Makeing a form displayed or not based on a variable

BRONIC · #1 (**permalink**) Apr 19th, 2007, 04:23

Hey i was wondering if someone could help me with my script to make it so that if you have logged in then the login form on the main page disapears and when you log out it is visible?

Here is the code:

<?php
if(!$_SESSION['name']){
you are logged in.<br>
view you profile <a href="Profile.php">here</a><br>
or logout <a href="logout.php">here</a>.
} else {
//This is where i need to make the form invisible
}
?>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<title>Untitled Document</title>
<style type="text/css">

</style></head>

<body>
<div align="left"></div>
//this is the form i need to make invisible when !$_SESSION['name'] is used in the php script at the top
<form name="login" method="post" action="Login.php?action=login">
<p align="center">Login</p>
<p align="center">Username:
<input type="text" name="username">
Password:
<input type="password" name="password">
</p>
<p align="center">
<input type="submit" name="Submit" value="Login">
</p>
<p align="center">Don't have a username? Register <a href="Register.php">here </a></p>
</form>

Please help me

Thanks
BRONIC

Accurax · #2 (**permalink**) Apr 19th, 2007, 09:37

if(isset($_SESSION['logged_in'])){

Put whatever u want to display here when the user is logged in

}
else {

put the logged out content here

}

**The above assumes your using a session to track the users status.

Hope this helped

BRONIC · #3 (**permalink**) Apr 19th, 2007, 11:40

Is this what you mean?

<?php
if(isset($_SESSION['name'])){
you are logged in.<br>
view you profile <a href="Profile.php">here</a><br>
or logout <a href="logout.php">here</a>
}
else {
<form name="login" method="post" action="Login.php?action=login">
<p align="center">Login</p>
<p align="center">Username:
<input type="text" name="username">
Password:
<input type="password" name="password">
</p>
<p align="center">
<input type="submit" name="Submit" value="Login">
</p>
<p align="center">Don't have a username? Register <a href="Register.php">here </a></p>
</form>
}
?>

masonbarge · #4 (**permalink**) Apr 19th, 2007, 13:44

Quote:

Originally Posted by BRONIC

Is this what you mean?

<?php
if(isset($_SESSION['name'])){
you are logged in.<br>

First thing - you need to get the matter inside {} in echo/print format.

Headed your way: Parse error: syntax error, unexpected T_STRING

Accurax · #5 (**permalink**) Apr 19th, 2007, 15:52

to print anything to the screen.. tables divs.. in fact any html tag use this

Code: Select all

<?php
echo "
tags and other stuff here
";
?>

oh... and if you need to use "" inside the echo statement you need to escape them as follows:

Code: Select all

<?php
echo "<a href=\"url\">Text to be displayed</a>
";
?>

BRONIC · #6 (**permalink**) Apr 20th, 2007, 04:00

Never mind i fixed it this morning and it works like this:

<?php
session_start();
header("Cache-control: private");
if (!$_SESSION['name']){
print "<form name='login' method='post' action='Login.php?action=login'>
<p align='center'>Login</p>
<p align='center'>Username:
<input type='text' name='username'>
Password:
<input type='password' name='password'>
</p>
<p align='center'>
<input type='submit' name='Submit' value='Login'>
</p>
<p align='center'>Don't have a username? Register <a href='Register.php'>here </a></p>
</form>";
}
else {
print "<p align='center'>you are logged in.</p>
<p align='center'>view you profile <a href='Profile.php'>here</a></p>
<p align='center'>or logout <a href='logout.php'>here</a></p>";
}
?>

Thanks for your help though it is greatly appreciated!!

-BRONIC

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Makeing a form displayed or not based on a variable