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  • Login window language has changed after update on Snow Leopard

    - by Leonardo
    After the latest update (beginning of Dec. 2012) on Snow Leopard my login window language has changed. My system is set to be in english but the login windows now is in french. Currently I am in France so I suspect that during the update something has revealed my location and installed something in the locale of the current location. It doesn't seem fair to me, I want to switch it back to english but I couldn't find a way to do it.

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  • Telnet disconnected on login failure

    - by LinuxPenseur
    Hi, I have two servers running Linux 2.6.9 kernel. When i telnet to server1 , the telnet will be disconnected only after 3 login failures. While, when i telnet to server2, telnet will be disconnected after the first login failure. I expect server2 to behave the same way as server1 when telnet is executed. Kindly tell me where i should tweak my servers so that i could achieve what i expect. Thanks, LinuxPenseur

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  • hardening a server: disallow password-login for sudoers and log unusual ips

    - by Fabian Zeindl
    Two question regarding sudo-login into an ubuntu-system (debian tips welcome as well): Is it possible to require sudoers on my box to only login with publickey-authentication? Is it possible to log which ip sudoers log in from and check that for "unusual activity" or take actions? I'm thinking about temporarily removing sudo-rights if sudoers don't log in from whitelisted IPs. Or is that too risky to be exploited?

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  • windows 2008 server login screen remote desktop

    - by Marki
    how can I access a GRAPHICAL login screen (like on earlier Windows versions, or like VNC) instead of the Username/Password prompt via RDP? I'm asking because we have several servers where we have no passwords but the shutdown/reboot option is given via group policy on the login screen. Thus, instead of having to go the hardware console I'd like to be able to perform reboot/shutdown actions via terminal services. Regards, Marki

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  • Is this PHP/MySQL login script secure?

    - by NightMICU
    Greetings, A site I designed was compromised today, working on damage control at the moment. Two user accounts, including the primary administrator, were accessed without authorization. Please take a look at the log-in script that was in use, any insight on security holes would be appreciated. I am not sure if this was an SQL injection or possibly breach on a computer that had been used to access this area in the past. Thanks <?php //Start session session_start(); //Include DB config require_once('config.php'); //Error message array $errmsg_arr = array(); $errflag = false; //Connect to mysql server $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } //Select database $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } //Function to sanitize values received from the form. Prevents SQL injection function clean($str) { $str = @trim($str); if(get_magic_quotes_gpc()) { $str = stripslashes($str); } return mysql_real_escape_string($str); } //Sanitize the POST values $login = clean($_POST['login']); $password = clean($_POST['password']); //Input Validations if($login == '') { $errmsg_arr[] = 'Login ID missing'; $errflag = true; } if($password == '') { $errmsg_arr[] = 'Password missing'; $errflag = true; } //If there are input validations, redirect back to the login form if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; session_write_close(); header("location: http://tapp-essexvfd.org/admin/index.php"); exit(); } //Create query $qry="SELECT * FROM user_control WHERE username='$login' AND password='".md5($_POST['password'])."'"; $result=mysql_query($qry); //Check whether the query was successful or not if($result) { if(mysql_num_rows($result) == 1) { //Login Successful session_regenerate_id(); //Collect details about user and assign session details $member = mysql_fetch_assoc($result); $_SESSION['SESS_MEMBER_ID'] = $member['user_id']; $_SESSION['SESS_USERNAME'] = $member['username']; $_SESSION['SESS_FIRST_NAME'] = $member['name_f']; $_SESSION['SESS_LAST_NAME'] = $member['name_l']; $_SESSION['SESS_STATUS'] = $member['status']; $_SESSION['SESS_LEVEL'] = $member['level']; //Get Last Login $_SESSION['SESS_LAST_LOGIN'] = $member['lastLogin']; //Set Last Login info $qry = "UPDATE user_control SET lastLogin = DATE_ADD(NOW(), INTERVAL 1 HOUR) WHERE user_id = $member[user_id]"; $login = mysql_query($qry) or die(mysql_error()); session_write_close(); if ($member['level'] != "3" || $member['status'] == "Suspended") { header("location: http://members.tapp-essexvfd.org"); //CHANGE!!! } else { header("location: http://tapp-essexvfd.org/admin/admin_main.php"); } exit(); }else { //Login failed header("location: http://tapp-essexvfd.org/admin/index.php"); exit(); } }else { die("Query failed"); } ?>

