Why am I getting all the messages from this newsgroup delivered to my
mailbox?
I subscribed to the newsgroup -- or at least that is all I wanted.
Thanks.
-Doug
- Original Message -
From: Jason Soza [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Sent: Tuesday, June 11, 2002 9:36 AM
Subject: Re: [PHP] Different Problem [Re: Passing a Variable to PHP Via the
URL]
Rather than just setting globals on in php.ini, try this:
Your printf() line is:
printf(Variables: %s\nbr, $HTTP_GET_VARS[id]);
And your if() statement is:
if($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\nbr, $HTTP_GET_VARS[id]);
This line works - there IS a variable named 'id' in my page and it
has the correct value, 1.
if ($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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