Good day,

You get this message because the mysql_db_query statement failed, and that
your code didn't check to see that it did or not before proceeding to the
next statement.

And, this statement fails because you have not even connected to the
database yet.

At the risk of sounding condescending, I would request that you look at
PHP's documentation on how to use the MySQL wrapper functions (
http://www.php.net/manual/en/ref.mysql.php ).  There you can find good
documentation and a number of useful examples (the two best things a coder
could ask for).

Your error message is also in PHPs FAQ list.
http://www.php.net/manual/en/faq.databases.php .

Hope this helps,

============================
Darren Gamble
Planner, Regional Services
Shaw Cablesystems GP
630 - 3rd Avenue SW
Calgary, Alberta, Canada
T2P 4L4
(403) 781-4948


-----Original Message-----
From: cosmin laslau [mailto:[EMAIL PROTECTED]]
Sent: Friday, March 15, 2002 4:14 PM
To: [EMAIL PROTECTED]
Subject: [PHP] Driving me nuts, need one second of your time


<?
$query = "SELECT * from mytable";
$result = mysql_db_query("db", $query);

while ($myarray = mysql_fetch_array($result))
{
$title = $myarray[title];
echo "$title<br>";
}
?>

Can someone PLEASE tell me why the coding above gives the following error:


Warning: Supplied argument is not a valid MySQL result resource in 
/var/web/somesite.com/html/index.php on line 5


It's absurd. It's driving me nuts. I'm about to introduce my computer to the

pavement 40 feet below.

Thanks in advance for whoever sees what I am sure is a glaring and obvious 
flaw in the coding. I've been looking at it for an hours and just can't get 
anything from where I'm standing, maybe a different perpective will help.



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