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  • Cannot login to ISCSI Target - hangs after sending login details

    - by Frank
    I have an ISCSI target volume, to which i am trying to connect using CentOS Linux server. Everything works fine, but cannot its stuck at login. Here are the steps i am performing: [[email protected] ~]# iscsiadm -m node -l iscsiadm: could not read session targetname: 5 iscsiadm: could not find session info for session20 iscsiadm: could not read session targetname: 5 iscsiadm: could not find session info for session21 iscsiadm: could not read session targetname: 5 iscsiadm: could not find session info for session22 iscsiadm: could not read session targetname: 5 iscsiadm: could not find session info for session23 iscsiadm: could not read session targetname: 5 iscsiadm: could not find session info for session30 iscsiadm: could not read session targetname: 5 iscsiadm: could not find session info for session31 iscsiadm: could not read session targetname: 5 iscsiadm: could not find session info for session78 iscsiadm: could not read session targetname: 5 iscsiadm: could not find session info for session79 iscsiadm: could not read session targetname: 5 iscsiadm: could not find session info for session80 iscsiadm: could not read session targetname: 5 iscsiadm: could not find session info for session81 Logging in to [iface: eql.eth2, target: iqn.2001-05.com.equallogic:0-8a0906-ab4764e0b-55ed2ef5cf350a66-neon105, portal: 10.10.1.1,3260] (multiple) After this step, its stucks, waits for some time and then gives this output: Logging in to [iface: iface1, target: iqn.2001-05.com.equallogic:0-8a0906-ab4764e0b-55ed2ef5cf350a66-neon105, portal: 10.10.1.1,3260] (multiple) iscsiadm: Could not login to [iface: eql.eth2, target: iqn.2001-05.com.equallogic:0-8a0906-ab4764e0b-55ed2ef5cf350a66-neon105, portal: 10.10.1.1,3260]. My iscsi.conf is this: node.startup = automatic node.session.timeo.replacement_timeout = 15 # default 120; RedHat recommended node.conn[0].timeo.login_timeout = 15 node.conn[0].timeo.logout_timeout = 15 node.conn[0].timeo.noop_out_interval = 5 node.conn[0].timeo.noop_out_timeout = 5 node.session.err_timeo.abort_timeout = 15 node.session.err_timeo.lu_reset_timeout = 20 node.session.initial_login_retry_max = 8 # default 8; Dell recommended node.session.cmds_max = 1024 # default 128; Equallogic recommended node.session.queue_depth = 32 # default 32; Equallogic recommended node.session.iscsi.InitialR2T = No node.session.iscsi.ImmediateData = Yes node.session.iscsi.FirstBurstLength = 262144 node.session.iscsi.MaxBurstLength = 16776192 node.conn[0].iscsi.MaxRecvDataSegmentLength = 262144 discovery.sendtargets.iscsi.MaxRecvDataSegmentLength = 32768 node.conn[0].iscsi.HeaderDigest = None node.session.iscsi.FastAbort = Yes Also, in access control, i have given full access to Any IP, Any CHAP user and fixed iscsi initiator name. With same access level, all other volumes on rest of servers are working, except this one.

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  • How to wake from screensaver directly to the login screen in Ubuntu

    - by torbengb
    (a.k.a. How to switch users without entering password, part II) (see part I) I've got Ubuntu 9.10 with a user account for my wife, and one for myself. "Wake-from-screensaver" should result in "choose user" without having to enter any password. I know how to do that in Windows, but I'm not good with Linux (yet). Part of this was answered in my earlier question which helped me get past the login screen without passwords (after booting, and after choosing "switch user"), but once the screensaver kicks in and I wake it up again, the system does not present the "choose user" screen. Instead, it either turns off the screensaver and presents the desktop of the most recent user, or (if the screensaver is set to lock the screen) prompt for the user's password (which can be handily surpassed by clicking the "switch user" button and choosing the same user again). So, the login ("choose user") screen has been dealt with. How do I make the (any) screensaver return to the login screen at wake, rather than to the current user's desktop? Windows can do this, I'm sure Linux can too - but how?

