Stupid If/Then kickin' my ****

Duell

It's late and I'm only getting more frustrated, which is making it worse.

I have two variables and if either one == x, I want to set a third variable...

In essence this:

if ( (ISSET($Feb27,$Feb28))&&(($Feb27 or $Feb28)=="NA") )
{
$FORM3="skip_it";
}

But it isn't working.

Any help would be much appreciated. I'm going to go and get a couple hours sleep now.

Thanks!

stuartc1

try this, should work.

if (isset($feb1) && isset($feb2)) {
   if ($feb1 == "NA" || $feb2 == "NA") {
    $form3 = "skipit";
   }
}

Duell

Originally posted by stuartc1
try this, should work.
if (isset($feb1) && isset($feb2)) {
   if ($feb1 == "NA" || $feb2 == "NA") {
    $form3 = "skipit";
   }
}
[/B]

THANK YOU! A much better && cleaner way to do it! 🙂

austingecko

I'm having similar frustrations with a similar problem.

Simple, but what am I missing here:

				if ($usertype != "1" || $usertype !="2")
		    echo "$usertype";
                die('');
				}

When $usertype is 1 the above should be passed over, but it isn't.

If I just do ($usertype != "1") it works. So, I must be doing the || incorrectly.

Any suggestions? Thanks in advance.
btw, my php formatting doesn't look like the php formatting above. I tried to make a look prettier, but it kept getting uglier. 🙁

itguysam

YOu need to bracket the two commands with a {} and the || needs to be && because a number cant be two values at once so one of them will always return true. You want them both to return true to show that it is not 1 or 2.
Try this:

if ($usertype != "1" && $usertype !="2"){
            echo "$usertype";
                die('');
}
}//You may not need this bracket I'm not sure if it is attached
//to something higher up in your program

austingecko

oops, yes the bracket is there. I must have accidently deleted while trying to make my code pretty here.

Thanks. I don't understand why this doesn't work though...

itguysam

I didnt look hard enought check my post again I edited it.

austingecko

Oh ok. That works. 🙂

I didn't use && because I assumed, incorrectly, that $usertype would have to 1 & 2 when I wanted it to be 1 or 2.

Your explanation makes sense. Usertype will always be true. Thanks again