PHP, switch case returned undefined variable - php

Question

Hy,
I got switch case inside function when but when i call it, i got error Undefined Variable and i don't know why (i use PHP 8)
private function getIDFromBrand() {
switch ($this->brand) {
case "Niky":
$id = 1;
break;
case "Pumo":
$id = 4;
break;
case "Coke":
if ($this->typecoke== 0) {
$id = 2;
} else {
if ($this->typecoke== 1) {
$id = 3;
}
}
break;
case "Tomato":
$id = 5;
break;
case "Riles":
$id = 6;
break;
case "TEST":
$id = 7;
break;
}
return $id; // Error Undefined variable $id
}
When i declare $id at the top of my function, like
$id = null
or
$id = 0
The switch doesn't update it, so it will return null or 0, it will return the declared value.

Answer 1

Your switch statement has no default branch, so if $this->brand is, say, "Stack Overflow", it will not run any of the statements, and $id will never be set.
See the PHP manual for the switch statement:
A special case is the default case. This case matches anything that wasn't matched by the other cases.
Similarly, if $this->brand is "Coke", but $this->typecoke is, say, 42, it will not match either of the conditions in that branch.
switch ($this->brand) {
case "Niky":
$id = 1;
break;
case "Pumo":
$id = 4;
break;
case "Coke":
if ($this->typecoke== 0) {
$id = 2;
} elseif ($this->typecoke== 1) {
$id = 3;
} else {
$id = -1; // WAS PREVIOUSLY NOT SET
}
break;
case "Tomato":
$id = 5;
break;
case "Riles":
$id = 6;
break;
case "TEST":
$id = 7;
break;
default:
$id = -1; // WAS PREVIOUSLY NOT SET
break;
}