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  • Skip new Windows 7 user selection and go to login prompt

    - by Doltknuckle
    We've begun our migration to windows 7 and we ran into an interesting issue. When we hit "CTRL+ALT+DEL" we are brought to the "User selection Screen". Normally, this screen will have an icon for every local user for the machine. These machines are Domain members with "Fast User Switch" disabled so no user names are listed only the "Other User" option. If you click "Other User" or hit enter, the system moves on to the normal login screen where it prompts for user name and password. Here's the issue: We want to find a way to skip over the part where a user selects "Other User". We essentially want the system to always assume that we always want "Other User" and to go directly to the login screen when a user hits "CTRL+ALT+DEL". What I find odd is that the "Other User" doesn't show up until we've had more than one domain user log in. Right after we re-image the machine, the login process goes directly to the user credential prompt. Anyone have any ideas?

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  • login form whith java/sqlite

    - by tuxou
    hi I would like to create a login form for my application with the possibility to add or remove users for an sqlite database, i have created the table users(nam, pass) but i can't unclud it in my login form, it someone could help me this is my login code: import java.awt.*; import java.awt.event.*; import javax.swing.*; public class login extends JFrame { // Variables declaration private JLabel jLabel1; private JLabel jLabel2; private JTextField jTextField1; private JPasswordField jPasswordField1; private JButton jButton1; private JPanel contentPane; // End of variables declaration public login() { super(); create(); this.setVisible(true); } private void create() { jLabel1 = new JLabel(); jLabel2 = new JLabel(); jTextField1 = new JTextField(); jPasswordField1 = new JPasswordField(); jButton1 = new JButton(); contentPane = (JPanel)this.getContentPane(); // // jLabel1 // jLabel1.setHorizontalAlignment(SwingConstants.LEFT); jLabel1.setForeground(new Color(0, 0, 255)); jLabel1.setText("username:"); // // jLabel2 // jLabel2.setHorizontalAlignment(SwingConstants.LEFT); jLabel2.setForeground(new Color(0, 0, 255)); jLabel2.setText("password:"); // // jTextField1 // jTextField1.setForeground(new Color(0, 0, 255)); jTextField1.setSelectedTextColor(new Color(0, 0, 255)); jTextField1.setToolTipText("Enter your username"); jTextField1.addActionListener(new ActionListener() { public void actionPerformed(ActionEvent e) { jTextField1_actionPerformed(e); } }); // // jPasswordField1 // jPasswordField1.setForeground(new Color(0, 0, 255)); jPasswordField1.setToolTipText("Enter your password"); jPasswordField1.addActionListener(new ActionListener() { public void actionPerformed(ActionEvent e) { jPasswordField1_actionPerformed(e); } }); // // jButton1 // jButton1.setBackground(new Color(204, 204, 204)); jButton1.setForeground(new Color(0, 0, 255)); jButton1.setText("Login"); jButton1.addActionListener(new ActionListener() { public void actionPerformed(ActionEvent e) { jButton1_actionPerformed(e); } }); // // contentPane // contentPane.setLayout(null); contentPane.setBorder(BorderFactory.createEtchedBorder()); contentPane.setBackground(new Color(204, 204, 204)); addComponent(contentPane, jLabel1, 5,10,106,18); addComponent(contentPane, jLabel2, 5,47,97,18); addComponent(contentPane, jTextField1, 110,10,183,22); addComponent(contentPane, jPasswordField1, 110,45,183,22); addComponent(contentPane, jButton1, 150,75,83,28); // // login // this.setTitle("Login To Members Area"); this.setLocation(new Point(76, 182)); this.setSize(new Dimension(335, 141)); this.setDefaultCloseOperation(WindowConstants.EXIT_ON_CLOSE); this.setResizable(false); } /** Add Component Without a Layout Manager (Absolute Positioning) */ private void addComponent(Container container,Component c,int x,int y,int width,int height) { c.setBounds(x,y,width,height); container.add(c); } private void jTextField1_actionPerformed(ActionEvent e) { } private void jPasswordField1_actionPerformed(ActionEvent e) { } private void jButton1_actionPerformed(ActionEvent e) { System.out.println("\njButton1_actionPerformed(ActionEvent e) called."); String username = new String(jTextField1.getText()); String password = new String(jPasswordField1.getText()); if(username.equals("") || password.equals("")) // If password and username is empty Do this { jButton1.setEnabled(false); JLabel errorFields = new JLabel("You must enter a username and password to login."); JOptionPane.showMessageDialog(null,errorFields); jTextField1.setText(""); jPasswordField1.setText(""); jButton1.setEnabled(true); this.setVisible(true); } else { JLabel optionLabel = new JLabel("You entered "+username+" as your username. Is this correct?"); int confirm =JOptionPane.showConfirmDialog(null,optionLabel); switch(confirm){ // Switch Case case JOptionPane.YES_OPTION: // Attempt to Login user jButton1.setEnabled(false); // Set button enable to false to prevent 2 login attempts break; case JOptionPane.NO_OPTION: // No Case.(Go back. Set text to 0) jButton1.setEnabled(false); jTextField1.setText(""); jPasswordField1.setText(""); jButton1.setEnabled(true); break; case JOptionPane.CANCEL_OPTION: // Cancel Case.(Go back. Set text to 0) jButton1.setEnabled(false); jTextField1.setText(""); jPasswordField1.setText(""); jButton1.setEnabled(true); break; } // End Switch Case } } public static void main(String[] args) { JFrame.setDefaultLookAndFeelDecorated(true); JDialog.setDefaultLookAndFeelDecorated(true); try { UIManager.setLookAndFeel("com.sun.java.swing.plaf.windows.WindowsLookAndFeel"); } catch (Exception ex) { System.out.println("Failed loading L&F: "); System.out.println(ex); } new login(); }; }

