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  • How can I cache a Subversion password on a server, without storing it in unencrypted form?

    - by Zilk
    My Subversion server only provides access via HTTPS; support for svn+ssh has been dropped because we wanted to avoid creating system users on that machine just for SVN access. Now I'm trying to provide a way for users to cache their passwords for a while, without leaving them stored on the filesystem in unencrypted form. This is no problem for Gnome or KDE users, because they can use gnome-keyring and kwallet, respectively. IIRC, TortoiseSVN has a similar caching mechanism, too. But what about users on a non-GUI system? Some context: in this case, we have a development/testing server where one project has been checked out into the Apache htdocs directory. Development for this project is almost complete, and only minor text/layout changes are performed directly on this server. Nevertheless, the changes should be checked into the repository. There's no kwallet and no gnome-keyring on this system, and the ssh-agent can't help because the repository is accessed via https instead of svn+ssh. As far as I know, that leaves them the choice of entering the password every time they talk to the SVN server, or storing it in an insecure way. Is there any way to get something like what gnome-keyring and kwallet provide in a non-GUI environment?

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  • How to create column of type password in gridview?

    - by Preeti
    Hi, I am creating an application in which user selects files and provides credentials to open that file. For that i have created three columns in a gridview. User enters password in password column. I want to display '*' in place of characters like we can create a textbox of password type. I have tried this code on 'GridView_CellClick' event : if (GridView.Columns[e.ColumnIndex].HeaderText == "Password") { txtPassword[e.RowIndex] = new TextBox(); txtPassword[e.RowIndex].Name = "txtPassword"+e.RowIndex; txtPassword[e.RowIndex].PasswordChar = '*'; txtPassword[e.RowIndex].Visible = true; txtPassword[e.RowIndex].TextChanged += new if (GridView.CurrentCell.Value == null) txtPassword[e.RowIndex].Text = ""; else txtPassword[e.RowIndex].Text = GridView.CurrentCell.Value.ToString(); txtPassword[e.RowIndex].Location = GridView.GetCellDisplayRectangle(e.ColumnIndex, e.RowIndex + 1, false).Location; txtPassword[e.RowIndex].Size = GridView.GetCellDisplayRectangle(e.ColumnIndex, e.RowIndex + 1, false).Size; txtPassword[e.RowIndex].Visible = true; txtPassword[e.RowIndex].Focus(); } But in above solution characters are displayed. How can i solve this problem???

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  • What's the difference between the input type "text" and "password" in an html form?

    - by Domingo
    Hi everybody, this question might seem stupid, but here's the situation: I'm trying to create an auto login page for my mail using jquery's post request, but it's not working, it works with all other pages except with webmail. So, trying to figure out what was wrong, I recreated the login form, here's the code: <form id="form1" name="form1" method="post" action="https://login.hostmonster.com/"> <label>User <input type="text" name="login" id="user" /> </label> <label>Pass <input name="password" type="password" id="pass" /> </label> <input name="doLogin" type="submit" id="doLogin" value="Login"> </form> The strange thing is when you change the input type of pass to text, the form doesn't work! I can't figure out why. Anyway, if you can tell me what's the real difference between the input type text and password (and not what it says everywhere on the net that the only difference is that when you type stars appear instead of characters) I would appreciate it. Also, do you think this is affecting my jquery's post? Here's the code for it: $j.post('https://login.hostmonster.com/', { login: '[email protected]', password: 'xxx' }, function(data, text){ if (text=='success') { alert('Success '+data); } else { alert('Failed'); } }); Thanks a lot! Regards, D

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  • How to reset the postgres super user password on mac os x

