while statement in drop down menu?

[garyed]

I'm having a problem with populating a second drop down menu that needs to be identical to the first from a mysql database.

I have about 20 items in the DB & I only want to show the first 9.

My code is:

      <select name="a_type" style="width: 175px;"> <option value="" selected="selected"> None selected</option> 
      <?php
$i=1;
while (($_produce = mysql_fetch_array($result_produce)) && ($i <=9 ))
{
$produce_type=$_produce["type"];
$produce_id=$_produce["id"];
$i++;
?>

<option value="<?php echo $produce_id; ?>"><?php echo $produce_id; ?> - <?php echo $produce_type; ?> </option>
<?php
}
?>
        </select>

 

That works fine in my first drop down menu but the second drop down menu with the same exact code will show items starting at 11 & higher

instead of 1 to 9. I haven't tried closing the DB & the reopening it which I assume would work but there are a lot of other variables & things I'd

have to check to make sure they don't get effected so I'd rather not.  Is there a simple way to solve this? 

 

[wildteen88]

You need to reset the internal data pointer. Look into using mysql_data_seek.

[garyed]

I appreciate the help but I've been messing with it all day & still haven't gotten it.

I can get the pointer back to whatever number I put in the argument but I just get that same number 8 times.

I even tried reset() & that helped a little populating the drop down but it locked up the database.

If  anyone could show me how to fix my code that would be greatly appreciated. 

 

[wildteen88]

Make sure you are using mysql_data_seek() before your second dropdown menu. Do not place it inside the while loop.

[garyed]

I finally got it!!

I was using ($_produce,0) as the argument when I needed to use ($result_produce,0).

Thanks