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  • Android: prevent user from coming back to login page after logging in

    - by user522559
    Hi all, I have a login page in my android app. I want to prevent the user after logging in to come back to the login page. Also, when the user reopens the app, if he has logged in before, I want to go directly to the main page without having to go to the login page. What the best way of doing that? One idea I thought of is to save the login parameters in a cookie, and then, when the app is reopens, if the cookie contains some login information, I inflate the main page, otherwise, I inflate the login page. Is this the best way of doing it? Thanks,

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  • Magento set Store Id - customer login - but still logged out

    - by user3564050
    I've got an overridden AccountController in which i set the current store to an other as currently running (example: Customer is in website default and store default, going to login page, click login, my loginPostAction sets the store to id "2" (on website 2) and then executes the parent code loginPostAction. The store is set, of course, but after the login and the redirect to home, the customer is not logged in anymore... Customer-sendlogindata-myaccountcontroller sets store-original account controller logs in without errors (cause $session customer is set)-redirect to home-customer is not logged in anymore... i set the store with Mage::app()-setCurrentStore($id); . And in index.php i've got an extra where the store is set to the right id (2) too and this works... but the customer is not logged in anymore.. is that an issue with the session cause different websites ? I don't want to globally share customer.. each website has his own customers, but every customer has to be able to login on default store. AccountController.php overridden: public $Website_Ids = array( array("code" => "gerstore", "id" => "3", "website" => "ger"), array("code" => "ukstore", "id" => "2", "website" => "uk"), array("code" => "esstore", "id" => "4", "website" => "es"), array("code" => "frstore", "id" => "5", "website" => "fr") ); public function loginPostAction() { $login = $this->getRequest()->get('login'); if(isset($login['username'])) { $found = null; foreach($this->Website_Ids as $WebsiteId) { $customer = Mage::getModel('customer/customer'); $customer->setWebsiteId($WebsiteId['id']); $customer->loadByEmail($login['username']); if(count($customer->getData()) > 0) { $found = $WebsiteId; } } if($found != null && Mage::app()->getStore()->getId() != $found['id']) { /* found, so set currentstore to id */ Mage::app()->setCurrentStore($found['id']); $_SESSION['current_store_b2b'] = $found; } /* not found, doesn't matter cause mage login exception handling */ } parent::loginPostAction(); } Index.php : session_start(); $session = $_SESSION['current_store_b2b']; if($session != null || $session != "") { Mage::app()->setCurrentStore($session['id']); Mage::run($session['code'], 'store'); } else { /* Store or website code */ $mageRunCode = isset($_SERVER['MAGE_RUN_CODE']) ? $_SERVER['MAGE_RUN_CODE'] : ''; /* Run store or run website */ $mageRunType = isset($_SERVER['MAGE_RUN_TYPE']) ? $_SERVER['MAGE_RUN_TYPE'] : 'store'; Mage::run($mageRunCode, $mageRunType); } Whats the matter ? Thanks.

