php switch between range of numbers does not work - php

Question

I have this PHP function for changing color between 2 numbers:
function color_switch($number){
switch (true){
case $number == range(1 , 3):
$color = "progress-bar-danger";
break;
case $number == range(3 , 5):
$color = "progress-bar-warning";
break;
case $number == range(5 , 6):
$color = "progress-bar-default";
break;
case $number == range(6 , 8):
$color = "progress-bar-success";
break;
case $number == range(8, 10):
$color = "progress-bar-success";
break;
}
return $color;
}
But in action this function does not work for me. How should I fix this ?

Answer 1

You are comparing range() which is an array, and $number is integer, which is invalid,
Change your function something like,
function color_switch($number) {
switch ($number) { // switching the function argument
case $number <= 3 : // if less than three, execute case
$color = "progress-bar-danger";
break;
case $number <= 5 :
$color = "progress-bar-warning";
break;
case $number <= 6 :
$color = "progress-bar-default";
break;
case $number <= 8 :
$color = "progress-bar-success";
break;
case $number <= 10 :
$color = "progress-bar-success";
break;
}
return $color;
}

Answer 2

Your utilization of switch is incorrect and your utilization of range() is too.
Your parameter of switch should be the variable you evaluate.
Range() will return an array containing the range.
So the correct code is better :
function color_switch($number) {
switch ($number){
case in_array($number, range(1 , 3)):
$color = "progress-bar-danger";
break;
case in_array($number, range(3 , 5)):
$color = "progress-bar-warning";
break;
case in_array($number, range(5 , 6)):
$color = "progress-bar-default";
break;
case in_array($number, range(6 , 8)):
$color = "progress-bar-success";
break;
case in_array($number, range(8 , 10)):
$color = "progress-bar-success";
break;
}
return $color;
}