php switch for a null string - php

Question

For the following php program with a switch statement, why '' give me $vSS=2 instead of =1?
Quite strange to me. I am using PHP 5.5.9.
I can add case '': to resolve the problem, but I am curious why PHP give $vSS=2 instead of
$vSS=1. Is it normal or a bug?
<?php
R(15); // 1 ok
R(''); // why give me 2
R(40); // 2 ok
R(70); // 3 ok
#
function R($SS){
switch($SS){
case $SS<=20: $vSS=1;break;
case ($SS>20 and $SS<=49.9): $vSS=2; // why here?
if($SS == '') echo "DEBUG: SS is a null string.<br>\n";
break;
case ($SS<=100 and $SS>49.9): $vSS=3; break;
default:$vSS=0 ;
}
echo "DEBUG:(SS/vSS) $SS:$vSS\n";
}
?>
------ RESULT
DEBUG:(SS/vSS) 15:1
DEBUG: SS is a null string.<br>
DEBUG:(SS/vSS) :2
DEBUG:(SS/vSS) 40:2
DEBUG:(SS/vSS) 70:3

Answer 1

You don't understand how switch works. It compares the value in switch($SS) with each of the case values, it doesn't just test each case. So
switch ($SS) {
case $SS<=20:
is similar to:
if ($SS == ($SS<=20))
The reason the second case is being executed is because ($SS > 20 && $SS <= 49.9) is false, and false is considered equal to zero or an empty string.
You shouldn't use switch for what you're doing, you should use if/then/elseif/else:
if ($SS <= 20) {
$vSS = 1;
} elseif ($SS <= 49.9) {
$vSS = 2;
} else {
$vSS = 0;
}

Answer 2

#Barmar is right, expression in case() is compared to switch(something_here) but you don't have to cahnge your all your code to if/elsif/elsif/.../... logic. Just change switch() statement to true
switch(true) { // <-- this part only
case $SS<=20:
$vSS=1;
break;
case ($SS>20 and $SS<=49.9):
$vSS=2; // why here?
// must not be here
// if($SS == '') echo "DEBUG: SS is a null string.<br>\n";
break;
case ($SS<=100 and $SS>49.9):
$vSS=3;
break;
case $SS=='': // you can check it here
echo "DEBUG: SS is a null string.<br>\n";
break;
default:
$vSS=0 ;
}