! - sign adds a negative meaning to the statement.
Thus,
if(!isset($_SESSION['uid'])) means = If (NOT isset($_SESSION['uid'])))
you know that if(statement) will work if "statement" returns a "true"
value...so if $_SESSION['uid'] is already set(has some value),
isset($_SESSION['uid']) will also return a true value...
so, by writting if(!isset()) { ...some_actions... }, some_actions will be
executed only if uid session is not set (doesn't contain any value).
Hope this helps,
Muhammed Mamedov
----- Original Message -----
From: "Kieran Hood" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Monday, December 22, 2003 1:26 PM
Subject: [PHP-DB] Database Array Question
> I'm new to PHP and was wondering if there's any better way of retrieving
> the records from the query other than using an array (This query will only
> return 1 row):
>
> $auth = mysql_query("SELECT username, userlevel, password, email FROM
> users WHERE id = '".$_SESSION['uid']."'");
> while ($row = mysql_fetch_array($auth))
> {
> $password = $row['password'];
> }
>
>
> Also, can anyone tell me what the relevance of adding '!' is to an if
> statement eg.
>
> This:
> if(!isset($_SESSION['uid']))
>
> Instead of this:
> if(isset($_SESSION['uid']))
>
>
> Thanks
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