PHP - proper check if $_POST['variable'] is posted

Question

I want to check if $_POST['submit'] is posted.

My original code was:

if ($_POST['submit']) { }

But I have a PHP notice with this code - "Undefined index: submit in..."

So to remove the notice I have to write this:

if (isset($_POST['submit'])) { }

But this is pointless because $_POST array is global and it return always true. Also if I want to check if $_POST['submit'] is not 0 without PHP notice I have to write this:

if (isset($_POST['submit']) && $_POST['submit'] != 0) { }

In this particular case I prefer:

if ($_POST['submit']) {}

But here I get the PHP notice.

So which way is the most proper/accepted?

Thank you

Answer 1

isset($_POST['submit']) checks if the submit key is set in the $_POST array. It doesn't just check whether the $_POST array exists and is therefore not "pointless". If you want to check whether the value is not falsey (== false), which includes 0, without triggering an error, that's what empty is for:

if (!empty($_POST['submit']))

which is the same thing as

if ($_POST['submit'])

but without triggering a notice should the value not exist.

See The Definitive Guide To PHP's isset And empty for an exhaustive explanation.

Answer 2

Try

if ($_SERVER['REQUEST_METHOD']=='POST')
{
  //do 
}

Answer 3

As per my understanding, It should be like below:

if (isset($_SERVER) && isset($_SERVER['REQUEST_METHOD']) && $_SERVER['REQUEST_METHOD']=='POST'){
  # if method is post, code goes here.
}

and if you are sure that your method is POST for sure. and you have data post in $_POST you can use code like below:

if (isset($_POST['submit']) && $_POST['submit'] != '') {# I think, '' instead of 0
  # if data is posted, code goes here.
}

I usually prefer $_POST.

Answer 4

$_POST[] checks to see if a variable is submitted and not the form name.

if ( isset($_POST['name']) ) {
    // work here
}