Rather than just setting globals on in php.ini, try this:

Your printf() line is:
printf("Variables:  %s\n<br>", $HTTP_GET_VARS["id"]);

And your if() statement is:
if($id) {}

Where is $id set? It's probably not. $HTTP_GET_VARS["id"] doesn't set 
$id. If you want the "id" variable in your if(), you need:
if($HTTP_GET_VARS["id"]) {}

Or alternately do:
$id = $HTTP_GET_VARS["id"]
if($id) {}

Although that's not necessary. Also, try using $_GET["id"], as 
$HTTP_GET_VARS[] has been deprecated in newer versions.


Jason Soza

----- Original Message -----
Date: Tuesday, June 11, 2002 8:16 am
Subject: [PHP] Different Problem [Re: Passing a Variable to PHP Via the 

> I found out that in fact PHP is creating a variable with the name 
> and value I'm passing through a URL from the querystring.  But 
> it's still not working as planned.
> The url server/test.php?id=1 creates the following results in my code:
> printf("Variables:  %s\n<br>", $HTTP_GET_VARS["id"]);
> This line works - there IS a variable named 'id' in my page and it 
> has the correct value, 1.
> if ($id) {}
> This fails.  If I test 'id' instead of '$id' then it works but my 
> page doesn't seem to equate 'id=1' with the presence of $id.
> $result = mysql_query("SELECT * FROM employees WHERE id=$id",$db);
> This doesn't work - again it seems $id isn't being treated 
> properly.  I get this error:
> Warning: mysql_fetch_array(): supplied argument is not a valid 
> MySQL result resource
> If I hardwire my page with the line '$id=1;' before the if 
> statement and the query everything works.
> So why isn't the variable from my URL being treated properly?
> Jesse

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