if statement says true to a null variable - php

Question

What is happening here? Is it to late for my brain?
I have a php file containing exaclty the code below.
<?php
$foo = 'foo';
$bar = null;
switch ($foo) {
default:
case 'foo':
if ($bar) {
echo 'true';
} else {
echo 'false';
}
}
Prints true when run in the browser but false when run on command line... HOW? I noticed when I remove the line default: it works, but how should this whole switch statement be related to this? It's still a simple if (null) { doing this anyway }
PHP 7.0.13
Apache/2.4.18
And yes... I cleared my browser cache, hit ctrl+f5... I even changed the scripts file name.
UPDATE:
After making changes to this simple file... (just adding whitespaces to the end) and hitting f5 in the browser it says false once but than true again... no matter what I do. I really don't get it.
UPDATE:
My PHP version just got updated from 7.0.13 to 7.0.15. Guess what... still the same output:
Apache/Browser: true
Console: false

Answer 1

For this you can use isset
<?php
$foo = 'bar';
$bar = null;
switch ($foo) {
default:
case 'bar':
if (isset($bar)) {
echo "true";
die();
} else {
echo "false";
die();
}
}
This code will echo false.
You could also use a double not operator:
<?php
$foo = 'bar';
$bar = null;
switch ($foo) {
default:
case 'bar':
if (!!$bar) {
echo "true";
die();
} else {
echo "false";
die();
}
}
The problem with your default case
In the code you provided (and the code I re-quoted), you will be hitting case 'bar' every time, no matter what $foo is set to.
<?php
$foo = 'notbar';
$bar = null;
switch ($foo) {
default:
case 'bar':
if (!!$bar) {
echo "true";
die();
} else {
echo "false";
die();
}
}
This code will still echo false.
What you should do is include a break in the default case.
<?php
$foo = 'notbar';
$bar = null;
switch ($foo) {
default:
echo "default";
break;
case 'bar':
if (!!$bar) {
echo "true";
die();
} else {
echo "false";
die();
}
break;
}
This code will echo default and nothing else.