldnt. Paste the first line of your key in your "ServerB" authorized_keys file WITHOUT the last 2 characters, then type yourself the missing characters from line 1 and the first one from line 2, this will prevent adding a "new line" between the first and second line of the key. Repeat with the 3d line.

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  • Permission denied (publickey,gssapi-with-mic,password) ssh error

    - by zentenk
    Heads up I'm a noob with linux and networking. I set up a ubuntu server and I have a static ip for my network. When I try to connect to the server at home (external), it prompts me to log in. I supply the correct password (or incorrect pw), I get the error Permission denied, please try again. and after 3 times I get Permission denied (publickey,gssapi-with-mic,password) I am however able to connect with SSH from another computer in the same network with ssh < internal ip of server > I'm connecting with mac os x and my config file is vanilla. Note: During installation of ubuntu it says I don't have a default route or something while doing auto network configuration, but I ignored it and continued the installation, could this be the problem? EDIT: I have tried the below, I have nothing in hosts.allow and also iptables shows the ports that I have allowed, which is 22. I checked the auth.log, and there is nothing when I connect to it remotely (even when it says permission denied). I have tried connecting to it internally and the correct authentication logs show. Any idea whats wrong?

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  • Implement Semi-Round-Robin file which can be expanded and saved on demand

    - by ircmaxell
    Ok, that title is going to be a little bit confusing. Let me try to explain it a little bit better. I am building a logging program. The program will have 3 main states: Write to a round-robin buffer file, keeping only the last 10 minutes of data. Write to a buffer file, ignoring the time (record all data). Rename entire buffer file, and start a new one with the past 10 minutes of data (and change state to 1). Now, the use case is this. I have been experiencing some network bottlenecks from time to time in our network. So I want to build a system to record TCP traffic when it detects the bottleneck (detection via Nagios). However by the time it detects the bottlenecking, most of the useful data has already been transmitted. So, what I'd like is to have a deamon that runs something like dumpcap all the time. In normal mode, it'll only keep the past 10 minutes of data (Since there's no point in keeping a boat load of data if it's not needed). But when Nagios alerts, I will send a signal in the deamon to store everything. Then, when Naigos recovers it will send another signal to stop storing and flush the buffer to a save file. Now, the problem is that I can't see how to cleanly store a rotating 10 minutes of data. I could store a new file every 10 minutes and delete the old ones if in mode 1. But that seems a bit dirty to me (especially when it comes to figuring out when the alert happened in the file). Ideally, the file that was saved should be such that the alert is always at the 10:00 mark in the file. While that is possible with new files every 10 minutes, it seems like a bit dirty to "repair" the files to that point. Any ideas? Should I just do a rotating file system and combine them into 1 at the end (doing quite a bit of post-processing)? Is there a way to implement the semi-round-robin file cleanly so that there is no need for any post-processing? Thanks Oh, and the language doesn't matter as much at this stage (I'm leaning towards Python, but have no objection to any other language. It's less of an issue than the overall design)...

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  • Value isnt being saved in the strings

    - by Raptrex
    I'm trying to make a class where I put a key and value into the put method which puts the key in the k string array and value into the v string array, however it is not being saved in the array when I do get or display. For example: put(dan,30) get(dan) returns null display returns null null 10 times. Anyone know whats wrong? public class Memory { final int INITIAL_CAPACITY = 10; String[] k = new String[INITIAL_CAPACITY]; String[] v = new String[INITIAL_CAPACITY]; int count = 0; public Memory() { count = 0; } public int size() { return count; } public void put(String key, String value) { int a = 0; boolean found = false; for (int i = 0; i < k.length; i++) { //System.out.println("key is " + key.equals(k[i])); if (key.equalsIgnoreCase(k[i])) { v[i] = value; found = true; } if (found) break; a++; } //System.out.println(a == k.length); if (a == k.length); { k[count] = key; v[count] = value; //System.out.println(k[count] + " " + v[count]); count++; //System.out.println(count); } } public String get(String key) { String output = "a"; for(int i = 0; i < k.length; i++) { if(!key.equalsIgnoreCase(k[i])) { output = null; } else { output = v[i]; return output; } } return output; } public void clear() { for (int i = 0; i < k.length; i++) { k[i] = null; v[i] = null; } count = 0; } public void display() { for (int i = 0; i < k.length; i++) { System.out.println(k[i] + " " + v[i]); } } }

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  • saved image in the picturebox shows no preview