    - by Andrew Barinov
    I installed postgres on my mac running 10.6.8 and I would like to reset the password for the postgres user (I believe this is the super user password) and then restart it. All the directions I found do not work because I think my user name is not recognized by pg as having authority to change the password. (I am on the admin account of my mac) Here is what I tried: Larson-2:~ larson$ psql -U postgres Password for user postgres: psql (9.0.4, server 9.1.2) WARNING: psql version 9.0, server version 9.1. Some psql features might not work. Type "help" for help. postgres=# ALTER USER postgres with password 'mypassword' postgres-# \q and for restart I did: Larson-2:~ larson$ su postgres -c 'pg_ctl -D /opt/local/var/db/postgresql84/defaultdb/ restart > Which didn't work, as the password remained the same as it was before. Can someone provide directions for doing this and for making sure it's recognized by PG? Update I went ahead and edited the pg_hba.conf file located in /Library/PostgreSQL/9.1/data and set the settings as follows: # TYPE DATABASE USER ADDRESS METHOD # "local" is for Unix domain socket connections only local all all trust # IPv4 local connections: host all all 127.0.0.1/32 trust # IPv6 local connections: host all all ::1/128 trust However, like before, the password stayed the same after I changed it. I am not sure what further steps I can take from here.

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  • ssh - "Connection closed by xxx.xxx.xxx.xxx" - using password

    - by Michael B
    I attempted to create an new user account that I wish to use to log in using ssh. I did this (in CentOs): /usr/sbin/adduser -d /home/testaccount -s /bin/bash user passwd testaccount This is the error I receive when trying to log in via ssh: ~/.ssh$ ssh -v [email protected] OpenSSH_5.1p1 Debian-5ubuntu1, OpenSSL 0.9.8g 19 Oct 2007 debug1: Reading configuration data /etc/ssh/ssh_config debug1: Applying options for * debug1: Connecting to xxx.xxx.xxx [xxx.xxx.xxx.xxx] port 22. debug1: Connection established. debug1: identity file /home/user/.ssh/identity type -1 debug1: identity file /home/user/.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/user/.ssh/id_dsa type -1 debug1: Remote protocol version 2.0, remote software version OpenSSH_4.3 debug1: match: OpenSSH_4.3 pat OpenSSH_4* debug1: Enabling compatibility mode for protocol 2.0 debug1: Local version string SSH-2.0-OpenSSH_5.1p1 Debian-5ubuntu1 debug1: SSH2_MSG_KEXINIT sent debug1: SSH2_MSG_KEXINIT received debug1: kex: server->client aes128-cbc hmac-md5 none debug1: kex: client->server aes128-cbc hmac-md5 none debug1: SSH2_MSG_KEX_DH_GEX_REQUEST(1024<1024<8192) sent debug1: expecting SSH2_MSG_KEX_DH_GEX_GROUP 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 /home/user/.ssh/known_hosts:8 debug1: ssh_rsa_verify: signature correct debug1: SSH2_MSG_NEWKEYS sent debug1: expecting SSH2_MSG_NEWKEYS debug1: SSH2_MSG_NEWKEYS received debug1: SSH2_MSG_SERVICE_REQUEST sent debug1: SSH2_MSG_SERVICE_ACCEPT received debug1: Authentications that can continue: publickey,gssapi-with-mic,password debug1: Next authentication method: gssapi-with-mic debug1: Unspecified GSS failure. Minor code may provide more information No credentials cache found debug1: Unspecified GSS failure. Minor code may provide more information No credentials cache found debug1: Unspecified GSS failure. Minor code may provide more information debug1: Next authentication method: publickey debug1: Offering public key: /home/user/.ssh/id_rsa debug1: Authentications that can continue: publickey,gssapi-with-mic,password debug1: Trying private key: /home/user/.ssh/identity debug1: Trying private key: /home/user/.ssh/id_dsa debug1: Next authentication method: password testaccount@xxx's password: Connection closed by xxx.xxx.xxx.xxx The "connection closed" message appeared immediately after entering the password (if I enter the wrong password it waits and then prompts for another password) I am able to log in from the same computer using other accounts that had been setup previously. When logged into the remote machine I am able to do 'su testaccount' Thanks for your time.

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  • How to deecode your ODI encoded password in SDK

    - by tina.wang
    Someone asked me he want to use SDK to create ODI repository, but latest 11g API in SDK use plain password parameter. But he don't want to use plain text for security reason. So he want to transfer an encoded password, then decode it inside his code. He ask me whether there is a way.  After some investigating, I find com.sunopsis.dwg.DwgObject class has a static method snpsDecypher(String), it can satisfy his requirement. But seems this method is deprecated, I am trying to find the new replaced method. 

