trouble w/select menu from array

[Anon]

Hi there,

I have an array that populates a drop down menu. The user makes his choice and submits the form moving on to the next page.

The problem is that no matter what choice the user makes, the form always passes the last choice in the drop down menu from the previous page.

The code I'm using is below. Any ideas of how to fix this so the choice made by the user is the info that's passed to the next page?

Thanks very much,

Rick

---

<select name="GamesInProgress">
<?

/****************************************
This portion of the script sorts and loops through the array to populate the drop down menu
****************************************/

sort($PlayArray);
foreach ($PlayArray as $PlayInProgress) {
echo '<option>' . $PlayInProgress . '</option>';
}

$ContinuedGame = Array();
$ContinuedGame = explode ("-",$PlayInProgress);
$Category=trim($ContinuedGame[0]);
$Topic=trim($ContinuedGame[1]);

?>
</select>

[junebug]

add in the value attribute to the option tag:

<option value="1">one</option>

[rickdeet]

I tried it and it didn't work.

Here's what I tried:

---

<select name="GamesInProgress">
<?
/**************************************
This portion of the script sorts and loops through the array to populate the drop down menu
***************************************/

sort($PlayArray);
foreach ($PlayArray as $PlayInProgress) {
echo '<option value="1">' . $PlayInProgress . '</option>';
}

$ContinuedGame = Array();
$ContinuedGame = explode ("-",$PlayInProgress);
$Category=trim($ContinuedGame[0]);
$Topic=trim($ContinuedGame[1]);
?>
</select>

---

[junebug]

the option values need to be different. now they are all the value "1". try this:

for($count = 0; $PlayInProgress = each($PlayArray); $count++)
{
print "<option value=\"$count\">$PlayInProgess</option>";
}