Question

I am having trouble with my IF statement, it is always TRUE although this is incorrect. I’m using an OR operator as there are two possible scenarios I want to capture in the IF statement.

The array string ad_status is “1” but using the below -3 is returned, I’m expecting the IF to be FALSE. If I remove the OR and second statement from the IF, the result of the IF is correct.

What have I done wrong? Thanks.

    if(($getadstatus['ad_status'] != "1" || $getadstatus['ad_status'] != "4"))
    {
        return -3;
        exit;
    }

Additional:
What I want to do is exit the function (not seen in full here) if ad_status does not equal 1 or 4. If it equals any other value other than 1 or 4, the IF statement should return TRUE and exit.
ad_status could be any value from 0 to 4.

Answer 1

What you are saying is that any value that is not 1 OR is not 4 should return true.

For ‘1’ you get the statement

if( 1 != 1 || 1 != 4)

which translates to

if( false || true )

which is ofcourse true.

What you need is:

if(!($value == 1 || $value==4))

which is the same as (de Morgan’s law)

if($value != 1 && $value != 4)