Re: Inner Join or 2nd Query...?

Richard Lynch wrote:

On Thu, May 12, 2005 12:17 am, Mark Sargent said:

Hi All, ok, this revised code produces the error below, <?php

You took out the mysql_connect completely??? Don't do that. Sorry, was following what you suggested b4, no..?

$db = mysql_connect("localhost", "root", "grunger"); Don't connect to the database twice. That's the most expensive (time-wise) thing in your whole script, probably.

mysql_select_db("status",$db); $maker_result = mysql_query("SELECT Makers.maker_id Makers.maker_detail FROM Makers",$db);

if (!$maker_result) die(mysql_error()); Ok, will add that

$maker_num = mysql_num_rows($maker_result);    //Line 40

?> <select name="slct_maker"> <?php for ($i=0; $i<$maker_num; $i++){ $maker_myrow=mysql_fetch_array($maker_result); $maker=mysql_result($maker_result,$i,"maker_detail"); $maker_id=mysql_result($maker_result,$i,"maker_id"); echo "<option value=\"$maker_id\">$maker</option><br>"; } ?> </select> *Warning*: mysql_num_rows(): supplied argument is not a valid MySQL result resource in */var/www/html/products.php* on line *40 Cheers. Mark Sargent. * -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php

Cheers.

Mark Sargent.