drop down list and $myrow["value"]

melrse

Hi,
I am having a problem with displaying the file value from mysql table in a form during update mode in a drop down list.

When I display the form for possible update, I want to show the value currently in the table as the first option value and then the remaining list in case the person wants to select the new value and I can't get it to work.

For example:

<SELECT NAME="Trigger">
???? <option value="<?=$myrow["Trigger"]?>how do i do this??</option>
<option value='1'> list 1 </option>
<...
<..

Can someone please help me with how to do this?

Thanks,
Melissa

ccalzaretta

your going to need this line
$num_results = mysql_num_row($sql_result)
<?
for ($i=0; $i <$num_results; $i++)
{
$row = mysql_fetch_array($sql_result);
$varid2 = $row["ProdID"];
?>
<option value=<%echo $varid2%>>
<?
echo $row["ProdCat"];
?>
</option>
<?
}
?>

you need to move
$row = mysql_fetch_array($sql_result);
to where i have it

melrse

Whew, I better wait to get into work tomorrow to look at my php book.
Thanks,
melissa

Anon

Hay it is really easy..
It is a for loop that just loops until end of record.

That is why you need to get a record count with
$mysql_num_row
it works well
you can copy and paste the code I sent you for the last post