严格标准的PHP错误

Question

I have a small issue with my script.

I'm getting Strict Standards: Only variables should be passed by reference in

if( $checkDNS && ($domain = end(explode('@',$email, 2))) )

Answer 1

From the PHP manual :

This array is passed by reference because it is modified by the function. This means you must pass it a real variable and not a function returning an array because only actual variables may be passed by reference.

So you must use a variable in the end function:

$domain = explode('@',$email, 2);
if( $checkDNS && ($domain = end($domain)) )

Answer 2

From the manual:

mixed end ( array &$array )

end takes the array by reference and move the internal pointer. Your array is the function output, so its unable to correctly modify the array by reference.

Answer 3

Like the message says, end expects a variable because its parameter is a reference.

But since PHP 5.4 you can dereference arrays like that:

$domain = explode('@',$email, 2)[1];

Assuming that $email always contains @ . You should assure that beforehand, otherwise end(...) would give you unexpected results too.