First value in array not found with array_search( )

[Claus Mygind]

I create an array like this

$dispArray = array ("FIRST","LAST","name","ID","DIRECTPHON","MOBILE","MRMRS","myKey");
 

I search on the array like this:

if(array_search ( $key , $searchArray )){....}
 

Using firephp, I verified that the $key value of "FIRST" was included in the row set I returned.

Also if I move FIRST into another position in the array the value is found in the $dispArray. But then the new value that is first in the array is not found.

Very puzzling. Any suggestion?

[Claus Mygind]

The error in my thinking (I guess) was, I thought the "if" statement returned true or false which of course I now know it does not. So this seems to have corrected the problem.

if(array_search ( $key , $dispArray ) === false )
 

I can test for false as it will return false if not found but will return the position in the array if found.

which now begs the question, is there a way to test for "Not false" so as to avoid this odd construct.

if(array_search ( $key , $searchArray ) === false )
{
  //do nothing
}else{
...the code I want to execute ....
}
 

[Phelim Cavlan]

This had me stumped for a bit too.
I found that a negation in the if statement worked to test for the opposite case - therefore

if (!($myvar===fal se)){
//code
}