Querying one field for display in a drop-down box, but storing another field based...

[medar]

I have the following snippet of code that grabs Usernames and displays them based on whether they are active. This code places the result in a selectable drop-down box so you can click the box, scroll to the name you want, select it - and the name stores in the "name" field of a separate table.

<?
$result = mysql_query("SELECT UNum, UName FROM Users WHERE UActive='Yes' ORDER BY UName");
if ($myrow = mysql_fetch_array($result)) {
echo "<select name=name>";
do {
printf("<option>%s</option>\n", $myrow["UName"]);
} while ($myrow = mysql_fetch_array($result));
echo "</select>";
}
?>

Now this is probably easy, but I could not find a similar question here on the boards. I know this is possible in MSAccess...but how do I display the UName field so that you see the names in the drop down box...but STORE the UNum (User-ID field) in the separate table?

So I could select someone's name...and when I hit enter, their User ID would store in the separate table.

Thanks in advance for any help.

[sgoku01]

Like this:

<? 
$result = mysql_query("SELECT UNum, UName FROM Users WHERE UActive='Yes' ORDER BY UName"); 
if ($myrow = mysql_fetch_array($result)) { 
echo "<select name=name>"; 
while ($myrow = mysql_fetch_array($result))
{
echo "<option value=".$myrow['UNum'].">".$myrow['UName']."</option>";
} 
echo "</select>"; 
} 
?>

(sorry, didn't know how to use the sprintf-command 🙂)