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posting a selection from a dropdown box to a variable

Posted on 2004-08-27
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Last Modified: 2008-03-06
Hi-
This code displays a dropdown box containing a list of all my current clients.  I need the selected clients id number to post to $referedby.  

Any ideas?

$sql = "SELECT id, name FROM results";
$res_cat = mysql_query($sql); // handle to category set

<select name="clientid">
<option value="">-</option>
<?php while ($row = mysql_fetch_assoc($res_cat)) { ?>        
<option value='<?php echo $row['id']; ?>'><?php echo $row['name']; ?></option>              
<?php } ?>
</select>
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Question by:livegirllove
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3 Comments
 
LVL 27

Accepted Solution

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Diablo84 earned 500 total points
ID: 11918253
on your processing page

$referedby = $_POST['clientid'];

im not 100% sure thats what you meant, if not could you elaborate further please

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LVL 1

Author Comment

by:livegirllove
ID: 11918330
Exactly what I needed.  I wasn't thinking clearly ;-)  
I'm a newbie.  selectname = clientid means the selected name gets posted as clientid.  

thanks for the quick response.
0
 
LVL 27

Expert Comment

by:Diablo84
ID: 11918347
no problem :)

we all start off somewhere, these things will become more obvious to you as you learn, good luck!

|)iablo
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