also try to:
echo mysql_error();
after the select-query, it will tell you what went wrong.
Thomas
On Fri, 17 May 2002 18:45:07 +0100
[EMAIL PROTECTED] (Randum Ian) wrote:
> 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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