Selecting from a drop down menu

[Anon]

Hi Guys,
Let's say i have the following drop down menu:

<select name="building_status">
<option selected> ----------------------- Select ------------------------</option>
<option value="1">Melbournian</option>
<option value="2">Regency Towers</option>
<option value="3">Melbourne</option>
<option value="4">Sydney</option>
<option value="5">Adelaide</option>
<option value="6">Brisbane</option>
<option value="7">Global</option>
</select>

In the mySQL DB i have a field called "building_status". This holds the values from the above drop down menu.
What i'm wanting to do is create an edit page where the option (name from the list) is selected depending on the value associated to it.
Can anyone tell me how to go about this?

Cheers,
maccam

[vasse]

You could retrieve building_status into the var $status and then do if-statements to print out "selected".

Something like this:

print "<option value='1'";
if($status == 1) { print " selected" };
print ">Melbournian</option>";
..
..
and so on

Maybe this isn't the best way to go at it but it works... (was working on something similar just now)

Hope that helps
Vasse

[vasse]

Oops, a small error in the code... Misplaced the ";". This should work though, it does the job for me.

print "<option value='1'";
if($status == 1) { print " selected"; }
print ">Melbournian</option>";

[pnoeric]

--or you can put the word 'selected' on an array with the key = value and then put the PHP inline with html, which looks a little nicer imho:

<option value='1'<?=$selected[1]?>>Melbournian</option>

<option value='2'<?=$selected[2]?>>Victorian</option>

<option value='3'<?=$selected[3]?>>Delorian</option>

easier to copy-and-paste too...?

best
Eric

[pnoeric]

to clarify - the code BEFORE you display the options would be

unset($selected);
$selected[$status] = ' selected';

then the code above.

best
Eric