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  • Login or Register (Ruby on rails)

    - by DanielZ
    Hello stackoverflow, I'm working on an Ruby on Rails application (2.3.x) and i want to make a form that lets the user login or register. I want to do this in the same form. I have a JS function that replaces the form elements like this: Login form: <% form_for @user do |f| %> <div id="form"> <%= f.label :email, "E-mail" %> <%= f.text_field :email %> <%= f.label :password, "Password" %> <%= f.password_field :password %> <%= link_to "I don't have an account, "#", :id => "changeForm"%> <%= f.submit "Login" %> </div> <% end %> The id "changeForm" triggers a JS function that changes the form elements. So if you press the url the html looks like this: <% form_for @user do |f| %> <div id="form"> <%= f.label :name, "Name" %> <%= f.text_field :name %> <%= f.label :email, "E-mail" %> <%= f.text_field :email %> <%= f.label :password, "Password" %> <%= f.password_field :password %> <%= f.label :password_confirmation, "Password confirmation" %> <%= f.password_field :password_confirmation %> <%= link_to "I do have an account, "#", :id => "changeForm"%> <%= f.submit "Register" %> </div> <% end %> I added the neccesary validations to my user model: class User < ActiveRecord::Base attr_reader :password validates_presence_of :name, :email, :password validates_format_of :email, :with => /\A([^@\s]+)@((?:[-a-z0-9]+\.)+[a-z]{2,})\Z/i validates_confirmation_of :password But what happens when you fill in the email / password you get the errors that the name is missing and that the password fields aren't confirmed. So i could do some nasty programming in my user model like this: #if password_conf or the name are present the user has tried to register... if params[:user][:password_confirmation].present? || params[:user][:name].present? #so we'll try to save the user if @user.save #if the user is saved authenticate the user current_session.user = User.authenticate(params[:user]) #if the user is logged in? if current_session.user.present? flash[:notice] = "succesvully logged redirect_to some_routes_path else #not logged in... flash[:notice] = "Not logged in" render :action => "new" end else #user not saved render :action => "new" end else #So if the params[:user][:password_confirmation] or [:user][:name] weren't present we geuss the user wants to login... current_session.user = User.authenticate(params[:user]) #are we logged_in? if current_session.user.present? flash[:notice] = "Succesvully logged in" redirect_to some_routes_path else #errors toevoegen @user.errors.add(:email, "The combination of email/password isn't valid") @user.errors.add(:password," ") render :action => "new" end end end Without validations this (imho dirty code and should not be in the controller) works. But i want to use the validates_presence_of methods and i don't want to slap the "conventions over configurations" in the face. So another thing i have tried is adding a hidden field to the form: #login form <%= f.hidden_field :login, :value => true %> # and ofcourse set it to false if we want to register. And then i wanted to use the method: before_validation before_validation_on_create do |user| if params[:user].login == true #throws an error i know... validates_presence_of :email, :password validates_format_of :email, :with => /\A([^@\s]+)@((?:[-a-z0-9]+\.)+[a-z]{2,})\Z/i else validates_presence_of :name, :email, :password validates_format_of :email, :with => /\A([^@\s]+)@((?:[-a-z0-9]+\.)+[a-z]{2,})\Z/i validates_confirmation_of :password end end But this doesn't work because i can't access the params. And login isn't a attribute for the user object. But i thought that in this way i could validate the email and password params if the user wants to login. And all the other attrs if the user want to register. So all i could think of doesn't work how i want it to work. So my main goal is this: 1 form for login/register with the use of the validation methods in the user model. So if we want to login but don't fill in any information = give validation errors. And if the user wants to login but the email/password combination doens't match give the "@user.errors.add(:email, "the combination wasn't found in the db...")". And the same goes for user register... Thanks in advance!

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  • Users Password does not reset after successful login at the console but works fine with SSH

    - by jnbbender
    The title says it all. I have my unsuccessful login attempts set to three. I purposefully fail logging in 2x, then when I SSH into the box successfully the 3rd time my count drops back to zero; exactly what should happen. But at the console I get failed login attempts EVEN for my successful login attempts. I am using RHEL 5.6 and no I am not able to upgrade. Here is my system-auth file: auth required pam_env.so auth required pam_tally.so onerr=fail deny=3 per_user auth sufficient pam_unix.so nullok try_first_pass auth requisite pam_succeed_if.so uid >= 500 quiet auth required pam_deny.so account required pam_unix.so account required pam_tally.so account sufficient pam_succeed_if.so uid < 500 quiet account required pam_permit.so password requisite pam_cracklib.so try_first_pass retry=3 password sufficient pam_unix.so md5 shadow nullok try_first_pass use_authtok password required pam_deny.so session optional pam_keyinit.co revoke session required pam_limits.so session [success=1 default=ignore] pam_succeed_if.so service in crond quiet use_uid session required pam_unix.so I have tried adding reset after and in place of per_user in the auth required pam_tally.so field. Nothing seems to work and I don't know why SSH is working just fine. Any ideas?

