Posted by Richard Lynch on 05/13/05 02:47

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.

> 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());

> $maker_num = mysql_num_rows($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.
> *
>
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