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  • How does a web browser save passwords?

    - by marcus
    How do current web browsers (or mobile mail clients and any software in general) save user passwords? All answers about storing passwords say we should store only hashes, not the password themselves. But I'm having a hard time searching the web trying to find the best techniques to store passwords when we know we will need them in plain text later on — without storing them in plain text, without using a weak encryption (known key) and without asking the user for a master password. Any ideas?

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  • How to host a site in another site - with little or no coding

    - by tunmise fasipe
    SUMMARY: All of these happens on Site A User visits site A User enter username and password User click on Login Button User authenticated on Site B behind the scene User is shown a page on Site A that contains his/her profile from Site B as layout/styled from Site B User can click links in the Profile page that links to other area in Site B Meaning: Session has to be maintained somehow I have web application where I store users' password and username. If you logon to this site, you can login with the password and username to have access to your profile. There is another option that requires you to login to my site from your site and have your profile displayed within your site. This is because you might already have a site that your clients know you with. This link is close to what I want to do: http://aspmessageboard.com/showthread.php?t=235069 A user on Site A login to Site B and have the information on site B showing in site A. He should not know whether Site B exists. It should be as if everything is happening in Site A This latter part is what I don't know to implement. I have these ideas: Have a fixed IFrame within your site to contain my site: but I am concerned about size/layout since different clients have different layout/size for their content section. I am thinking of how to maintain session too A webservice: I don't know how feasible this is since the Password and ID are on my server. You may have to send them back and forth. It means client would have to code with my API. But I am not just returning data, I have to show them a page that contains the profile details OpenID, Single-SignOn: Just guessing - but the authentication and data resides on my server. there is nothing to access on your side in this case Examples: like login into facebook within my site and still be able to do post updates, receive notifications Facebook implement some of these with IFrame e.g. the Like button *NOTE: * I have tested the IFrame option. It worked but I still have to remove my site specific content like my page Banner, Side Navigation etc. I was able to login normally as if I was actually on the site. This show my GUI but - style sheet was missing - content not styled with CSS - Any relative url won't work. It would look for that resource relative to the current server. Unless I change links to absolute - Clicking on the LogIn button produces this error: The state information is invalid for this page and might be corrupted. UPDATE: I was reading about REST webservice few days ago and I got this idea: What about the idea of returning an XML from a webservice [REST or SOAP] and providing an XSLT (that I can provide) to display it. Thus they won't have to do much coding?

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  • Why does key-based ssh fail even after setting up the authorized_keys file on the remote host?

    - by Brad Grissom
    These details don't matter but I am on a Ubuntu 12.04 machine and I want to ssh into my RaspberryPi without a password. I followed the standard procedure for setting up ssh without a password: local $ ssh-keygen -t rsa (hit enter for defaults to the questions) local $ scp ~/.ssh/id_rsa.pub matt@raspihost:~/.ssh/authorized_keys I logged onto the raspihost and checked all my permissions on ~/.ssh/ and on the authorized_keys file itself. It was still not working!

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  • Broken Sudo - sudo: parse error in /etc/sudoers near line 23

    - by Robert Fáber
    I am getting this error: sudo: parse error in /etc/sudoers near line 23 sudo: no valid sudoers sources found, quitting sudo: unable to initialize policy plugin I was trying to disable password authentication so I don't have to type password every time I want to install something, but I probably changed it in a not very good way. I am a newbie to Ubuntu, I got sick of Windows :) So far I've found some people suggesting booting in single user mode, but I'm afraid of messing things up more.

