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. *
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Cheers.
Mark Sargent.
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