using List() with array PHP, Why won't this work?

Question

<?php
$query = mysql_query("SELECT `uid`,`name` FROM `location`");
while($result = mysql_fetch_assoc($query)) {
    list($location_id, $location_name) = $result;
?>
<option value="contactLocation.php?view_id=<?php echo $location_id; ?>"><?php echo $location_name; ?></option>
<?php
}
?>

Can someone help me and tell me why this isn't working? I'm trying to put the uid and name columns into those two variables in the list.

sigh Do you really need to be asked what "isn't working" is supposed to mean? — Ignacio Vazquez-Abrams
– Ignacio Vazquez-Abrams, Commented Nov 1, 2013 at 23:21
usually I just do $result['uid'] and $result['name'] but i want to put this all in one array to seperate variables. I seen this method used before but not exactly sure if this is right. It looks correct. $result is an array. edit: The variables are empty? — TheCryptKeeper
– TheCryptKeeper, Commented Nov 1, 2013 at 23:23

Ry- · Accepted Answer · 2013-11-01 23:25:27Z

1

You’re fetching an associative array. You can use mysql_fetch_row or mysql_fetch_array instead for list() assignment.

Better yet, upgrade to PDO! (Or MySQLi!)

answered Nov 1, 2013 at 23:25

Ry-♦

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2 Comments

TheCryptKeeper Over a year ago

Thanks, it seems mysql_fetch_assoc doesn't work with list() for php for some reason. Cheers thank you for your time and help.

Ry- Over a year ago

@TheCryptKeeper: Yep, see the notes section in the docs; it only works with numeric arrays.

AbraCadaver · Accepted Answer · 2013-11-01 23:33:08Z

0

You can also assign it right from the fetch:

<?php
$query = mysql_query("SELECT `uid`,`name` FROM `location`");
while(list($location_id, $location_name) = mysql_fetch_array($query)) {
?>
   <option value="contactLocation.php?view_id=<?php echo $location_id; ?>"><?php echo  $location_name; ?></option>
<?php
}
?>

answered Nov 1, 2013 at 23:33

AbraCadaver

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