[SOLVED] Displaying a username when a user is logged in, HELP!

SauloA · February 12, 2007

I've created a login in page and I want the login form to appear when a user isn't logged in and hide the login form and displaying the username when the user is logged in. I think I'm close to solving the problem but I'm probably missing some code.

Here's my code. I'd appreciate the help.

<?php require_once('../Connections/connBlog.php'); ?>

<?php

// *** Validate request to login to this site.

if (!isset($_SESSION)) {

session_start();

}

$loginFormAction = $_SERVER['PHP_SELF'];

if (isset($_GET['accesscheck'])) {

$_SESSION['PrevUrl'] = $_GET['accesscheck'];

}

if (isset($_POST['username'])) {

$loginUsername=$_POST['username'];

$password=$_POST['password'];

$MM_fldUserAuthorization = "al_levelid";

$MM_redirectLoginSuccess = "home.php";

$MM_redirectLoginFailed = "failed.php";

$MM_redirecttoReferrer = true;

mysql_select_db($database_connBlog, $connBlog);

$LoginRS__query=sprintf("SELECT user_id, user_sname, user_pass, al_levelid FROM user_tbl WHERE binary user_sname='%s' AND binary user_pass='%s'",

get_magic_quotes_gpc() ? $loginUsername : addslashes($loginUsername), get_magic_quotes_gpc() ? $password : addslashes($password));

$LoginRS = mysql_query($LoginRS__query, $connBlog) or die(mysql_error());

$LoginArray= mysql_fetch_array($LoginRS); //Creates the array

$loginFoundUser = mysql_num_rows($LoginRS);

if ($loginFoundUser) {

$loginStrGroup = mysql_result($LoginRS,0,'al_levelid');

//declare three session variables and assign them

$_SESSION['MM_Username'] = $loginUsername;

$_SESSION['MM_UserID'] = $LoginArray['user_id'];

$_SESSION['MM_UserLevel'] = $LoginArray['al_levelid'];

$_SESSION['MM_UserGroup'] = $loginStrGroup;

if (isset($_SESSION['PrevUrl']) && true) {

$MM_redirectLoginSuccess = $_SESSION['PrevUrl'];

}

header("Location: " . $MM_redirectLoginSuccess );

}

else {

header("Location: ". $MM_redirectLoginFailed );

}

?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">

<head>

<title>Blog Login</title>

</head>

<body>

<?php if ($_SESSION['MM_Username'] == 0) { ?>

<form ACTION="<?php echo $loginFormAction; ?>" METHOD="POST" name="frmLogin" id="frmLogin">

<table>

Login to the site

</caption>

<tr>

<th align="right" scope="row">Username:</th>

</tr>

<tr>

<th align="right" scope="row">Password:</th>

</tr>

<tr>

</tr>

</table>

</form>

<?php } else { ?>

<p>Welcome <?php echo $_SESSION['MM_Username']; ?></p>

<?php } ?>

</body>

</html>

GingerRobot · February 12, 2007

Well whats actually happening at the moment? Also do you think you could modify your post and put [ code] [ /code] tags around it so we get syntax highlighting; makes it a lot easier to read.

One little problem i can spot, however, is this part:

if (!isset($_SESSION)) {

session_start();

}

You should put session_start() before any attempt to do anything at all with sessions. As far as i can tell, the way you've set it up at the moment will probably prevent the sessions from being set when the person trys to log in.

so just change it to session_start(); on its own. No need for the if statement.

SauloA · February 12, 2007

The problem isn't in the:

if (!isset($_SESSION)) {

session_start();

}

The problem is most likely in this region near the bottom:

<?php if ($_SESSION['MM_Username'] == 0) { ?>
<form ACTION="<?php echo $loginFormAction; ?>" METHOD="POST" name="frmLogin" id="frmLogin">
 <table>
   <caption>
     Login to the site
   </caption>
   <tr>
     <th align="right" scope="row">Username:</th>
     <td><input name="username" type="text" id="username" size="10" maxlength="10" /></td>
   </tr>
   <tr>
     <th align="right" scope="row">Password:</th>
     <td><input name="password" type="password" id="password" size="10" maxlength="10" /></td>
   </tr>
   <tr>
     <th align="right" scope="row"> </th>
     <td><input type="submit" name="Submit" value="Login" /></td>
   </tr>
 </table>
</form>
<?php } else { ?>
<p>Welcome <?php echo $_SESSION['MM_Username']; ?></p>
<?php } ?>

Balmung-San · February 12, 2007

The problem isn't in the:

if (!isset($_SESSION)) {

session_start();

}

Could you explain where you got to this conclusion? If I remember correctly $_SESSION is a predefined superglobal variable, meaning it's always set. If you're so confident that it's not wrong, please humor us by replacing it with just session_start().

As well, what output are you getting? We have no idea what's wrong if you don't tell us what output you get.

SauloA · February 12, 2007

I change:

if (!isset($_SESSION)) {

session_start();

}

into:

if (!isset($_SESSION)) {

session_start()

}

and I get an error on my page. When I change it back, I no longer have the error.

Balmung-San · February 12, 2007

You just removed the semi-colon, and PHP requires the semi-colon.

Change:

if(!isset($_SESSION)) {

session_start();

}

into:

session_start();

The Little Guy · February 12, 2007

To make sure that the session actually has a value in it, try this:

print_r($_SESSION);

The value you could be storing in the variable could be white space, so it is actually printing out something, but since it is possibly a space, then that would be what your seeing.

SauloA · February 12, 2007

Well, I changed:

if(!isset($_SESSION)) {

session_start();

}

into:

session_start();

and I get no errors and the page works fine. But, that wasn't the issue I was trying to solve. I want the login form to appear when a user isn't logged in. If a user is logged in then the form should be hidden and "Welcome username" should appear.

SauloA · February 12, 2007

I'm essentially trying to check is there is anything in my session variable. If my session variable is empty then it should display the login form. If there is something in the session variable then the login form should no longer display and "Welcome 'username' " should appear. My " if " statement begins here:

<?php if ($_SESSION['MM_Username'] == 0) { ?>
<form ACTION="<?php echo $loginFormAction; ?>" METHOD="POST" name="frmLogin" id="frmLogin">
 <table>
   <caption>
     Login to the site
   </caption>
   <tr>
     <th align="right" scope="row">Username:</th>
     <td><input name="username" type="text" id="username" size="10" maxlength="10" /></td>
   </tr>
   <tr>
     <th align="right" scope="row">Password:</th>
     <td><input name="password" type="password" id="password" size="10" maxlength="10" /></td>
   </tr>
   <tr>
     <th align="right" scope="row"> </th>
     <td><input type="submit" name="Submit" value="Login" /></td>
   </tr>
 </table>
</form>
<?php } else { ?>
<p>Welcome <?php echo $_SESSION['MM_Username']; ?></p>
<?php } ?>

but I don't know how to do what I explained above.

Jessica · February 12, 2007

if(isset($_SESSION['MM_Username']) && $_SESSION['MM_Username']){

print $_SESSION['MM_Username'];

}else{

//Login form.

}

SauloA · February 12, 2007

That code worked perfectly jesirose. So far most or all of the problems I've posted on PHP Freaks have been solved by you. Thanks a bunch.

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[SOLVED] Displaying a username when a user is logged in, HELP!

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