Hi
I am using the following code to get users to enter their username and password, it is then checked and a new page is displayed with their username. My question is, how can I display the rest of their details from that session, I have googled and read many articles, but I am still not any futher forward, any help would be much appreciated. -
<?php
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session_start();
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$errorMessage = '';
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if (isset($_POST['txtUsername']) && isset($_POST['txtPassword'])) {
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include 'includes/config.php';
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include 'includes/opendb.php';
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$username = $_POST['txtUsername'];
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$password = md5($_POST['txtPassword']);
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// check if the user id and password combination exist in database
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$sql = sprintf("SELECT username
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FROM phpbb_users
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WHERE username = '%s'
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AND user_password = '%s'",$username,$password);
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$result = mysql_query($sql)
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or die('Query failed. ' . mysql_error());
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if (mysql_num_rows($result) == 1) {
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// the user id and password match,
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// set the session
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$_SESSION['db_is_logged_in'] = true;
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$_SESSION['username'] = $username;
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// after login we move to the main page
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header('Location:main.php');
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exit;
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} else {
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$errorMessage = 'Sorry, wrong user id / password';
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}
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include 'includes/closedb.php';
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}
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?>
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<html>
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<head>
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<title>Basic Login</title>
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<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
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</head>
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<body>
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<?php
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include 'includes/header.php';
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include 'includes/navbar.tpl';
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if ($errorMessage != '')
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?>
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<p> <strong><font color="#990000"><?php echo $errorMessage; ?></font></strong></p>
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<form method="post" name="frmLogin" id="frmLogin">
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<table width="400" border="1" align="center" cellpadding="2" cellspacing="2" bordercolor="#000000">
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<tr>
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<td width="150">User Name</td>
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<td><input name="txtUsername" type="text" id="txtUsername"></td>
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</tr>
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<tr>
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<td width="150">Password</td>
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<td><input name="txtPassword" type="password" id="txtPassword"></td>
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</tr>
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<tr>
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<td width="150"> </td>
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<td>
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<input type="submit" name="btnLogin" value="Login">
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</div></td></tr>
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</table>
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</form>
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</body>
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</html>
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Make use of the available formatting tags in the forum Use CODE tags around your code as appropriate: - [code]..code goes here..[ /code]
- [code=php] ..php code goes here.. [ /code]
- [code=c] ..C like code goes here.. [ /code]
- [code=html] ..html code goes here.. [ /html]
9  3545 
Here is what is coded for the main page -
<?php
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session_start();
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include 'includes/header.php';
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include 'includes/navbarout.tpl';
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// is the one accessing this page logged in or not?
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if (!isset($_SESSION['db_is_logged_in'])
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|| $_SESSION['db_is_logged_in'] !== true) {
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// not logged in, move to login page
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header('Location: login.php');
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exit;
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}
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echo sprintf("Welcome to the Revenge's members area, you are logged in as %s",$_SESSION['username']);
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?>
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ak1dnar 1,584
Recognized Expert Top Contributor Thread Title changed : Ajaxrand
From >>> Help for a n00b, whilst using sessions
To >>> User Login Script with Session
ak1dnar 1,584
Recognized Expert Top Contributor
You have created two session Variables so far,Here. -
// set the session
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$_SESSION['db_is_logged_in'] = true;
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$_SESSION['username'] = $username;
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You Can use this $_SESSION['username'] to get the other records for the logged user.
Other_page.php -
session_start();
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$Logged_in_user = $_SESSION['username'];
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Now the the $Logged_in_user is here in this variable.
re-use it for any data manipulation,If this user name is a Qnique One
Thanks !
Thank you for the reply, I understand, what you are saying, however, how can extract the data from multiple tables with that one variable
kovik 1,044
Recognized Expert Top Contributor
Thank you for the reply, I understand, what you are saying, however, how can extract the data from multiple tables with that one variable
Depends on how the tables are related. Typically, all tables dealing with a user are somehow related by the user id. - SELECT `users`.`username`, `profiles`.`dateJoined` FROM `users` LEFT JOIN (`profiles`) ON (`profiles`.`userid` = `users`.`id`) WHERE `users`.`username` = $foo;
Understand now, however the query I used looks like the :
[PHP]$query = "SELECT phpbb_users.username, phpbb_users.user_from, phpbb_users.user_email, phpbb_ranks.rank_image
FROM phpbb_users
LEFT JOIN (phpbb_ranks)
ON (phpbb_users.user_rank = phpbb_ranks.rank_id)
WHERE phpbb_users.username = $Logged_in_user";[/PHP]
When I run it, this error appears, which I believe means that it's not connecting to the DB:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in
If I change the variable $Logged_in_user to phpbb_users, everybodies details appear, so any ideas whats wrong with the query?
HOOAH, done it thank you for all your help, I understand now, I chnaged the query to
[PHP]$query = sprintf("SELECT phpbb_users.username, phpbb_users.user_from, phpbb_users.user_email, phpbb_ranks.rank_image
FROM phpbb_users
LEFT JOIN (phpbb_ranks) ON (phpbb_users.user_rank = phpbb_ranks.rank_id)
WHERE phpbb_users.username = '%s'",$Logged_in_user);[/PHP]
ak1dnar 1,584
Recognized Expert Top Contributor
ditch,
Since you have asked unrelated Question within the same thread, regarding PHP based calender I have to Split the thread. Please Read Why New Thraed Location:
Thanks ,
.
ak1dnar 1,584
Recognized Expert Top Contributor
Copied Contents from Splitted thread -Ajaxrand
By the way, the only problem with the first query was that you forgot the single quotes around the value.
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