PHP IF Then Print

Watcher

Ok i am beyond new, i had my php friend (who is wacked out over gf and cant help) to write me a few scripts that gather info. I am trying to modify a form he made for me. I want to be able to have users fill out a form and check a box and IF the box is checked then print specific text, if it is not then continue to next if statement. Am i on the right track because code appears to be missing?

<? $zone=$_POST['zone']; if($zone == "") { print " stuff, stuff

" }
{ ; die(); } ?>

<? $zone2=$_POST['zone2']; if($zone2 == "")
{ print "blahblah blah" }
{ ; die(); } ?>

NogDog

A checkbox value will only appear in the $POST array if it was checked, so you can simply do this check in the code:

<?php
if(isset($_POST['zone']))
{
  print "Stuff...........";
}
elseif(isset($_POST['zone2']))
{
  print "Blah...........";
}
?>

If you want both "zones" to print if both checkboxes are checked, then change the elseif to just another if.

Watcher

Wow thanks alot for the info i have implemented this and it works like a charm! Now my guru i have yet another question.

In this script prior to printing the current example above, I have the program sending me an e-mail with the notifications of what the user is trying to use.
ex is code before html
// requested sites
$body .= "</tr></table> \n Requested Sites to work:"; if(isset($POST['zone1']))
{
print " ZONE1 / "; } if(isset($POST['zone2']))
{
print "ZONE2 / "; } if(isset($_POST['zone3']))
{
print "ZONE3\n \n "; }

however only the "Requested sites to work" text appears in my notification e-mail and the other info is displayed at the top of the webpage instead of within my e-mail, your help with this matter is much appreciated

NogDog

You probably want to replace the print commands with more $body .= commands.