Form select option help

[Anon]

Hello:

I have created a form that verifies required variables once the form has been submitted. I have done it this way for 2 reasons:
1. I don't want to use Javascript to check on the clientside the required fields have been filled in (ie) A dialog box with info on what hasn't been filled in.
2. I don't want to have them direct to another page and have to click the back button or use something to the effect like javascript:history.back(1) to return to the form to fix information required.

The problem that I am having with the code below is that the select menu is selecting everything as "selected". (ie) If I select Option 1 when I hit the Submit Info button Option 3 becomes the default select item. Any ideas on what I can do so the $MenuSelect variable isn't coming up true for every menu item?

Thanks,
Troy

Below is the snippet just for the menu section.

<form method="post" ACTION="<?echo$PHP_SELF?>">
<select name="MenuSelect">
<option>-- Select Item--</option>
<option<? if ($upload && $MenuSelect="Option 1") { echo " selected";}?> value="Option 1">Option 1</option>
<option<? if ($upload && $MenuSelect="Option 2") { echo " selected";}?> value="Option 2">Option 2</option>
<option<? if ($upload && $MenuSelect ="Option 3") { echo " selected";}?> value="Option 3">Option 3</option>
</select>
<input type="submit" name="upload" value="Submit Information">

[Anon]

<option<? if ($upload && $MenuSelect="Option 1") { echo " selected";}?> value="Option 1">Option 1</option>

Try using a '==' comparison in your if statement. Thus:

<option<? if ($upload && $MenuSelect == "Option 1") { echo " selected";}?> value="Option 1">Option 1</option>

This should do the trick. It was making number 3 the default because your browser saw all your options as 'selected' and therefore left it as the last one in the option list. If you view the old browser source you'll see all options as 'selected'. Viewing the source is a valuable tool for debugging.

Good luck,
Steven Livanes

[Anon]

You can actually omit the $upload part out of your if statement, because $MenuSelect will never have a value until after you hit submit

[Anon]

Thanks to all!

That did the trick. I figured it was something that simple.

Thanks,
Troy