    - by Nivas
    Hi iam new to C# and have loaded the image in the picture box using menustrip and have displayed some text using picturebox_Paint and label. now i tried to save the image (with image and text) using save event from the menustrip. in the saved location the file shows as no preview avaliable and when i tried to open the file it shows out of memory. can any one say where iam going worng.... my coades private void openToolStripMenuItem_Click(object sender, EventArgs e) { string file = ""; OpenFD.FileName = ""; OpenFD.Title = "open image"; OpenFD.InitialDirectory = "C"; OpenFD.Filter = "JPEG|.jpg|Bmp|.bmp|All Files|..*"; if (OpenFD.ShowDialog() == DialogResult.OK) { file = OpenFD.FileName; pictureBox1.Image = Image.FromFile(file); sz = pictureBox1.Size; a=sz.Width; b= sz.Height; } private void pictureBox1_MouseMove(object sender, MouseEventArgs e) { switch (e.Button) { case MouseButtons.Left: { rect = new Rectangle(rect.Left, rect.Top, e.X - rect.Left, e.Y - rect.Top); this.Invalidate(); y = flag.e; Application.DoEvents(); break; } } } private void pictureBox1_MouseDown(object sender, MouseEventArgs e) { rect = new Rectangle(e.X, e.Y, 0, 0); this.Invalidate(); } private void pictureBox1_Paint(object sender, PaintEventArgs e) { using (Pen pen = new Pen(Color.Red, 2)) e.Graphics.DrawRectangle(pen, rect); //e.Graphics.DrawString(label1.Text, label1.Font, new // SolidBrush(label1.ForeColor), label1.Left - pictureBox1.Left, label1.Top - pictureBox1.Top); if (label1.TextAlign == ContentAlignment.TopLeft) { e.Graphics.DrawString(label1.Text, label1.Font, new SolidBrush(label1.ForeColor), label1.Bounds); } else if (label1.TextAlign == ContentAlignment.TopCenter) { SizeF size = e.Graphics.MeasureString(label1.Text, label1.Font); float left = ((float)this.Width + label1.Left) / 2 - size.Width / 2; RectangleF rect1 = new RectangleF(left, (float)label1.Top, size.Width, label1.Height); e.Graphics.DrawString(label1.Text, label1.Font, new SolidBrush(label1.ForeColor), rect1); } else { SizeF size = e.Graphics.MeasureString(label1.Text, label1.Font); float left = (float)label1.Width - size.Width + label1.Left; RectangleF rect1 = new RectangleF(left, (float)label1.Top, size.Width, label1.Height); e.Graphics.DrawString(label1.Text, label1.Font, new SolidBrush(label1.ForeColor), rect1); } label1.Top = rect.Top; label1.Left = rect.Left; label1.Width = rect.Width; label1.Height = rect.Height; } private void saveToolStripMenuItem_Click(object sender, EventArgs e) { SaveFileDialog SaveFD1 = new SaveFileDialog(); //string Sd_file = ""; SaveFD1.FileName = ""; SaveFD1.InitialDirectory = "C"; SaveFD1.Title = "save file Name"; SaveFD1.Filter= "JPG|.jpg|Bmp|.bmp"; if (SaveFD1.ShowDialog() != DialogResult.Cancel) { System.IO.Stream filename = (System.IO.FileStream)SaveFD1.OpenFile(); if (SaveFD1.Filter == "JPG") pictureBox1.Image.Save(SaveFD1.FileName); //pictureBox1.Image.Save (filename, System.Drawing.Imaging.ImageFormat.Jpeg); else if (SaveFD1.Filter == "Bmp") { //pictureBox1.Image.Save(filename, System.Drawing.Imaging.ImageFormat.Bmp); } filename.Close(); } }

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  • SQL Server 2005 Express: upgrading to SP3 in mixed-mode installations

    - by Jeroen Pluimers
    I'm having trouble upgrading SQL Server 2005 Express SP1 to SP3. The SP1 install uses mixed mode authentication (so there is an sa password). This is the message I get: TITLE: Microsoft SQL Server Setup ------------------------------ None of the selected features can be installed or upgraded. Setup cannot proceed since no effective change is being made to the machine. To continue, click Back and then select features to install. To exit SQL Server Setup, click Cancel. For help, click: http://go.microsoft.com/fwlink?LinkID=20476&ProdName=Microsoft+SQL+Server&ProdVer=9.00.4035.00&EvtSrc=setup.rll&EvtID=SQLSetup90&EvtType=28108 ------------------------------ BUTTONS: OK ------------------------------ The link then tells me To continue you must provide a strong sa password. I tried some searching, and found something about BPAClient.dll, but this batch-file does not fix it: mkdir "%ProgramFiles%\Microsoft SQL Server\90\Setup Bootstrap\BPA\BPAClient" copy "%ProgramFiles%\Microsoft SQL Server\90\Setup Bootstrap\BPA\bin\BPAClient.dll" "%ProgramFiles%\Microsoft SQL Server\90\Setup Bootstrap\BPA\BPAClient\" So I think the clue is the strong in the link above. Am I on the right track? Where do I find more information on the strongness of an sa password? --jeroen (who will adjust the question when he has dug further)

