The mysql_select_db() function is going to return a handle. You need to
capture that handle and use it as the second parameter in your mysql_query()
function. Although this is not required in certain cases where the server
automatically knows which database to work with, chances are thats whats
giving you the error.
Use like this..
$db = mysql_select_db();
mysql_query($query, $db);
-Kevin
----- Original Message -----
From: "Randum Ian" <[EMAIL PROTECTED]>
To: "PHP" <[EMAIL PROTECTED]>
Sent: Friday, May 17, 2002 11:45 AM
Subject: [PHP] mysql error
> Hi all,
>
> I have got this code but I can get it to work, it doesnt make sense!
>
> <snip>
> <?php
>
> $host="localhost";
> $username="user";
> $password="pass";
> $database="admin";
>
> $mysqlconnect = mysql_connect($host, $username, $password) or die
("Cannot
> connect to mySQL server, check it is running");
>
> mysql_select_db($database, $mysqlconnect) or die ("Cannot connect to the
> database, check it is set right");
>
> $result = mysql_query("SELECT * FROM user");
>
> $count = mysql_num_rows($result);
>
> $counted = number_format($count);
>
> echo $counted;
>
> ?>
> <snip>
>
> but I am getting this error - "Warning: Supplied argument is not a valid
> MySQL result resource in /usr/home/e/a/eagadmin/public_html/main.php on
line
> 14" but I dont understand why!
>
> Can anyone help?
>
> Regards, Ian.
> ---
> Randum Ian
> DJ / Reviewer / Webmaster, DancePortal (UK) Limited
> [EMAIL PROTECTED]
> http://www.danceportal.co.uk
> DancePortal.co.uk - Global dance music media
>
>
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