Page 2 - Drop Down Menu - populated from a mysql database

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<?php
$result = mysql_query("SELECT type FROM category") or die(mysql_error()); while ($row = mysql_fetch_array($result)) {
$type=$row["type"];
$options.= '<option value="'.$row['type'].'">'.$row['type'].'</option>';
};?>





<SELECT NAME=catid>
<OPTION VALUE=0>Choose</OPTION>
<?php echo $options; ?>
</SELECT>

I have a similar problem, except:

I have 2 tables, one called users which consists of id, name, distance (which in this instance say distance is set at 500)
I then have another table called location, this has id, name, distance in it and lets say for example this data
id=1, location=london, Distance=50

What i need to do is when a user logs in, they select the location from the drop down menu which is populated from the table, this then in turn takes 50 (distance) from the users distance (leaving 450)

Has anyone got a code example I could study to try and work this out?

thanks for sharing

I'm new to php myself but use something similar on my site when doing a search and use the following code assuming the 3 dropdown boxes are named a, b and c

Great work ! I will keep following this thread!!

I have a similar situation.

I have a database table with all 50 states. I am asking users to pick their 3 favorite states in order.
The first pulldown is easy, using this method.
The second pulldown needs to eliminate the selection of the first pulldown. And the third pulldown needs to eliminate the selections of the first and second pulldowns.