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  • Cannot log into Windows XP Embedded after changing computer name

    - by bignis
    Hi everyone, I purchased a tablet pc running Windows XP Embedded. The tablet was used in a medical clinic on a domain. For illustrative purposes, say the computer name was "COMPLEXCOMPUTERNAME". There was an administrator account, so I changed the password on account "COMPLEXCOMPUTERNAME\Administrator" to a blank password. I logged out and logged in successfully with the blank administrator password when the log-in dialog said "Log in to COMPLEXCOMPUTERNAME (this computer)". Next I renamed the computer from COMPLEXCOMPUTERNAME to SIMPLECOMPUTERNAME, which required a reboot. I did so, and I can't log in anymore. The log in screen still just says "Log in to COMPLEXCOMPUTERNAME (this computer)", but the account "COMPLEXCOMPUTERNAME\Administrator" no longer works. I suspect that this is because the computer has been renamed to SIMPLECOMPUTERNAME and it can no longer find the account. The "Log in to" dropdown can't be typed in, so I can't change the computer name Windows is trying to log into. I fear that I'm stuck. Is there a way I can get Windows to log into the computer name that I chose? Thanks! -Mike

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  • Get Safari to use different autocompletion on different URLs on same hostname

    - by Luke404
    I have a webserver publishing different services over the same SSL VirtualHost, the two most commonly used being PhpMyAdmin and Cacti. These (and others) use 'cookie' style authentication, asking user and password in an HTML form (thus not using HTTP Authentication). Being on the same hostname, the Safari browser didn't manage too well stored passwords: if I login to one app with user foo, and then go to app two it would propose me user foo and its password in the login form. Changing just the username to bar used to be sufficient to let Safari autocomplete the correct password in its form field. Annoying, but I could live with it - usernames are short and easy to remember when compared to the passwords we use. After the update to safari5 this seems to be no longer true: if I store in safari (actually user keychain on OSX) credentials for https://www.foobarbaz.com/app1 AND credentials for https://www.foobarbaz.com/app2 there seem to be no way for it to autocomplete both based on the url. Even editing the keychain to add the path (it will store only the hostname by default) does not help. Is there anything I can do to let it work the way I want while still keeping everything on one hostname? Modifying anything server side is of course possible, but I can't switch apps to HTTP Auth (and not every one will support it anyway) to use different 'realms'.

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  • MySQL stopped asking for passwords

    - by BlaM
    I'm currently experiencing a weird problem with one of my MySQL database servers: It stopped asking for passwords when I try to access the database from local with the mysql command line tool. I need a valid admin username. I also still need a password for remote access (i.e. from another IP). And I need a password when I - for example - access the database from a PHP script. But when I try to access the database from local host/commandline it will let me straight in to the data with my administrative users. They (admin users) have passwords set - and as I mentioned - I still need to specify those when I try to access the data via PHP. Changing the password didn't help. Non-Administrative users need to specify their passwort, but that doesn't really help if they can get anywhere with "mysql -u root" (or another admin user account name). (System Debian Linux Lenny, MySQL 5.0.51a) Any ideas? Anything that explains this behaviour? I don't understand how this can happen.

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  • Saving in mongoDb with Mongoose, unexpected elements saved

