function not evaluating correctly

[rofl90]

Heres my function, it always returns 0...

 

function user_type($page) {
$user = $logged_user;
$level = mysql_query("SELECT * FROM users WHERE user='$user'");
$level_a = mysql_fetch_array($level);
$final_level = $level_a['usergroup'];
$check_level = mysql_query("SELECT * FROM userstructure WHERE name='$final_level'");
$check_level_a = mysql_fetch_array($check_level);
if ( $check_level_a[''.$page.''] == '1' ) {
	return "1";
} else {
	return "0";
}
}

 

Heres how I use it in a backend page: ex. testimonials:

 

if(user_type("testimonials") == "0") {
header("Location: notallowed.php");
exit();
}

[ohdang888]

try putting this:

if ( $check_level_a['$page'] == '1' ) {

 

instead of this:

if ( $check_level_a['page'] == '1' ) {

 

or is page is a vriable, do this:

if ( $check_level_a['{$page}'] == '1' ) {

[rofl90]

if ( $check_level_a['{$page}'] == '1' ) {

 

doesn't work, and yes, it's a variable acquired through

 

user_type("myVar") == 0 //etc

 

strange ;S Have I got the wrong syntax?

[rofl90]

Btwe I've used ''$var'' and it works..

[kenrbnsn]

The correct syntax for

<?php
if ( $check_level_a[''.$page.''] == '1' ) {
?>

is

<?php
if ( $check_level_a[$page] == '1' ) {
?>

 

Ken

[rofl90]

still doesn't work with that

[Jeremysr]

In this line:

 

$user = $logged_user;

 

$logged_user doesn't exist in that scope. If it exists outside the function you should use:

 

global $logged_user;
$user = $logged_user;

[rofl90]

thanks it worked!!!

 

thanks so much that was hours i took to fix it... and i didn't. thankyou so much!