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  • Login on VMWare XP guest machine keeps locking user AD account

    - by mark
    Environment: Windows 2003 AD Host: Windows 7 Pro VMWare Guest: Windows XP I use a normal user but I also have AD Admin rights (via another user). I use my AD account to login on the host as well on the aforementioned guest system. They don't share their profiles, so no problems here. I had reason to change my user (AD) password. When I did this, the guest was suspended but my user was logged in. A few days after my password change I resumed the guest. I was able to work but couldn't access networked mapped drives. I logged out and tried to log in again. At this point I realized that I initially was logged in with a user from a point before I changed my password. I logged in again with the new password, but then things went bad. I was able to successfully log in to my XP guest, however once that was completed, my AD user account got locked. This now also affected my user on the host. I was able to unlock the account, but there is still this problem: I log in via my new password into the guest and then my AD account gets locked. I'm successfully logged into the guest, but I can't access network shares from the AD server. If I don't unlock my account on the AD server, I will get further problems with my AD user. I tried multiple things, none worked: removed XP guest from AD, deleted all users, even my XP AD user profile on the guest, added machine to the AD, logged in - log in successful, account locked I resumed an older state of my guest (sometimes from the last year even) but the problem still persists. I tried this with disabled networking when the old machine state is resumed and so on, but no luck. It seems to me, although only my account is locked, this is somehow connected to the guest machine itself. I really want to avoid re-installation. This guest image was my old workstation which I virtualized once I moved to W7 pro and thus is still very valueable or me. I can work locally on the guest once logged in, but I can't access any network shares which is a problem. thanks

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  • wubi would not uninstalling

    - by swaleh hussein
    I installed wubi 12.04 with a username and password. It installed successfully but when I tried logging in, it said my password was incorrect. I then decided to uninstall but it kept giving me this error message: An error occurred: Permission denied For more information, please see the log file: c:\users\optimus\appdata\local\temp\wubi-12.04-rev266.log how can I then uninstall wubi from my laptop? Kindly assist.

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  • stuck on "preparing..." when restoring from deja-dup backup

    - by Dan
    I'm trying to restore my deja-dup backup from a certain date. However during restore after selecting the date to restore from i get a "restoring... "preparing"..." window that just seems stuck there doing nothing forever (past 1/2 hour). There was a point when i was prompted for the "encryption password" but i don't remember it, so i just entered one. I never got any error if the password i entered was not accepted.

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  • How to handle encryption key conflicts when synchronizing data?

    - by Rafael
    Assume that there is data that gets synchronized between several devices. The data is protected with a symmetric encryption algorithm and a key. The key is stored on each device and encrypted with a password. When a user changes the password only the key gets re-encrypted. Under normal circumstances, when there is a good network connection to other peers, the current key gets synchronized and all data on the new device gets encrypted with the same key. But how to handle situations where a new device doesn’t have a network connection and e.g. creates its own new, but incompatible key? How to keep the usability as high as possible under such circumstances? The application could detect that there is no network and hence refuse to start. That’s very bad usability in my opinion, because the application isn’t functional at all in this case. I don’t consider this a solution. The application could ignore the missing network connection and create a new key. But what to do when the application gains a network connection? There will be several incompatible keys and some parts of the underlying data could only be encrypted with one key and other parts with another key. The situation would get worse if there would be more keys than just two and the application would’ve to ask every time for a password when another object that should get decrypted with another key would be needed. It is very messy and time consuming to try to re-encrypt all data that is encrypted with another key with a main key. What should be the main key at all in this case? The oldest key? The key with the most encrypted objects? What if the key got synchronized but not all objects that got encrypted with this particular key? How should the user know for which particular password the application asks and why it takes probably very long to re-encrypt the data? It’s very hard to describe encryption “issues” to users. So far I didn’t find an acceptable solution, nor some kind of generic strategy. Do you have some hints about a concrete strategy or some books / papers that describe synchronization of symmetrically encrypted data with keys that could cause conflicts?

