PHP Code:

<?php
        //info list
        include "../conn.php";
        $id=$_GET['id'];
        echo 'id is '.$id.'<br />';
        $sql=mysql_query("SELECT * FROM work WHERE id='$id'");
        if(!$sql)
            { echo mysql_error(); }
        echo '<li>'.mysql_result($sql, 1).'<li>';
        ?>

I keep getting the following result

id is 1

Warning: mysql_result() [function.mysql-result]: Unable to jump to row 1 on MySQL result index 7 in /home/inspin/public_html/work/details.php on line 13

I dont know why, because id comes up as 1, and there is a row in the work table which has an id of 1!

Any reason your using mysql_result function.

If your resultset is expected to only return one row then the first and only row would be referenced with a 0 (numbering starts at 0). A 1 would be referencing a row that might not exist so the error.

mysql_result($sql, 0)//also needs third parameter(field name) as your selecting *all

You'd be better using one of the mysql_fetch functions. http://uk2.php.net/manual/en/functio...etch-assoc.php

Ah thanks it's been a while since I've used the mysql_result() function and had forgot how to use it properly!