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  • Login screen restarts while entering password

    - by Shane L
    I am having a problem that only occurred after installing the fglrx proprietary driver through the additional drivers app. The exact same issue is described in this question, however it's closed. Must login twice before entering Unity; first login screen has graphical anomalies When I boot up my computer, once I get to lightdm login screen I will start typing my password, and upon the entering of two of my password characters "n2" the screen will start displaying some corruption in the very top edge of the screen, then after I hit the number 2, lightdm seems to crash and it will restart and everything is fine, I can login from that point. So the issue is, every time I reboot, I have to login once, allow lightdm to crash and restart, then login as normal. The issue did not occur prior to installing fglrx, and has happened through several reinstalls.

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  • SSH is not working .. Password promt is not coming

    - by Sumanth Lingappa
    I am not able to SSH into my ubuntu server since yesterday. I am not using any keyless or public key method.. Its simple SSH with username and password everytime.. However I can do a VNC session running on my ubuntu server.. But I am afraid that if the vnc session goes out, I wont be having any way to login to the server.. My ssh-vvv output is as below.. sumanth@sumanth:~$ ssh -vvv user@serverIP OpenSSH_6.6.1, OpenSSL 1.0.1f 6 Jan 2014 debug1: Reading configuration data /etc/ssh/ssh_config debug1: /etc/ssh/ssh_config line 19: Applying options for * debug2: ssh_connect: needpriv 0 debug1: Connecting to 172.16.2.156 [172.16.2.156] port 22. debug1: Connection established. debug1: identity file /home/sumanth/.ssh/id_rsa type -1 debug1: identity file /home/sumanth/.ssh/id_rsa-cert type -1 debug1: identity file /home/sumanth/.ssh/id_dsa type -1 debug1: identity file /home/sumanth/.ssh/id_dsa-cert type -1 debug1: identity file /home/sumanth/.ssh/id_ecdsa type -1 debug1: identity file /home/sumanth/.ssh/id_ecdsa-cert type -1 debug1: identity file /home/sumanth/.ssh/id_ed25519 type -1 debug1: identity file /home/sumanth/.ssh/id_ed25519-cert type -1 debug1: Enabling compatibility mode for protocol 2.0 debug1: Local version string SSH-2.0-OpenSSH_6.6.1p1 Ubuntu-2ubuntu2 debug1: Remote protocol version 2.0, remote software version OpenSSH_5.9p1 Debian-5ubuntu1 debug1: match: OpenSSH_5.9p1 Debian-5ubuntu1 pat OpenSSH_5* compat 0x0c000000 debug2: fd 3 setting O_NONBLOCK debug3: load_hostkeys: loading entries for host "172.16.2.156" from file "/home/sumanth/.ssh/known_hosts" debug3: load_hostkeys: found key type ECDSA in file /home/sumanth/.ssh/known_hosts:5 debug3: load_hostkeys: loaded 1 keys debug3: order_hostkeyalgs: prefer hostkeyalgs: [email protected],[email protected],[email protected],ecdsa-sha2-nistp256,ecdsa-sha2-nistp384,ecdsa-sha2-nistp521 debug1: SSH2_MSG_KEXINIT sent debug1: SSH2_MSG_KEXINIT received debug2: kex_parse_kexinit: [email protected],ecdh-sha2-nistp256,ecdh-sha2-nistp384,ecdh-sha2-nistp521,diffie-hellman-group-exchange-sha256,diffie-hellman-group-exchange-sha1,diffie-hellman-group14-sha1,diffie-hellman-group1-sha1 debug2: kex_parse_kexinit: [email protected],[email protected],[email protected],ecdsa-sha2-nistp256,ecdsa-sha2-nistp384,ecdsa-sha2-nistp521,[email protected],[email protected],[email protected],[email protected],[email protected],ssh-ed25519,ssh-rsa,ssh-dss debug2: kex_parse_kexinit: aes128-ctr,aes192-ctr,aes256-ctr,arcfour256,arcfour128,[email protected],[email