PHP If Statement if (!$_POST) - What does !$_POST mean?

Question

PHP If Statement if (!$_POST) - What does !$_POST mean?

For instance, in some legacy code I'm looking at, the code reads:

    <form action="paypalorders.php" method="POST">
         <input type="hidden" name="orderList" value="' . $orderList . '">
         <input type="submit" value="Archive">
    </form>';
if (!$_POST) {
    file_put_contents('/orders.txt', $output_line1);
}

I've looked at a bunch of other threads and haven't seen this exact question asked, so I'm trying to find out what it is. I've seen threads that format it like this:

if(!empty($_POST)) {

But not quite the same as what I'm looking for. Is it the same thing, just shorthand? I'm not sure, which is why I'm asking. I've Googled around and looked at a handful of threads and I'm still not sure.

Thank you.

Possible duplicate of [Exclamation mark in front of variable - clarification needed](https://stackoverflow.com/questions/13029921/exclamation-mark-in-front-of-variable-clarification-needed) — Scuzzy, Aug 15 '18 at 21:39
Comments are not for extended discussion; this conversation has been [moved to chat](https://chat.stackoverflow.com/rooms/178101/discussion-on-question-by-james-anderson-php-if-statement-if-post-what-do). — Bhargav Rao, Aug 15 '18 at 23:51

Don't Panic · Accepted Answer · 2018-08-15T21:57:21.213

The ! (not) logical operator returns true if the argument on its right-hand side is not true. It forces the argument to be evaluated as a boolean. In this case, when the $_POST array is evaluated as a boolean it will evaluate as true if it is not empty and false if it is. (See converting to boolean.)

if (!$_POST) { should be a safe way to detect whether or not anything is in $_POST if you want to do that. empty isn't necessary in that case because superglobals are always set, and since you aren't referring to a specific key, you don't need to worry about an undefined index notice.

I think it's also worth mentioning that if the only point of the check is to see what type of request was sent, it is probably better to just check the request method directly, because !$_POST does not mean the request wasn't a post, since a post request can be empty.

score 2 · Answer 2 · answered Aug 15 '18 at 21:41

Since $_POST is an array, if it's empty his value is null, so if(!$_POST) would look like this:

if(!null){
  //code 
}

The following code returns true or false, but the objective of both is the same.

if(!empty($_POST)){
  //code
}

Hope it helps you!

score 2 · Answer 3 · answered Aug 15 '18 at 21:45

(bool)$array evaluates to true if $array contains elements, and false if it is empty.

Since $_POST is an array, !$_POST returns true if $_POST is empty.

Another way to interpret this, you are performing conditional tasks for the case where this page was not reached through a HTTP POST method.

score 0 · Answer 4 · answered Aug 15 '18 at 21:45

when you cast an array to bool, it will be cast to bool(false) if the array is empty (eg if count($arr)===0), or bool(true) otherwise. ! casts whatever it's checking to bool. because $_POST always exists, if (!$_POST) { and if(empty($_POST)) { both do the exact same thing, they check if $_POST is empty. it's not even about being legacy code, this is still perfectly valid for 7.3.0. the difference between the 2 approaches will only become apparent when you're checking variables that may not exist, !$arr will trow an Notice: Undefined variable-error if $arr doesn't exist, empty($arr) will not.

PHP If Statement if (!$_POST) - What does !$_POST mean?

4 Answers4