i wrote a really basic script sucessfully to echo the url variable to the
screen... will go from there. thanks beau.

The single quotes around the $id variable in the select statement worked
wonderfully. thanks John.


-----Original Message-----
From: John W. Holmes [mailto:holmes072000@;charter.net]
Sent: Thursday, 7 November 2002 1:20 PM
To: 'Gavin Amm'; 'Php-Db (E-mail)'
Subject: RE: [PHP-DB] Warning - newby question -- $_GET

> I'm trying to call a script with the "?=" after the php script name in
> URL, but can't seem to pick up the variable to use in my script...
> any help would be greatly appreciated :)
> I'm calling the script with this url:
> http://localhost/Data/test/dbconnect2c.php?id=TestPage
> The section of the script i'm using is:
> $id = $_GET['id'];
> $sql="SELECT * FROM content where title=$id";
> $result=mysql_query($sql,$db);

Well, for your example, $id is going to equal "TestPage", so you are
making an invalid query: where title=TestPage. If you echo
mysql_error(), it'll probably say something about unknown column

At the very least, use quotes:

where title='$id'

> Error message returned:
> Warning: mysql_fetch_array(): supplied argument is not a valid MySQL
> result
> resource in...

This means your query failed. Echo mysql_error() to determine why. Get
used to using mysql_error for good debugging.

---John Holmes...

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