protected],[email protected],aes128-cbc,3des-cbc,blowfish-cbc,cast128-cbc,aes192-cbc,aes256-cbc,arcfour,[email protected] debug2: kex_parse_kexinit: aes128-ctr,aes192-ctr,aes256-ctr,arcfour256,arcfour128,[email protected],[email protected],[email protected],aes128-cbc,3des-cbc,blowfish-cbc,cast128-cbc,aes192-cbc,aes256-cbc,arcfour,[email protected] debug2: kex_parse_kexinit: [email protected],[email protected],[email protected],[email protected],[email protected],[email protected],[email protected],[email protected],[email protected],hmac-md5,hmac-sha1,[email protected],[email protected],hmac-sha2-256,hmac-sha2-512,hmac-ripemd160,[email protected],hmac-sha1-96,hmac-md5-96 debug2: kex_parse_kexinit: [email protected],[email protected],[email protected],[email protected],[email protected],[email protected],[email protected],[email protected],[email protected],hmac-md5,hmac-sha1,[email protected],[email protected],hmac-sha2-256,hmac-sha2-512,hmac-ripemd160,[email protected],hmac-sha1-96,hmac-md5-96 debug2: kex_parse_kexinit: none,[email protected],zlib debug2: kex_parse_kexinit: none,[email protected],zlib debug2: kex_parse_kexinit: debug2: kex_parse_kexinit: debug2: kex_parse_kexinit: first_kex_follows 0 debug2: kex_parse_kexinit: reserved 0 debug2: kex_parse_kexinit: ecdh-sha2-nistp256,ecdh-sha2-nistp384,ecdh-sha2-nistp521,diffie-hellman-group-exchange-sha256,diffie-hellman-group-exchange-sha1,diffie-hellman-group14-sha1,diffie-hellman-group1-sha1 debug2: kex_parse_kexinit: ssh-rsa,ssh-dss,ecdsa-sha2-nistp256 debug2: kex_parse_kexinit: aes128-ctr,aes192-ctr,aes256-ctr,arcfour256,arcfour128,aes128-cbc,3des-cbc,blowfish-cbc,cast128-cbc,aes192-cbc,aes256-cbc,arcfour,[email protected] debug2: kex_parse_kexinit: aes128-ctr,aes192-ctr,aes256-ctr,arcfour256,arcfour128,aes128-cbc,3des-cbc,blowfish-cbc,cast128-cbc,aes192-cbc,aes256-cbc,arcfour,[email protected] debug2: kex_parse_kexinit: hmac-md5,hmac-sha1,[email protected],hmac-sha2-256,hmac-sha2-256-96,hmac-sha2-512,hmac-sha2-512-96,hmac-ripemd160,[email protected],hmac-sha1-96,hmac-md5-96 debug2: kex_parse_kexinit: hmac-md5,hmac-sha1,[email protected],hmac-sha2-256,hmac-sha2-256-96,hmac-sha2-512,hmac-sha2-512-96,hmac-ripemd160,[email protected],hmac-sha1-96,hmac-md5-96 debug2: kex_parse_kexinit: none,[email protected] debug2: kex_parse_kexinit: none,[email protected] debug2: kex_parse_kexinit: debug2: kex_parse_kexinit: debug2: kex_parse_kexinit: first_kex_follows 0 debug2: kex_parse_kexinit: reserved 0 debug2: mac_setup: setup hmac-md5 debug1: kex: server->client aes128-ctr hmac-md5 none debug2: mac_setup: setup hmac-md5 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 ea:4e:15:52:15:dd:6b:09:d4:36:cb:14:2d:c3:1b:7a debug3: load_hostkeys: loading entries for host "172.16.2.156" from file "/home/sumanth/.ssh/known_hosts" debug3: load_hostkeys: found key type ECDSA in file /home/sumanth/.ssh/known_hosts:5 debug3: load_hostkeys: loaded 1 keys debug1: Host '172.16.2.156' is known and matches the ECDSA host key. debug1: Found key in /home/sumanth/.ssh/known_hosts:5 debug1: ssh_ecdsa_verify: signature correct debug2: kex_derive_keys debug2: set_newkeys: mode 1 debug1: SSH2_MSG_NEWKEYS sent debug1: expecting SSH2_MSG_NEWKEYS debug2: set_newkeys: mode 0 debug1: SSH2_MSG_NEWKEYS received debug1: Roaming not allowed by server debug1: SSH2_MSG_SERVICE_REQUEST sent debug2: service_accept: ssh-userauth debug1: SSH2_MSG_SERVICE_ACCEPT received debug2: key: /home/sumanth/.ssh/id_rsa ((nil)), debug2: key: /home/sumanth/.ssh/id_dsa ((nil)), debug2: key: /home/sumanth/.ssh/id_ecdsa ((nil)), debug2: key: /home/sumanth/.ssh/id_ed25519 ((nil)),