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  • Can't login after upgrading to Windows 8.1

    - by flatline
    This afternoon, I upgraded my work laptop from Windows 8 to Windows 8.1. I had previously had a local account, but after the upgrade, it prompted me to enter my windows account credentials, which I had set up beforehand at some point. I entered my password and clicked next, went through another screen or two, grew tired with the process, and clicked whatever the equivalent button to "skip this step" that I was presented with. Now I can't log in. Not with my (previous) local account password, and not with my windows account password. It's a Dell with biometric identification, which I had set up previously, so I put my finger on the reader and it complained that I couldn't use that fingerprint because I had changed my password. But, I hadn't wittingly changed my password at all. I assume that what happened is that, by entering my credentials, my local account was tied to the Windows account, but because I cancelled the process partway through, something went wrong and I cannot log in. A few questions: 1) How do I log in with my windows account credentials? Should LOCALMACHINENAME\username, which was my previous login method, still work for the Windows account? When I booted to safemode it prompted me with WindowsAccount\myemailaddress, which allowed me to login there, but the regular login doesn't accept the '@' symbol. 2) Is there any way to make that account local-only again? I can't find any way of doing it. 3) I managed to enable the local administrator account and get back into the box; failing all else, is there a quick way to migrate my old profile over to a new user?

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  • Fb.Login() does not popup OAuth Dialog?

    - by cnz81
    I'm study fb api recently and I use fb api to do login, however, when I use Fb.Login() method, and scope the permission that I needed. I found the problem that when I use account ( this account is also the fb app admin ) to login, everything is be work including login dialog and oauth dialog , but when I use another account ( General account ) the login dialog is work but OAuth Dialog is not work . I've tried another account ( General account ), it had the same situation, Only when I use the account which is also the fb app owner account would be work. Any Ideas ? Below is the code... FB.login(function (response) { if (response.authResponse) { alert('success !') ; } else { alert(' faild ! '); } }, { scope: "publish_actions" });

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  • 13.10 Saucy login issues; black screen, loop, and freeze

    - by user135598
    Once upgraded to I 13.10 I was not able to get past the login screen. I had the black screen. Also; I can not log on with low graphics mode from recovery either. It makes no difference if I try default graphics driver or not. Then after running sudo install -f from recovery root prompt I got a login loop. I have purged fglrx, fglrx-legacy, and nvidia-current. I updated my repository with xorg-edgers and reinstalled nvidia-current. Now it semi-freezes at the login screen when I try to log on as my normal user. I say 'semi' because I can still use my mouse to click on the upper right hand Ubuntu logo and Shut Down or Restart the PC. I still cannot log on with my user name, but I can through the Guest login. While logged in as Guest I added a new user account with administrative privileges. I CAN log into this account without problem and from here am able to see that my .dmrc file in my original account reads: [Desktop] Session=XBMC I have changed 'Session=XBMC' to 'Session=ubuntu' and rebooted, but to no avail. The file resets itself and makes a backup of my changes. Any help would be greatly appreciated.

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  • Help a Beginner with a PHP based Login System

    - by Brian Lang
    I'm a bit embarrassed to say, but I've run into issue with creating a PHP based login system. I'm using a site template to handle the looks of the the login process, so I will spare you the code. Here is my thought process on how to handle the login: Create a simple login.php file. On there will be a form whose action is set to itself. It will check to see if the submit has been clicked, and if so validate to make sure the user entered a valid password / username. If they do, set a session variable save some login info (username, NOT password), and redirect them to a restricted area. If the login info isn't valid, save an error message in a session variable, display error message giving further instruction, and wait for the user to resubmit. Here is a chunk of what I have - hopefully one of you experts can see where I've gone wrong, and give me some insight: if(isset($_POST['submit'])) { if(!empty($_POST['username']) AND !empty(!$_POST['password'])) { header("Location: http://www.google.com"); } else { $err = 'All the fields must be filled in!'; } } if($err) { $_SESSION['msg']['login-err'] = $err; } ? Now the above is just an example - the intent of the above code is to process user input, with the script validating simply that the user has given input for username and password. If they have, I would like them, in this case, to be redirected to google.com (for the sake of this example). If not, save an error message. Given my current code, the error message will display perfectly, however if the user submits and has something entered for the username and password, the page simply doesn't redirect. I'm sure this is a silly question, but I am a beginner, and well, to be honest, a bit buzzed right now. Thanks so much!

