Posted by Mark Sargent on 05/12/05 10:17

Hi All,

ok, this revised code produces the error below,

<?php
mysql_select_db("status",$db);
$maker_result = mysql_query("SELECT Makers.maker_id Makers.maker_detail
FROM Makers",$db);
$maker_num = mysql_num_rows($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.
*

• Next in forum: Recursive function - nothing in return values!
• Prev in forum: Re: [PHP] Inner Join or 2nd Query...?
• Thread view: Re: [PHP] Inner Join or 2nd Query...?

[Reply to this message]