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.
HTH,
Jason Soza
----- Original Message -----
From: <[EMAIL PROTECTED]>
Date: Tuesday, June 11, 2002 8:16 am
Subject: [PHP] Different Problem [Re: Passing a Variable to PHP Via the
URL]
> 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
>
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
