Question

Oh well, maybe it’s my ignorance, but PHP documentation about include statement says take care when comparing return value:

<?php
// won't work, evaluated as include(('vars.php') == 'OK'), i.e. include('')
if (include('vars.php') == 'OK') {
    echo 'OK';
}

// works
if ((include 'vars.php') == 'OK') {
    echo 'OK';
}
?>

But this is what i’ve experienced; boolean false in case of fail, int 1 if success:

$exists = 'Found.php';
var_dump((include $exists)); // Type is: int, value is: 1

$notexists = 'NotFound.php';
var_dump((include $notexists)); // Type is: boolean, value is: false

Is this my bad? Why returning value is inconsistent (meaning not always boolean, for example) and differs from PHP documentation?

Answer 1

Well this is exactly what the documentation states:

$bar is the value 1 because the include was successful. …
If the file can’t be included, FALSE is returned and E_WARNING is issued.

When it becomes inconsistent is when you actually return something in the included file,
e.g. in Found.php:

<?php
return 'Returned from Found.php';

And then

include 'Found.php'; // -> Returns "Returned from Found.php"

It is also worth reading up on how booleans evaluate in PHP:

http://php.net/manual/en/language.types.boolean.php