[Resolved] PHP IF or - problem

fean0r

I've been testing the below code and I seem to be getting a strange result.

$a="1";
if (($a == "1" || "2")) { echo "YES<-----<BR><BR>"; }
if (($a != "1" || "2")) { echo "NO<-----<BR><BR>"; }

When I run this code it returns both echo's

YES<-----

NO<-----

But surely it should only echo 1 of them? Perhaps my code is wrong so any help would be great.

Thanks.

Installer

Your code is wrong.

if (($a == "1") || ($a == "2"))

is right.

Your clause:

(($a == "1" || "2"))

resolves to true because even though "$a" is not equal to "1", "2" is cast to true because it's not equal to 0. So one of the "or" conditions is true, making the whole clause true.

fean0r

$a="1";
if (($a == "1") || ($a == "2")) { echo "Test Yes<---<BR><BR>"; }
if (($a != "1") || ($a != "2")) { echo "Test No<---<BR><BR>"; }

Thanks, I've used the code as you suggested but still get 2 results.....

Test Yes<---

Test No<---

Any other ideas?

fean0r

I've used an else statement which seems to be working fine.

laserlight

Next time resolve the conditionals:

($a == "1") || ($a == "2")
= (1 == 1) || (1 == 2)
= 1 || 0
= 1

($a != "1") || ($a != "2")
= (1 != 1) || (1 != 2)
= 0 || 1
= 1

So both expressions resolve to 1, a true value.
Using an else statement is the way to go, as you have discovered.

fean0r

Now it all makes sense, thanks laserlight.