[Ady]

'pages' Table:
http://img106.imageshack.us/img106/2240/sig2409sm.jpg

PHP Code:

<?

include("config.php");

$sql = "SELECT * FROM pages WHERE id='" . mysql_escape_string($_GET['user']) . "'"; 

$result3=mysql_query($sql);

$rows2=mysql_fetch_array($result3);

$user2=$rows2['id'];

$key2=$rows2['key'];

$title2=$rows2['title'];

$body2=$rows2['body'];

if($user==$user2 && $page==$key2){ echo ('<b>' . $title2 . '</b><br>' . $body2); };

?>

I'm sure I've done this right, but it doesn't work. I'm trying to make it so if you visit the page with ?user=#&page=#, it searches the table for rows with whatever $user and $page are. If they are both in the same row, then it displays 'title' and 'body' from that row. I hope someone understands what I'm talking about... any help is appreciated.

PS. Sorry my code is probably really messy, I'm not too good with PHP / MySQL.

EDIT: Fixed a little of the code.

[Nico]

The $page variable is not set, and the $user variable neither. At least can't I see them. And are you sure the $user variable is a number and not a name?

The table looks kinda weird too. The entries are duplicated.

[Ady]

Originally Posted by nico_swd

The $page variable is not set, and the $user variable neither. At least can't I see them. And are you sure the $user variable is a number and not a name?

The table looks kinda weird too. The entries are duplicated.

$page should be what the ?page= is in the URL, and the same with $user. $user is also definatley a number. To make sure, I've echoed the variable and it comes up with a number. The 'id' column in the table just represents what user the page is assigned to.

[Nico]

Try replacing $page with $_GET["page"] then. Or is the variable somewhere else set?

And I'd strongly recommend not using numbers in the var names. This is only confusing.

[Ady]

Originally Posted by nico_swd

Try replacing $page with $_GET["page"] then. Or is the variable somewhere else set?

And I'd strongly recommend not using numbers in the var names. This is only confusing.

I've changed $page to $_GET["page"] now, but still no luck. I think I need to do this too:

PHP Code:

$sql = "SELECT key FROM pages WHERE key='" . mysql_escape_string($_GET['page']) . "'"; 

mysql_fetch_array($sql); 

But it just comes up saying:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/smashroc/public_html/webedit/index.php on line 68

[Nico]

You forgot the mysql_query()

PHP Code:

$sql = mysql_query("SELECT key FROM pages WHERE key='" . mysql_escape_string($_GET['page']) . "'"); 

mysql_fetch_array($sql); 

[Ady]

Originally Posted by nico_swd

You forgot the mysql_query()

PHP Code:

$sql = mysql_query("SELECT key FROM pages WHERE key='" . mysql_escape_string($_GET['page']) . "'"); 

mysql_fetch_array($sql); 

Oh yeh lol. It still comes up with the same thing though.

[Nico]

Try this.

PHP Code:

$sql = mysql_query("SELECT * FROM pages WHERE key='" . mysql_escape_string($_GET['page']) . "'"); 

mysql_fetch_array($sql); 

[Ady]

Still doesn't work. It's starting to annoy me now lol. Thanks for your help so far anyway. I appreciate it a lot.

[Nico]

Ok, try this then:

PHP Code:

$sql = mysql_query("SELECT * FROM pages WHERE id='". $_GET["user"] ."' AND key='". $_GET["page"] ."'");

$row = mysql_fetch_array($sql); 

Now call your-site.com/?user=1&key=page1

This should work.