Select and Selected

All,
I have the following code:


<?php
$qrypic = "Select * from pictures where picture_id='$picid'";
$resultpic = mysql_query($qrypic);
$resultsetpic = mysql_fetch_array($resultpic);
?>

<select name="job">
<?php
$qryjobs = "Select job_id, name from jobs order by name ASC";
$resultjobs = mysql_query($qryjobs);
while($resultsetjobs = mysql_fetch_array($resultjobs)){
if($resulsetjobs['job_id'] == $resultsetpic['job_id']){
?>
<option value="<?php echo $resultsetjobs['job_id']; ?>" selected><?php echo $resultsetjobs['name']; ?></option>
<?php
}else{
?>
<option value="<?php echo $resultsetjobs['job_id']; ?>"><?php echo $resultsetjobs['name']; ?></option>
<?php
}
}
?>
</select>


It should make the row that matches selected but for some reason it's not working. Can anyone see what I'm doing wrong?

Thanks in advance.

What does your output look like in view source?

This is one problem .... see missing red letter ...

if($resultsetjobs['job_id'] == $resultsetpic['job_id']){

But I would also say ... unless there is only one $resultsetpic['job_id'],
it will never be equal. So, check and see what it is by using an echo ...
See what those array variables hold.

echo $resultsetpic['job_id'];
echo $resultsetjobs['job_id']."<br>";
if($resultsetjobs['job_id'] == $resultsetpic['job_id']){

That was it. Sorry for bothering you with stupid spelling errors.

Nice spot mlseim! Your eyesight is impeccable :p