PHP in_array returns false positive

Question

I'm checking whether a checkbox should be checked against a user preferences array of id's with:

if(in_array($category['id'], $checkedarray)){
  $checked = "checked='checked'";
}

the result of checked array looks something like:

array(43) {
[0]=>
string(2) "31"
[1]=>
string(2) "32"
[2]=>
string(2) "34"
[3]=>
string(2) "35"
}

In the above array example, 33 still returns true even though its not in there.

I've tried using:

(in_array($category['id'],$checkedarray, true)

The results were same as without the strict condition.

Any suggestions would be greatly appreciated.

The if block is missing a ) before the opening {. I'm guessing that was just an error in translation?
sorry that was a typo, i retyped here instead of copy/paste. I'll change it in a sec, but it does have proper brackets, else it'd returns errors.
This works fine as shown. Can you show us the exact case that caused the problem? I ran your code at a site called Ideone, you can upload an example there, too. ideone.com/FaMiw
Ok, just a silly double check since I can't see a possible else resetting $checked, you don't happen to be declaring $checked outside the loop and forgetting to reset it the next time around the loop?

Mikhail · Accepted Answer · 2012-02-15 17:40:56Z

up vote 2 down vote accepted

33 does NOT return true -- you're simply never resetting your $checked variable.

if (true) { $checked = "checked" };
if (false) { /* never gets executed */ }

add an else!

if (in_array(33, $checkedarray)) {
    $checked = "checked='checked'";
} else {
    $checked = "";
}

answered Feb 15 '12 at 17:40

Mikhail
2,886631

	I feel like a tool, made the assumption that a blank else condition pretty much did that. – Mike Feb 15 '12 at 17:46
	I had an else, but it was blank, as in }else{ //} – Mike Feb 15 '12 at 17:47
	Don't beat yourself up. The only reason I know this is because I've been there :) – Mikhail Feb 15 '12 at 18:04

1 Answer