Problem populating a dropdown box with MySQL query results (PHP/MySQL)

Question

so as the title states, using the following code I have got it populating the dropbox with a single result from the query, that result being the latest added in the table.

here is my code:

<?php 
$query = "SELECT * FROM units_tb WHERE user_id='$userid'";
$result = mysql_query($query) or die (mysql_error());
while($row = mysql_fetch_assoc($result)){
$aa = "<option value='{$row['unit_id']}'>{$row['unit_code']}</option>";
}
?>
<select name="t_unit"><? echo $aa; ?></select>

The odd thing is, I use this same code for another field, and it works, populating the dropdown with all the results, however in this case it only fills in the last unit code in the table and not all of which are attached to the particular user id.

I would appreciate anyones thoughts :D

thanks

as bensiu points out below, you are not appending to your string, you're overwriting each time. But if you're doing this multiple places, why isn't it a function that you pass an array to and get the HTML back from? — Erik
– Erik, Commented Mar 21, 2011 at 13:34
Do a sanity check... how many rows are being returned by the query? Run the exact query against mysql directly -- do you get the results you expect? — ybakos
– ybakos, Commented Mar 21, 2011 at 13:36
Is the user_id attribute of the units_tb table an int or a string? — ybakos
– ybakos, Commented Mar 21, 2011 at 13:37

bensiu · Accepted Answer · 2011-03-21 13:32:19Z

4

$aa .= "<option value='{$row['unit_id']}'>{$row['unit_code']}</option>";

add . before = and initiate $aa = ''; before while loop

answered Mar 21, 2011 at 13:32

bensiu

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2 Comments

ybakos Over a year ago

Good eye... a classic syntax mistake.

dapperwaterbuffalo Over a year ago

apologies, i do now feel like a fool. it was just the . thanks though!

j_freyre · Accepted Answer · 2011-03-21 13:33:55Z

0

<?php 
$query = "SELECT * FROM units_tb WHERE user_id='$userid'";
$result = mysql_query($query) or die (mysql_error());
$options = "";
while($row = mysql_fetch_assoc($result)){
    $options .= "<option value='{$row['unit_id']}'>{$row['unit_code']}</option>";
}
?>
<select name="t_unit"><? echo $options; ?></select>

should work. You forgot a . in your while loop

answered Mar 21, 2011 at 13:33

j_freyre

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1 Comment

bensiu Over a year ago

and $aa has not been initiated

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