Alexhoward

Here's the answer. include("whatever.php"); mysql_connect($host, $db, $pass) or die ("Could not connect to mysql because ".mysql_error()); mysql_select_db($db) or die ("Could not select database because ".mysql_error()); $mysite_username = $HTTP_COOKIE_VARS["sitename"]; $year=mysql_query("SELECT dob FROM `profile` WHERE username = '$mysite_username'"); while($year1 = mysql_fetch_array($year)) $test = $year1['dob']; list($a, $v, $c) = explode('-',$test); echo $c yipee! i solved one

it appears that it's not actually bringing anything back. yet if i do the same query in phpmyadmin, it works....?

Hi Andy, because when i do that i get the errors as above... also, i have just been testing the code, and it appears that is does not work correctly...? Thanks for taking the time to look at this

Hi Andy, Thanks for the reply I got it going like this, which i assume works, but now it's only single digits, e.g: 0 instead of 00 is there a way to make number_format double digits..? Thanks <?php $year=mysql_query("SELECT dob FROM $table WHERE username = '$mysite_username'"); while($year1 = mysql_fetch_array($year)) echo $year1['dob']; echo number_format(date($year1,"dd")); ?>

Hi guys, back again so soon, sorry. but i'm getting an error code from this piece for script that i don't understand: Undefined offset: 2 and : Undefined offset: 1 am i not doing this correctly...? <?php include("whatever.php"); mysql_connect($host, $db, $pass) or die ("Could not connect to mysql because ".mysql_error()); mysql_select_db($db) or die ("Could not select database because ".mysql_error()); $mysite_username = $HTTP_COOKIE_VARS["sitename"]; $year=mysql_query("SELECT dob FROM $table WHERE username = '$mysite_username'"); while($year1 = mysql_fetch_array($year)) $year1['dob']; list ($y, $sm, $d) = explode ('-', $year1); echo number_format($d); ?> Thanks for all your help,

Nice one! so if i just bang the php in there to grad a number from mysql, that's what it'll show! e.g. <option value="<?php ?>" selected="selected"><?php ?></option> however, this would put the number in twice...? for example php "selected", 1,2,3,4,... but i can live with that! cheers

Hi guys, hoping someone here might have an idea about this one... I looked around the internet but can't find a soulution... Basically does anyone know how to set a default value to a select drop down box? Thanks in advance