    - by guiomie
    When I write in my mongoDB with mongoose the operation is treated with success, my document is saved, but there is also all kind of weird other sutff written down. It seems to be mongoose code. What could cause this? I add stuff in a specific array with: resultReference.ref[arrayLocation].allEvents.push(theEvent); {id: 11, allEvents: [] } is the structure of a ref element, and I push theEvent in the allEvents array. I then resultReference.save() I use express, mongoose and mongoHQ for database. I tried on a local mongo server, and this annoyance is still there. I've print in my console the document to write before save() and non of this weird code is there. { id 11 allEvents [ 0 { _events { maxListeners 0 } _doc { _id {"$oid": "4eb87834f54944e263000003"} title "Test" allDay false start 2011-11-10 13:00:00 UTC end 2011-11-10 15:00:00 UTC url "/test/4eb87834f54944e263000002" color "#99CCFF" ref "4eb87834f54944e263000002" } _activePaths { paths { title "modify" allDay "modify" start "modify" end "modify" url "modify" color "modify" ref "modify" } states { init { } modify { title true allDay true start true end true url true color true ref true } require { } } stateNames [ 0 "require" 1 "modify" 2 "init" ] } _saveError null _validationError null isNew true _pres { save [ 0 function (next) { // we keep the error semaphore to make sure we don't // call `save` unnecessarily (we only need 1 error) var subdocs = 0 , error = false , self = this; var arrays = this._activePaths .map('init', 'modify', function (i) { return self.getValue(i); }) .filter(function (val) { return (val && val instanceof DocumentArray && val.length); }); if (!arrays.length) return next(); arrays.forEach(function (array) { subdocs += array.length; array.forEach(function (value) { if (!error) value.save(function (err) { if (!error) { if (err) { error = true; next(err); } else --subdocs || next(); } }); }); }); } 1 "function checkForExistingErrors(next) { if (self._saveError){ next(self._saveError); self._saveError = null; } else { next(); } }" 2 "function validation(next) { return self.validate.call(self, next); }" ] } _posts { save [ ] } save function () { var self = this , hookArgs // arguments eventually passed to the hook - are mutable , lastArg = arguments[arguments.length-1] , pres = this._pres[name] , posts = this._posts[name] , _total = pres.length , _current = -1 , _asyncsLeft = proto[name].numAsyncPres , _next = function () { if (arguments[0] instanceof Error) { return handleError(arguments[0]); } var _args = Array.prototype.slice.call(arguments) , currPre , preArgs; if (_args.length && !(arguments[0] === null && typeof lastArg === 'function')) hookArgs = _args; if (++_current < _total) { currPre = pres[_current] if (currPre.isAsync && currPre.length < 2) throw new Error("Your pre must have next and done arguments -- e.g., function (next, done, ...)"); if (currPre.length < 1) throw new Error("Your pre must have a next argument -- e.g., function (next, ...)"); preArgs = (currPre.isAsync ? [once(_next), once(_asyncsDone)] : [once(_next)]).concat(hookArgs); return currPre.apply(self, preArgs); } else if (!proto[name].numAsyncPres) { return _done.apply(self, hookArgs); } } , _done = function () { var args_ = Array.prototype.slice.call(arguments) , ret, total_, current_, next_, done_, postArgs; if (_current === _total) { ret = fn.apply(self, args_); total_ = posts.length; current_ = -1; next_ = function () { if (arguments[0] instanceof Error) { return handleError(arguments[0]); } var args_ = Array.prototype.slice.call(arguments, 1) , currPost , postArgs; if (args_.length) hookArgs = args_; if (++current_ < total_) { currPost = posts[current_] if (currPost.length < 1) throw new Error("Your post must have a next argument -- e.g., function (next, ...)"); postArgs = [once(next_)].concat(hookArgs); return currPost.apply(self, postArgs); } }; if (total_) return next_(); return ret; } }; if (_asyncsLeft) { function _asyncsDone (err) { if (err && err instanceof Error) { return handleError(err); } --_asyncsLeft || _done.apply(self, hookArgs); } } function handleError (err) { if ('function' == typeof lastArg) return lastArg(err); if (errorCb) return errorCb.call(self, err); throw err; } return _next.apply(this, arguments); } errors null } ] } ]

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  • Windows 7: Creating a password-protected task (NOT a programming question)

    - by Matthias
    Hello, I would like to configure a task like "child control software", so it would hibernate the pc at certain times. Is it possible to prevent modification (here: pausing) of a task through requiring the entering of the admin password to modify, EVEN THOUGH the currently-logged-in (and only) user is the admin account itself? (Do you know of any child control software that does NOT require an additional account yet is able to hibernate the system at certain times?) Thanks a lot! Matthias

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  • Wireless Password

    - by Campo
    I have a security policy question: I want to know how other admins handle the WIFI password in the office. Does everyone know it? Do you enter it in for the user or guest every time and keep it a secret. I am in camp 2. Just want to know what others do and their reasoning behind it.

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  • .htaccess password not working for all files

    - by hapalibashi
    My .htaccess on Rackspace looks like this: AuthType Basic AuthName "Restricted Area" AuthUserFile /path-to-htdocs/.htpasswd Require valid-user Now I would expect this to password protect the whole directory, however, it only protects files with .php extension! What is wrong with it? Is it something in the default http.conf that I cannot override? The path used to .htpasswd is correct as it accepts the user/passwd in the case of .php

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  • Accessing a Windows 7 print share without a password

    - by user101141
    In our network we have a Windows 7 print server. Users connect to this machine by typing \\server_name on their own workstations. The print server and the users` computers are members of Active Directory. In AD, only computers have accounts, users are using local accounts. Is it possible to configure Windows 7 so that it doesn't ask for login and password when a user tries to access it from computer which is member of domain?

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  • Supply username and password to mstsc

    - by Will
    In previous versions of the remote desktop client there were methods of passing in the password through various methods. Has anybody found a good method using the latest remote desktop client? I'm aware of LaunchRDP but that doesn't meet our needs. Perhaps somebody knows the algorithm so I can dynamically assemble RDP connection files?

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