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  • EPM 11.1.2.1 - Smartview client and HFM office provider

    - by user809526
    If your connection to the smartview provider is very slow, because the login part takes a long time (user directory slowness, ...), consider adding on the desktop side a Windows parameter: HKEY_CURRENT_USER\Software\Microsoft\Windows\CurrentVersion\InternetSettings\ ReceiveTimeout 300000 to avoid being prompted over and over again for username/password This is an addition to the support doc id: "Smart View 11.1.2.1 Keeps Prompting For Username And Password For Financial Management Provider [ID 1353294.1]"

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  • Script to check a shared Exchange calendar and then email detail

    - by SJN
    We're running Server and Exchange 2003 here. There's a shared calendar which HR keep up-to-date detailing staff who are on leave. I'm looking for a VB Script (or alternate) which will extract the "appointment" titles of each item for the current day and then email the detail to a mail group, in doing so notifying the group with regard to which staff are on leave for the day. The resulting email body should be: Staff on leave today: Mike Davis James Stead @Paul Robichaux - ADO is the way I went for this in the end, here are the key component for those interested: Dim Rs, Conn, Url, Username, Password, Recipient Set Rs = CreateObject("ADODB.Recordset") Set Conn = CreateObject("ADODB.Connection") 'Configurable variables Username = "Domain\username" ' AD domain\username Password = "password" ' AD password Url = "file://./backofficestorage/domain.com/MBX/username/Calendar" 'path to user's mailbox and folder Recipient = "[email protected]" Conn.Provider = "ExOLEDB.DataSource" Conn.Open Url, Username, Password Set Rs.ActiveConnection = Conn Rs.Source = "SELECT ""DAV:href"", " & _ " ""urn:schemas:httpmail:subject"", " & _ " ""urn:schemas:calendar:dtstart"", " & _ " ""urn:schemas:calendar:dtend"" " & _ "FROM scope('shallow traversal of """"') " Rs.Open Rs.MoveFirst strOutput = "" Do Until Rs.EOF If DateDiff("s", Rs.Fields("urn:schemas:calendar:dtstart"), date) >= 0 And DateDiff("s", Rs.Fields("urn:schemas:calendar:dtend"), date) < 0 Then strOutput = strOutput & "<p><font size='2' color='black' face='verdana'><b>" & Rs.Fields("urn:schemas:httpmail:subject") & "</b><br />" & vbCrLf strOutput = strOutput & "<b>From: </b>" & Rs.Fields("urn:schemas:calendar:dtstart") & vbCrLf strOutput = strOutput & "&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<b>To: </b>" & Rs.Fields("urn:schemas:calendar:dtend") & "<br /><br />" & vbCrLf End If Rs.MoveNext Loop Conn.Close Set Conn = Nothing Set Rec = Nothing After that, you can do what you like with srtOutput, I happened to use CDO to send an email: Set objMessage = CreateObject("CDO.Message") objMessage.Subject = "Subject" objMessage.From = "[email protected]" objMessage.To = Recipient objMessage.HTMLBody = strOutput objMessage.Send S

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  • WordPress - Emergency Access Without Admin Accounts

    In some cases, when you need to do something in a WordPress website, but all you have is only access to WordPress database and FTP, and you cannot get the admin password from the database because it's decrypted. All changes you have to make via some low level MySQL queries, it's hard and easy mistaken. Joost de Valk has written a script for emergency access to WordPress dashboard by changing admin password or creating new user.

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  • How to increment a value appended to another value in a shell script

    - by Sitati
    I have a shell script that executes a backup program and saves the output in a folder. I would like to update the name of the output folder every time the shell script run. In the end I want to have many files with different names like this: innobackupex --user=root --password=@g@1n --database="open_cart" /var/backup/backup_1 --no-timestamp And after running the shell script again: innobackupex --user=root --password=@g@1n --database="open_cart" /var/backup/backup_2 --no-timestamp

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  • Webdav stops working after a few seconds

    - by user214885
    With Ubuntu 13.10, I connect to two different OwnCloud installations and can browse for only a few seconds before it freezes the connection and stops working. When I try to reinstate the connection it fails to even ask for the password (Ubuntu was told to forget the password). I did check the webdav connection through Firefox on two computers and ES File Explorer on android. I know that this isn't a webdav problem and don't know what is happening in Ubuntu to stop being able to read the connection.