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  • Formatting and installing Ubuntu over Windows 7 without admin password

    - by Emiliano Pasqualetti
    A borrowed a friend workstation with Windows 7 and Ubuntu 10.10. I do not have the administrator password and I would like to format everything and install Ubuntu 12.04. I have put the iso on the usb using PendriveLinux and restarted the machine. At the boot I am presented with several options but those Ubuntu ones simply load my friends Ubuntu 10.10 OS (for which I have no password). edit: I fixed this by pressing F12 at the boot before it gave me that prompt (called GRUB). I clicked USB and finally install.

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  • Amazon EC2 Instance - How to find SQL Server Password

    - by Prashant
    Hi I have created an Amazon EC2 Instance that provides Windows Server 2008 with SQL Sever 2008 pre-installed. Now in order to use the SQL Server for creating databases, or restoring backups of the databases that I have on my local machine, I need the "sa" password for SQL Server 2008. I have tried using the following but with no luck: sa password "blank password" "same password as the admin password for my EC2 instance" Could someone please guide me as to how to get started with using the Amazon EC2 Datacenter with respect to the "sa" password. Thanks

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  • I can't do "sudo"

    - by Klevin92
    Let's describe it from the beginning: I was planning to re-enable the password requirement in LightDM for security reasons. But, since my PC's been sluggish these times, it FC'd the password setup when I was entering and now I can't enter it even with combinatorics. I have followed the tips in the Help page, but with all of them I have issues: I try to enter recovery mode (so that I type passwd and my name and change it), but it is a black screen just like my boot screen (because of nVidia graphic card compatibility issue), then I can't do anything I also tried the editting "shadow" file, but the guide talks about some commas that I just don't see where they are supposed to be. I even tried deletting the keyring file like it's said, but nothing happens (except that I lose the other passwords) So is there anything I can do to have my password back? (a bonus would be stopping all this sluggish, apps not responding, etc)

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  • How to create a asp.net membership provider hashed password manually?

    - by Anheledir
    I'm using a website as a frontend and all users are authenticated with the standard ASP.NET Membership-Provider. Passwords are saved "hashed" within a SQL-Database. Now I want to write a desktop-client with administrative functions. Among other things there should be a method to reset a users password. I can access the database with the saved membership-data, but how can I manually create the password-salt and -hash? Using the System.Web.Membership Namespace seems to be inappropriate so I need to know how to create the salt and hash of the new password manually. Experts step up! :)

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  • How can it be impossible to "decrypt" an MD5 hash?

    - by Rob
    I was reading a question about MD5, and it made me remember something that boggles me. Very simple question, and I'm sorry if it's not a good one. I just can't understand how you convert something to one thing using some algorithm, and there being no way to convert it back using the algorithm in reverse. So how is this possible? Also, since multiple strings can create the same MD5 hash, due to it being less data than the input string, how would any other hashing system be any better?

