object as exit（）参数

Question

Is this even valid? Because it's working.

if(empty($query)) {
    exit(View::forge('error', $data));
    //View::forge('error', $data) = object
}

According to PHP manual exit should only accept "string" or "int" as parameter.

So I tried:

if(empty($query)) {
    return View::forge('error', $data);  //will not display
}

    return View::forge('default', $data);  //displays

But it seems it ignores it and displays the default template, so I tried combining it with exit:

if(empty($query)) {
    return View::forge('error', $data);  //will not display
    exit;
}

return View::forge('default', $data);  //displays

But still the same result, what I want to know that is exit($obj) valid in fuelphp? Because it seems it's working.

What I want to do is if query is empty page would display error template instead of default template, any help would be appreciated, thanks!

Answer 1

If the object's result is a string or an integer then it fits the criteria set down in the manual. Whatever is within the exit brackets is executed as a PHP line; see the example below:

$a = 4;
$b = 17;
exit(print $a * $b);

this will output

68

so, on your question:

 exit(View::forge('error', $data));

If the returned result of View::forge is a string or other printable output then it fits the criteria set out by exit and is executed accordingly.

What will not work is a non-integer, non-string result such as:

$a[] = "trees";
$b[] = "cats";

exit(array_merge($a,$b));

Gives:

Notice: Array to string conversion on line 4

Due to this "conversion" I expect that if you pass an object to the exit function then it will probably try to use the __toString() class method, if available, (but I'm not certain on this yet).