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  • Trouble creating FTP in Server 2008

    - by Saariko
    I have been trying to create an FTP server on my new Server 2008. I have been following both (very detailed and highly published here guides) For setting up using IIS Manager http://learn.iis.net/page.aspx/321/configure-ftp-with-iis-7-manager-authentication/ and For anonymous FTP http://www.trainsignaltraining.com/windows-server-2008-ftp-iis7 I am able to log as an anonymous user. My need is to use a named user, so I need to use the IIS Manager. I get error 530 when trying to log as a user. Connected to 127.0.0.1. 220 Microsoft FTP Service User (127.0.0.1:(none)): ftpmanager 331 Password required for ftpmanager. Password: 530-User cannot log in. Win32 error: Logon failure: unknown user name or bad password. Error details: Filename: Error: 530 End Login failed. ftp> I can not learn from this message anything. My password is set to: 1234 (so I don't think I make a mistake here - testing purposes only ofc) Thank you. Note - I went over other posts on SE that I read, and couldn't get the result: IIS7 Windows Server 2008 FTP -> Response: 530 User cannot log in. FTP Error 530, User cannot log in, home directory inaccessible. Having trouble setting up FTP server on Windows Server 2008 EDIT I think I found some errors with the physical path. Going to Basic settings, and Test Connection on the physical path, gave me the following error: The server is configured to use pass-through authentication with a built-in account to access the specified physical path. However, IIS Manager cannot verify whether the built-in account has access. Make sure that the application pool identity has Read access to the physical path. If this server is joined to a domain, and the application pool identity is NetworkService or LocalSystem, verify that \$ has Read access to the physical path. Then test these settings again. I am not sure which/whom should get access to the Root folder !? I want to point out, I managed to login with a domain user (change authorization and authentication methods) but this is NOT the requested solution. I checked to make sure that the FTP, folders, access is working properly. I am bit lost here. ==== More tries: I have enabled another Allow rule for ALL Users. I still get the same error. It seems that it doesn't matter if i use a correct or wrong password, I still get Error 530.

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  • Restrict access to apache2 web root but allow it to subfolders

    - by razor7
    I need to restrict access by password to my web root apache test server (ie http://localhost) but allow access to subfolders (ie: http://localhost/testsite) I did create the .htpasswd and .htaccess, and put the .htaccess to web root (http://localhost) so when trying to access web root, it asks for user and pass, but so does in subfolders (ie: trying to access http://localhost/testite) I want to be asked for password on web root, but not on subfolders. Is that possible?

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  • Double lock screen Ubuntu 14.04

    - by Adam
    So I've got a brandynew System 76 laptop running Ubuntu 14.04, and if I close the lid, putting it to sleep, and reopen it, I am presented with the very nice new Unity lock screen. However, when I enter my password, and it succeeds, it then presents me with a second lock screen. Where I have to enter my password again, before finally being let into the desktop. Hash anyone else seen this sort of behavior?

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  • If my URL's are static, but then parsed by Javascript, does that make it crawlable?

    - by Talasan Nicholson
    Lets say I have a link: <a href="/about/">About Us</a> But in Javascript [or jQuery] catches it and then adds the hash based off of the href attribute: $('a').click(function(e) { e.preventDefault(); // Extremely oversimplified.. window.location.hash = $(this).attr('href'); }); And then we use a hashchange event to do the general 'magic' of Ajax requests. This allows for the actual href to be seen by crawlers, but gives client-side users with JS enabled an ajax-based website. Does this 'help' the general SEO issues that come along with hashtags? I know hashbangs are 'ok', but afaik they aren't reliable?

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  • How to Create Network File Shares with No Passwords in Windows 8

    - by Taylor Gibb
    We have all had to connect to a network share at some point only to have the authentication dialog pop up. There are many ways around it, for example mapping a network drive, but if you have a lot of users connecting to copy some files you may want to disable the password dialog instead of distributing your password. How To Delete, Move, or Rename Locked Files in Windows HTG Explains: Why Screen Savers Are No Longer Necessary 6 Ways Windows 8 Is More Secure Than Windows 7

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  • Make sudoers work with only certain parameter?