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  • Encrypting and decrypting single file with master password and passphrase

    - by Iori
    Last night my system shutdown unexpectedly and when i restart it, it gave me superblock error. will then i fixed a little but. i was able to retrive my encrypted files which i encrypted my pgp public key but i was not able to retrive my private key or public key now i have ecrypted file which cannot be open because i have lost my private key. is there any way that i can encrypt my file by providing only master password and passphrase and no physical key or private key and when i am on another computer i can easily open it my same master password and passphrase. Thanks in advance

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  • Shifting from no-www to www and browsers' password storage

    - by user1444680
    I created a website having user-registration system and invited my friend to join it. I gave him the link of no-www version: http://mydomain.com but now after reading this and this, I want to shift to www.mydomain.com. But there's a problem. I saw that my browser is storing separate passwords for mydomain.com and www.mydomain.com. So in my friend's browser his password must have been stored for no-www. That means after I shift to www and next time he opens the login page, his browser wouldn't auto-fill the username and password fields and there will also be an extra entry (of no-www) in his browser's database of stored passwords. Can this be avoided? Can I do something that will convey to browsers that www.mydomain.com and mydomain.com are the same website? I already have a CNAME record for www pointing to mydomain.com but it seems that search engines consider CNAME as alias but browsers consider them as different websites, I don't know why.

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  • sha1(password) encryption

    - by Jason
    Alright, so I tried to make my users info super secure by adding '" . sha1($_POST['password']) . "' when inserting their password when they register. THAT WORKS great, looking at the database, I have no clue what their password is. Now the problem is logging in. I'm running some tests and when I try to log in, the password 12345 doesn't match the encrypted password using "$password=sha1($_POST['mypassword']);" Any idea's why?

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  • Why is this by passing the SUDO password?

    - by John Isaacks
    I have a bash script I am using to automate a SVN checkout. The contents of the file were: #!/bin/bash cd /var/www-cake sudo svn checkout file:///usr/local/svn/bash_repo/repo/ Then when I double click the file it would ask me what to do, I would click the button "Run In Terminal" and then a terminal would pop up and ask me for the SUDO password. I would enter it, the script would execute and the terminal would close. I wanted to give some sort of indication that the script ran successfully so I edited my file to look like: #!/bin/bash cd /var/www-cake sudo svn checkout file:///usr/local/svn/bash_repo/repo/ echo "Head revision has been pushed to live server" I expected the terminal to now stay open and tell me the message afterwards. To my surprise it now opens and immediately closes. The script does execute and I no longer have to put in the SUDO password. Is this right? I do not understand why this is happening, seems like a security issue.

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  • Check whether Excel file is Password protected

    - by Torben Klein
    I am trying to open an Excel (xlsm) file via VBA. It may or may not be protected with a (known) password. I am using this code: On Error Resume Next Workbooks.Open filename, Password:=user_entered_pw opened = (Err.Number=0) On Error Goto 0 Now, this works fine if the workbook has a password. But if it is unprotected, it can NOT be opened. Apparently this is a bug in XL2007 if there is also workbook structure protection active. (http://vbaadventures.blogspot.com/2009/01/possible-error-in-excel-2007.html). On old XL2003, supplying a password would open both unprotected and password protected file. I tried: Workbooks.Open filename, Password:=user_entered_pw If (Err.Number <> 0) Then workbooks.open filename This works for unprotected and protected file. However if the user enters a wrong password it runs into the second line and pops up the "enter password" prompt, which I do not want. How to get around this?

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  • How do I access the data in JSON converted to hash by crack in ruby?

    - by Angela
    Here is the example from the crack documentation: json = '{"posts":[{"title":"Foobar"}, {"title":"Another"}]}' Crack::JSON.parse(json) => {"posts"=>[{"title"=>"Foobar"}, {"title"=>"Another"}]} But how do I actually access the data in the hash? I've tried the following: array = Crack::JSON.parse(json) array["posts"] array["posts"] shows all the values, but I tried array["posts"]["title"] and it didn't work. Here is what I am trying to parse as an example: {"companies"=>[{"city"=>"San Mateo", "name"=>"Jigsaw", "address"=>"777 Mariners Island Blvd Ste 400", "zip"=>"94404-5059", "country"=>"USA", "companyId"=>4427170, "activeContacts"=>168, "graveyarded"=>false, "state"=>"CA"}], "totalHits"=>1} I want to access the individual elements under companies....like city and name.

