This PHP if loop should be the most, strangest question here, but I do want to know

This first loop executes when $cnt=3 and the other other than $cnt=3. Whatever the value of $cnt only the first loop executes whether $cnt=3 or $cnt==3.

$ary = explode(".", $string);
开发者_如何学Go$cnt = count($ary);
if ($cnt="3") {
//executes when cnt=3
  $fnm = $d[0];
  $fnxt = $d[1].".".$d[2];
} else {
//executes when anything other than when cnt=3
   $fnm = $d[0];
   $fnxt = $d[1];
}

I might be missing something here, what exactly wrong here?

Thanks Jean

You're missing an = sign on the comparison. It should be:

if ($cnt == 3)

As it is, you're assigning 3 to $cnt, and since the assignment operator returns its value, the test becomes if (3), which of course always succeeds.

NB: count() returns an integer, which is why my version above compares against 3 rather than "3"

You are missing an "="

if ($cnt="3") {  // This is an assignment, which returns true.

This should be:

if ($cnt == "3") { // This is a comparison.

$cnt="3" assigns the value "3" to $cnt, and the expression as a whole evaluates to "3", which is true, which causes the if block to always be executed. In order to test whether $cnt is equal to "3", use the == operator: $cnt == "3".