Thanks. Since I'm using not

aharown07 — Thu, 06 Aug 2009 15:31:52 +0000

Thanks. Since I'm using not equal, the && is actually the right way to do it. When you're IFing on "doesn't equal" things get counterintuitive... so in the past I've ended up with an always true condition.
With "&&" (or equivalent) it works because any $type that isn't "contentype1" or "contentype2," will return true for both "doesn't equal" tests.

In short, when you're working "doesn't equal" you achieve "isn't either...or..." by coding "isn't... and isn't..."

aharown07 wrote: Conditional

greg — Thu, 06 Aug 2009 11:42:08 +0000

aharown07 wrote:

Conditional logic still confuses me a bit.

It can sometimes be tricky depending on what you are trying to do, and if you use equals or does not equal, and or etc. The combinations can get tricky.
Trial and error and testing with the code you're working with to see what it outputs is the best way.

You can simplify your code. If you think:
"&&" is more like "BOTH" (or "ALL" if more than two conditions)

With this:

<?php
if ($type != 'contenttype1' && $type != 'contenttype2'):
?>

Simply means for IF to be TRUE, BOTH conditions have to NOT EQUAL

If your $type variable was set to either "contenttype1" or "contenttype2", then it would return FALSE, as one of the conditions WOULD be equal
I.E. BOTH are "not" NOT EQUAL because one if them IS EQUAL

If however your $type variable was set to "contenttype5", then it would return TRUE, as BOTH conditions are NOT EQUAL to it

You are looking for the var to NOT be BOTH specific things.

------
Whereas using "OR"
"OR" is is more like "EITHER" (or "ANY" if more than two conditions)

The first code you posted:

<?php
if ($type != 'contenttype1' OR $type != 'contenttype2'):
?>

Simply means for IF to be TRUE, EITHER conditions has to NOT EQUAL

This actually would always return TRUE, as $type cannot be both contenttype1 and contenttype2
EG
If $type is set to "contenttype1" then it != "contenttype2" - TRUE
If $type is set to "contenttype2" then it != "contenttype1" - TRUE

With OR you usually use either two different variables to check (or more) or you don't use "does not equal".
EG

<?php
if ($type == 'contenttype1' OR $type == 'contenttype2'):
?>

Simply means for IF to be TRUE, EITHER one of the conditions has to be EQUAL

If your $type variable was set to either "contenttype1" or "contenttype2", then it would return TRUE, as one of the conditions WOULD be equal
If however your $type variable was set to "contenttype5", then it would return FALSE, as NEITHER conditions are EQUAL to it

You are looking for the var to be EITHER of two specific things.

---------
Again, it can get tricky.
If you are doing something that you cannot get your head around, try outputting something simple.

<?php
$type = "test text";
if ($type != 'contenttype1' && $type != 'contenttype2'){
echo "was true";
}else{
echo "was false";
}
?>

Won't that result in the

jamespvee — Thu, 06 Aug 2009 07:44:24 +0000

Won't that result in the condition having to satisfy both since it's an AND statement?

Many thanks! That makes

aharown07 — Tue, 28 Jul 2009 10:52:02 +0000

Many thanks! That makes sense to me now.

Code should be:
pr0gr4mm3r — Mon, 27 Jul 2009 22:57:56 +0000

Code should be:

`<?php if ($type != 'contenttype1' && $type != 'contenttype2'): ?>`

Thanks. Since I'm using not

aharown07 wrote: Conditional

Won't that result in the

Many thanks! That makes

Code should be: pr0gr4mm3r — Mon, 27 Jul 2009 22:57:56 +0000 Code should be: <?php if ($type != 'contenttype1' && $type != 'contenttype2'): ?>

Code should be:
pr0gr4mm3r — Mon, 27 Jul 2009 22:57:56 +0000

Code should be:

`<?php if ($type != 'contenttype1' && $type != 'contenttype2'): ?>`