Switch statement using strstr always validates as true? - php

Question

I have the following switch statement.
The URL contains a referral ID e.g twitter, facebook or an email e.g mail#mail.com. This is stored as $ref
I have the following switch statement:
switch ($ref) {
case "twitter":
echo "twitter";
break;
case "facebook":
echo "facbeook";
break;
case "blog":
echo "blog";
break;
case strstr($ref,'#'):
echo "email = ".$ref;
default:
echo "no referral found";
break;
}
However if URL is passed with nothing (e.g just www.mything.co.uk) then I wish to go to the default case.
Instead, I get the following output:
email = no referral found
Why does the default also include the text I set for case strstr($ref,'#') ?

Answer 1

OP question: "Why does the default also include the text I set for case strstr($ref,'#') ?"
Answer: there's no break; following the output, and thus falls through to the default case.
UPDATE: Addressing the issue of putting a statement within a case, I'm also including an easy work-around:
switch ($ref) {
case "twitter":
echo "twitter";
break;
case "facebook":
echo "facbeook";
break;
case "blog":
echo "blog";
break;
default:
if (strstr($ref,'#')) {
echo "email = ".$ref;
} else {
echo "no referral found";
}
break;
}

Answer 2

When $ref is an empty String, then strstr($ref,'#'); returns an empty string too, this is why the case strstr($ref,'#'): matches the switch input $ref.
The problem is, you can't even use a email validation function like
filter_var($ref, FILTER_VALIDATE_EMAIL)
That would return false in case of an empty input instead of an empty string, but switch does loose comparison, meaning that an "" == false would return true:
http://php.net/manual/en/types.comparisons.php#types.comparisions-loose
Thus the only solution I see is to use an if statement using the === operator:
if($ref == 'twitter') {
echo "twitter";
} else if($ref == 'facebook') {
echo "facbeook";
} else if($ref == 'blog') {
echo "blog";
} else if($ref === filter_var($ref, FILTER_VALIDATE_EMAIL)) {
echo "email = ".$ref;
} else {
echo "no referral found";
}

Answer 3

That's because your test is performed like if ($ref == strstr($ref, '#')), where strstr returns false which equals an empty string. You cannot really use dynamic comparisons in switch statements. Use if..else if you need that. Alternatively, abuse switch a bit:
switch (true) {
case $ref == 'twitter':
..
case strstr($ref, '#'):
..
}

Answer 4

That will work:
case (strstr($ref, '#') ? true : false):
But it's not really good of practice.