    - by Evan
    I'm trying to make my sudoers file allow a user to adjust the backlight without having to enter in the password. This is what I have: # User alias specification Cmnd_Alias ADJBL = /usr/bin/su -c "echo 150 >/sys/class/backlight/intel_backlight/brightness" # For our user.. ouruser HOME=(root) NOPASSWD:ADJBL .. but it doesn't seem to be working, I still get prompted for the password when I try and run that command with sudo. Apparently there is something I'm missing here, any ideas?

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  • Sending login form to an open-mesh acess point for authentication with my own RADIUS server

    - by PachinSV
    I have my RADIUS server up and running and a custom external captive portal. But I'm not sure: what information should I send to the Open-Mesh AP with my login form (it is necessary to encrypt the password?, because if I don't use a secret word to encrypt in my network configuration the RADIUS server complaints about it and in the log shows me some strange characters in the password) I don't know what to do with the "challenge" and "md" parameters in my login splash page. Thank you very much for your help.

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  • Why isn't passwordless ssh working?

    - by Nelson
    I have two Ubuntu Server machines sitting at home. One is 192.168.1.15 (we'll call this 15), and the other is 192.168.1.25 (we'll call this 25). For some reason, when I want to setup passwordless login from 15 to 25, it works like a champ. When I repeat the steps on 25, so that 25 can login without a password on 15, no dice. I have checked both sshd_config files. Both have: RSAAuthentication yes PubkeyAuthentication yes I have checked permissions on both servers: drwx------ 2 bion2 bion2 4096 Dec 4 12:51 .ssh -rw------- 1 bion2 bion2 398 Dec 4 13:10 authorized_keys On 25. drwx------ 2 shimdidly shimdidly 4096 Dec 4 19:15 .ssh -rw------- 1 shimdidly shimdidly 1018 Dec 4 18:54 authorized_keys On 15. I just don't understand when things would work one way and not the other. I know it's probably something obvious just staring me in the face, but for the life of me, I can't figure out what is going on. Here's what ssh -v says when I try to ssh from 25 to 15: ssh -v -p 51337 192.168.1.15 OpenSSH_5.9p1 Debian-5ubuntu1, OpenSSL 1.0.1 14 Mar 2012 debug1: Reading configuration data /etc/ssh/ssh_config debug1: /etc/ssh/ssh_config line 19: Applying options for * debug1: Connecting to 192.168.1.15 [192.168.1.15] port 51337. debug1: Connection established. debug1: identity file /home/shimdidly/.ssh/id_rsa type 1 debug1: Checking blacklist file /usr/share/ssh/blacklist.RSA-2048 debug1: Checking blacklist file /etc/ssh/blacklist.RSA-2048 debug1: identity file /home/shimdidly/.ssh/id_rsa-cert type -1 debug1: identity file /home/shimdidly/.ssh/id_dsa type 2 debug1: Checking blacklist file /usr/share/ssh/blacklist.DSA-1024 debug1: Checking blacklist file /etc/ssh/blacklist.DSA-1024 debug1: identity file /home/shimdidly/.ssh/id_dsa-cert type -1 debug1: identity file /home/shimdidly/.ssh/id_ecdsa type -1 debug1: identity file /home/shimdidly/.ssh/id_ecdsa-cert type -1 debug1: Remote protocol version 2.0, remote software version OpenSSH_5.9p1 Debian-5ubuntu1 debug1: match: OpenSSH_5.9p1 Debian-5ubuntu1 pat OpenSSH* debug1: Enabling compatibility mode for protocol 2.0 debug1: Local version string SSH-2.0-OpenSSH_5.9p1 Debian-5ubuntu1 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: sending SSH2_MSG_KEX_ECDH_INIT debug1: expecting SSH2_MSG_KEX_ECDH_REPLY debug1: Server host key: ECDSA 54:5c:60:80:74:ab:ab:31:36:a1:d3:9b:db:31:2a:ee debug1: Host '[192.168.1.15]:51337' is known and matches the ECDSA host key. debug1: Found key in /home/shimdidly/.ssh/known_hosts:2 debug1: ssh_ecdsa_verify: signature correct debug1: SSH2_MSG_NEWKEYS sent debug1: expecting SSH2_MSG_NEWKEYS debug1: SSH2_MSG_NEWKEYS received debug1: Roaming not allowed by server debug1: SSH2_MSG_SERVICE_REQUEST sent debug1: SSH2_MSG_SERVICE_ACCEPT received debug1: Authentications that can continue: publickey,password debug1: Next authentication method: publickey debug1: Offering RSA public key: /home/shimdidly/.ssh/id_rsa debug1: Authentications that can continue: publickey,password debug1: Offering DSA public key: /home/shimdidly/.ssh/id_dsa debug1: Authentications that can continue: publickey,password debug1: Trying private key: /home/shimdidly/.ssh/id_ecdsa debug1: Next authentication method: password

