Posted by Mark Sargent on 05/13/05 06:14

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.

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