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  • Validating Login / Changing User settings / Php Mysql

    - by Marcelo
    Hi everyone, my questions are about login, and changing already saved data. (Q1) 'Till now I've only saved input in the tables of the database (registration steps), now I need to check if the input (login steps), are the same of my table in database, in fact I have 3 types of users, then I'll have to check 3 kind of tables. Then if the input data matches with one of those 3 tables I will redirect the user to his specific area. I'm thinking about saved the submitted data $login=$_REQUEST['login']; and $password=$_REQUEST['password']; and compare with the login column in the database. Then if the login matches, I'll compare the password submitted with the one in the row, not in the column. But I don't know how to do this search and comparison,neither what to use. Then if both matches I'll redirect the user. Else I'll send an login error message. (this I know how to do) (Q2) What if need to change an already saved user ? For example to change an email address. My changing user's data web page is exactly the same like the registration user web page. Can I load the already saved options and values of registration (table user for example). Then the user will change whatever he thinks it's necessary, and then when he submits the new information, they would not create a new row in my table, but just be overwritten the old information? How can I do this? Sorry for any mistake in English, and Thanks for the attention.

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  • SQL SERVER – Importance of User Without Login – T-SQL Demo Script

    - by pinaldave
    Earlier I wrote a blog post about SQL SERVER – Importance of User Without Login and my friend and SQL Expert Vinod Kumar has written excellent follow up blog post about Contained Databases inside SQL Server 2012. Now lots of people asked me if I can also explain the same concept again so here is the small demonstration for it. Let me show you how login without user can help. Before we continue on this subject I strongly recommend that you read my earlier blog post here. In following demo I am going to demonstrate following situation. Login using the System Admin account Create a user without login Checking Access Impersonate the user without login Checking Access Revert Impersonation Give Permission to user without login Impersonate the user without login Checking Access Revert Impersonation Clean up USE [AdventureWorks2012] GO -- Step 1 : Login using the SA -- Step 2 : Create Login Less User CREATE USER [testguest] 9ITHOUT LOGIN WITH DEFAULT_SCHEMA=[dbo] GO -- Step 3 : Checking access to Tables SELECT * FROM sys.tables; -- Step 4 : Changing the execution contest EXECUTE AS USER   = 'testguest'; GO -- Step 5 : Checking access to Tables SELECT * FROM sys.tables; GO -- Step 6 : Reverting Permissions REVERT; -- Step 7 : Giving more Permissions to testguest user GRANT SELECT ON [dbo].[ErrorLog] TO [testguest]; GRANT SELECT ON [dbo].[DatabaseLog] TO [testguest]; GO -- Step 8 : Changing the execution contest EXECUTE AS USER   = 'testguest'; GO -- Step 9 : Checking access to Tables SELECT * FROM sys.tables; GO -- Step 10 : Reverting Permissions REVERT; GO -- Step 11: Clean up DROP USER [testguest]Step 3 GO Here is the step 9 we will be able to notice that how a user without login gets access to some of the data/object which we gave permission. What I am going to prove with this example? Well there can be different rights with different account. Once the login is authenticated it makes sense for impersonating a user with only necessary permissions to be used for further operation. Again this is very basic and fundamental example. There are lots of more points to be discussed as we go in future posts. Just do not take this blog post as a template and implement everything as it is. Reference: Pinal Dave (http://blog.sqlauthority.com) Filed under: PostADay, SQL, SQL Authority, SQL Query, SQL Security, SQL Server, SQL Tips and Tricks, T SQL, Technology

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  • Problems with login scripts on Terminal Server 2008

    - by discovery
    We are having issues with login scripts not running on Windows 2008 Terminal Server. This is a brand new implementation and they have never worked. The test user in question doesn't have any problems running login scripts on their workstation. I have tried logging into the server directly with their account, but still no scripts run. I have setup a test account with Domain Admins rights in the same OU as theirs and the scripts don't run. I can manually run the scripts from the SYSVOL\somedomain.com\Policies folder and they run fine. The Terminal 2008 Server is in a mixed 2003/2008 domain. The user can run the gpupdate on the server without error. I have also run the Group Policy Results for this user and the terminal server and everything looks good, no errors. Any suggestions?