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  • Even distribution through a chain of resources

    - by ClosetGeek
    I'm working on an algorithm which routes tasks through a chain of distributed resources based on a hash (or random number). For example, say you have 10 gateways into a service which distribute tasks to 1000 handlers through 100 queues. 10,000 connected clients are expected to be connected to gateways at any given time (numbers are very general to keep it simple). Thats 10,000 clients 10 gateways (producers) 100 queues 1000 workers/handlers (consumers) The flow of each task is client-gateway-queue-worker Each client will have it's own hash/number which is used to route each task from the client to the same worker each time, with each task going through the same gateway and queue each time. Yet the algorithm handles distribution evenly, meaning each gateway, queue, and worker will have an even workload. My question is what exactly would this be called? Does such a thing already exist? This started off as a DHT, but I realized that DHTs can't do exactly what I need, so I started from scratch.

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  • Why is the error, dd: /dev/rdisk1bs=1m: Operation not supported, popping up while trying to instal ubuntu on usb?

    - by Jesse S
    I am trying to install ubuntu onto my flash drive using the instructions from the website, http://www.ubuntu.com/download/help/create-a-usb-stick-on-mac-osx , and after step 8, the terminal asks for my password, which it accepts and then pops u this error message, dd: /dev/rdisk1bs=1m: Operation not supported. I have also tried making the last m in that statement capital and then the system does not ask me for my password but the error message still pops up. What is happening and why?

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  • Getting current time in milliseconds

    - by user90293423
    How to get the current time in milliseconds? I'm working on a hacking simulation game and when ever someone connects to another computer/NPC, a login screen popups with a button on the side called BruteForce. When BruteForce is clicked, what i want the program to do is, calculate how many seconds cracking the password is going to take based on the player's CPU speed but that's the easy part. The hard part is i want to enter a character in the password's box every X milliseconds based on a TimeToCrack divided by PasswordLength formula. But since i don't know how to find how many milliseconds have elapsed since the second has passed, the program waits until the CurrentTime is higher than the TimeBeforeTheLoopStarted + HowLongItTakesToTypeaCharacter which is always going to be a second. How would you handle my problems? I've commented the game breaking part. std::vector<QString> hardware = user.getHardware(); QString CPU = hardware[0]; unsigned short Speed = 0; if(CPU == "OMG SingleCore 1.8GHZ"){ Speed = 2; } const short passwordLength = password.length(); /* It's equal to 16 */ int Time = passwordLength / Speed; double TypeSpeed = Time / passwordLength; time_t t = time(0); struct tm * now = localtime(&t); unsigned short EndTime = (now->tm_sec + Time) % 60; unsigned short CurrentTime = 0; short i = passwordLength - 1; do{ t = time(0); now = localtime(&t); CurrentTime = now->tm_sec; do{ t = time(0); now = localtime(&t); }while(now->tm_sec < CurrentTime + TypeSpeed); /* Highly flawed */ /* Do this while your integer value is under this double value */ QString tempPass = password; tempPass.chop(i); ui->lineEdit_2->setText(tempPass); i--; }while(CurrentTime != EndTime);

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