in_array oddity

Check out this code:

// Start Code -------------
function test_in_array($val)
{
$a = array('key' => $val);
printf("in_array: %d, value:%s<BR>", in_array('key', $a), $a['key']);
}
test_in_array(0);
test_in_array(1);
// End Code ---------------

The output I get is:

in_array: 1, value:0
in_array: 0, value:1

Why does the second in_array() call fail???

Tom Barnes wrote:
> function test_in_array($val)
> {
> $a = array('key' => $val);
> printf("in_array: %d, value:%s<BR>", in_array('key', $a), $a['key']);
> }
> test_in_array(0);
> test_in_array(1);
> // End Code ---------------
>
> The output I get is:
>
> in_array: 1, value:0
> in_array: 0, value:1
>
> Why does the second in_array() call fail???

Well, the second call returns false because 'key' does not appear as a
value in the associative array. (Only as a key.) It also returns false
for values of 2, 3, etc.

I have no idea why the first call returns true, though... If you pass
true as the third parameter ("strict") to in_array, it will return false
as expected.

-- brion vibber (brion @ pobox.com)

Tom Barnes wrote:
> Check out this code:
>
> // Start Code -------------
> function test_in_array($val)
> {
> $a = array('key' => $val);
> printf("in_array: %d, value:%s<BR>", in_array('key', $a), $a['key']);
> }
> test_in_array(0);
> test_in_array(1);
> // End Code ---------------
>
> The output I get is:
>
> in_array: 1, value:0
> in_array: 0, value:1
>
> Why does the second in_array() call fail???


Wrong question! The right question is:

"Why does the first in_array() call return true?"


And the answer is:

Because 'key' is converted to numeric, to 0 (zero) and 0 (zero) *is* in
the $a array.



The second in_array() call tries to find a 'key' (or 0) but fails
because the value in the array is 1.






Try specifying the third parameter to the in_array() call ...

in_array('key', $a, true)



Happy Coding :-)

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nospam1978@yahoo.com (Tom Barnes) wrote in message news:<5af28966.0410111503.f6f243f@posting.google.c om>...
> Check out this code:
>
> // Start Code -------------
> function test_in_array($val)
> {
> $a = array('key' => $val);
> printf("in_array: %d, value:%s<BR>", in_array('key', $a), $a['key']);
> }
> test_in_array(0);
> test_in_array(1);
> // End Code ---------------
>
> The output I get is:
>
> in_array: 1, value:0
> in_array: 0, value:1
>
> Why does the second in_array() call fail???

I'm so stupid, for some reason I thought in_array() was searching for
keys. I should use array_key_exists() instead. Thanks Brion and Pedro.