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  • After restoring a SQL Server database from another server - get login fails

    - by Renso
    Issue: After you have restored a sql server database from another server, lets say from production to a Q/A environment, you get the "Login Fails" message for your service account. Reason: User logon information is stored in the syslogins table in the master database. By changing servers, or by altering this information by rebuilding or restoring an old version of the master database, the information may be different from when the user database dump was created. If logons do not exist for the users, they will receive an error indicating "Login failed" while attempting to log on to the server. If the user logons do exist, but the SUID values (for 6.x) or SID values (for 7.0) in master..syslogins and the sysusers table in the user database differ, the users may have different permissions than expected in the user database. Solution: Links a user entry in the sys.database_principals system catalog view in the current database to a SQL Server login of the same name. If a login with the same name does not exist, one will be created. Examine the result from the Auto_Fix statement to confirm that the correct link is in fact made. Avoid using Auto_Fix in security-sensitive situations. When you use Auto_Fix, you must specify user and password if the login does not already exist, otherwise you must specify user but password will be ignored. login must be NULL. user must be a valid user in the current database. The login cannot have another user mapped to it. execute the following stored procedure, in this example the login user name is "MyUser" exec sp_change_users_login 'Auto_Fix', 'MyUser'   NOTE: sp_change_users_login cannot be used with a SQL Server login created from a Windows principal or with a user created by using CREATE USER WITHOUT LOGIN.

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  • Logic to create common Serverlet3 Login

    - by user3696143
    I am using Servlet3 Login to Authenticate User in website I have these Login Website Normal Login(Fill the Sigup form) Facebook Login (From Facebook Id) Twitter Login (From Twitter) And I am already authenticate user by below code HttpServletRequest request = (HttpServletRequest) FacesContext.getCurrentInstance().getExternalContext().getRequest(); request.login(username, password); And it is working fine for Website Login as user gave his/her EMailId and password and it store in DB. Now I modified table and added more columns to save Facebookid in same user table and also password for Facebook login FacebookId work as a Password as well. Same I will do for Twitter But I want the same Servlet3 to authenticate user. How can I achieve it? And also added context.xml file inside META-INF folder <Realm localDataSource="true" debug="99" className="org.apache.catalina.realm.JDBCRealm" connectionName="user" connectionPassword="password" connectionURL="jdbc:mysql://localhost:3306/ ccc" digest="md5" driverName="com.mysql.jdbc.Driver" roleNameCol="role_name" userCredCol="password" userNameCol="email_id" userRoleTable="users_list" userTable="user_list_view" /> Also it is possible to check which query fired by realm entry?

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  • Windows 7 and Vista Home Networked-Unable to login from Vista to Windows 7

    - by Lynn
    I set up both Vista and Windows 7 on the same workgroup. I can view Windows 7 from Vista and vice versa. I can login into Vista from Windows 7. I am unable to login to Windows 7 from Vista. When I enter the Windows 7 User name and Password on Vista, the following information appears: Logon unsuccessful: Windows is unable to log you on. Be sure that your user name and password are correct. Both are correct. Do you have any idea how I can resolve this logon issue? Thank you, Lynn

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  • Screen magnifier wrecks login screen — how to get rid of it?

    - by Jonik
    In the Ubuntu / GNOME login screen, I tried some of the different accessibility settings out of curiosity. Big mistake! Enabling the "use screen magnifier" (or whatever) option breaks down the view horribly, and makes it impossible to even access the settings again: (Right click & View Image for full size) Yes, I tried to access every corner of the screen using the mouse, but there's just chaos everywhere. Fortunately, I can still log in (pressing "Esc" makes the normal login dialog appear on the left-hand monitor). My question is, how to disable the "magnifier" option outside of the login screen itself? (By editing some config file perhaps?) I don't care about getting the magnifier mode to work properly - just make it go away altogether, please.

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