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  • Check wether Excel file is Password protected

    - by Torben Klein
    I am trying to open an Excel (xlsm) file via VBA. It may or may not be protected with a (known) password. I am using this code: On Error Resume Next Workbooks.Open filename, Password:=user_entered_pw opened = (Err.Number=0) On Error Goto 0 Now, this works fine if the workbook has a password. But if it is unprotected, it can NOT be opened. Apparently this is a bug in XL2007 if there is also workbook structure protection active. (http://vbaadventures.blogspot.com/2009/01/possible-error-in-excel-2007.html). On old XL2003, supplying a password would open both unprotected and password protected file. I tried: Workbooks.Open filename, Password:=user_entered_pw If (Err.Number <> 0) Then workbooks.open filename This works for unprotected and protected file. However if the user enters a wrong password it runs into the second line and pops up the "enter password" prompt, which I do not want. How to get around this?

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  • How do I change the "Default" password

    - by Allwar
    hi, here's my problem. When I log in I'm beeing asked for a password connecting to the internet which is strange because I'm the only user and the admin. I have auto log in and it never asks for passwords, (except when i do administrative tasks.) I have looked in the menu (System -- preferences, but i can't find any topic that say's passwords or similar. the only place where I have found a way to change my password is my user ((System -- preferences -- user and groups). I did try to take screenshots but nothing happened, which also is strange. I'm running ubuntu 10.04 LTS Desktop on my Asus 1201n (12").

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  • Server asks for username and password on boot

    - by Jon
    I installed the Ubuntu server 12.04 easily, however, when I boot it asks for a username and password. I don't know my previous username or password from Windows XP. I am currently trying to install any version of Ubuntu I have in my arsenal to recover the PC, but all intents so far have failed. They worked on other computers but not on this one, the error I get is No DEFAULT or UI configuration directive found! I have tried all the solutions found on this site but to no avail. I don't have a Windows Ultimate boot disk and I'm out of CDs to burn. I can, however, make live USBs. Any Suggestions?

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  • Make password case unsensitive in shared ASP.Net membership tables web ap

    - by bill
    Hi all, i have two webapps.. that share ASP.Net membership tables. Everything works fine except i cannot remove case-sensitivity in one of the apps the way i am doing it in the other. in the non-working app void Login1_LoggingIn(object sender, LoginCancelEventArgs e) { string username = Login1.UserName.Trim(); if (!string.IsNullOrEmpty(username)) { MembershipUser user = Membership.GetUser(username); if (user != null) { // Only adjust the UserName if the password is correct. This is more secure // so a hacker can't find valid usernames if we adjust the case of mis-cased // usernames with incorrect passwords. string password = Login1.Password.ToUpper(); if (Membership.ValidateUser(user.UserName, password)) { Login1.UserName = user.UserName; } } } } is not working. the password is stored as all upper case. Converted at the time the membership user is created! So if the password is PASSWORD, typing PASSWORD allows me to authenticate. but typing password does not! Even though i can see the string being sent is PASSWORD (converted with toUpper()). I am at a complete loss on this.. in the other app i can type in lower or upper or mixed and i am able to authenticate. In the other app i am not using the textboxes from the login control though.. not sure if this is making the difference??

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  • How long can a hash left out in the open be considered safe?

    - by Xeoncross
    If I were to leave a SHA2 family hash out on my website - how long would it be considered safe? How long would I have before I could be sure that someone would find a collision for it and know what was hashed? I know that the amount of time would be based on the computational power of the one seeking to break it. It would also depend on the string length, but I'm curious just how secure hashes are. Since many of us run web-servers we constantly have to be prepared for the day when someone might make it all the way to the database which stores the user hashes. So, move the server security out of the way and then what do you have? This is a slightly theoretical area for many of the people I have talked with, so I would love to actually have some more information about average expectations for cracking.

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