$_POST as a function parameter [duplicate] - php

Question

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passing POST array to php function
(3 answers)
Closed 9 years ago.
I've been working on a project for a game (so you'll understand that I can't post the actual script), and I'm cleaning up the code and placing code into functions instead of just after the
if ($_POST['name'] { //code }
Is this the correct method of doing what I'm trying? (I tried, but it broke my system)
if ($_POST['action'] == "actionName") actionName($_POST); //calls function if the $_POST is present
function actionName($_POST) { //code }
Thanks in advance for correct answers!

Answer 1

$_POST is globally accessible. So you don't have to pass to your function:
if ($_POST['action'] == 'actionName') actionName();
function actionName() {
//code using $_POST
}
If you want your function to be able to do what it wants with any array (i.e. other than $_POST), then make the function take a parameter, but don't call it $_POST:
if ($_POST['action'] == 'actionName') actionName($_POST);
function actionName(parameters) {
//code using parameters
}

Answer 2

$_POST is a super-global. All PHP code, no matter the scope, automatically has access to this data structure.
Therefore, the following is fine:
if ($_POST['action'] == "actionName") {
actionName();
}
and then
function actionName() {
print_r($_POST);
// code
}

Answer 3

$_POST is already a superglobal variable so you do not need to pass it to a function. If the function needs the content of $_POST then make a copy and pass it to the function.
However, from your example, it seems to me that you need to pass $_POST['action'] to your function instead of $_POST.

Answer 4

Well in your example you need to have quotes around your string variables:
//calls function if the $_POST is present
if ($_POST['action'] == 'actionName') actionName($_POST);
function actionName($_POST) { /*code*/ }

Answer 5

Although $_POST is super-global, yes of course you can do that and is actually a good idea. This way you decouple your code and maybe you can reuse actionName in another part of the application.
However, don't name the parameter of the function $_POST (this is more confusing than helpful).
To avoid a bunch of if statements for every possible action, have a look at call_user_func [docs].
E.g. you could do:
$allowed_actions = array_fill_keys(array('actionName', ...), true);
if(isset($allowed_actions[$_POST['action']])) {
call_user_func($_POST['action'], $_POST);
}

Answer 6

You should move the function before the code that executes it along with what everyone else here is saying about the $_POST array. Try this if you want
function actionName($input) { // code and return }
if (isset($_POST['action'])) {
$action = $_POST['action'];
if ($action == 'actionName') {
$name = actionName($action);
}
}

Answer 7

$_POST is a super global (as other have mentioned), so you don't need to pass it in.
You can pass it in so that you decouple the function code from the environment in which it's running. For example, should you wish to initiate such an action as a result of something else in the code, you can pass in an array with the same key/values that $_POST would have to do so
If you do pass in $_POST to a function, name the function argument something else, hopefully something that indicates what the data represents
Consider that the function may not need all the key/values that $_POST normally contains, and if you might be better off defining a function that takes only the arguments it needs in order to function

Answer 8

If you are talking about dynamically calling a function, you could achieve it this way:
/*
Example Post
$_POST['functionName'] = 'doSomething';
$_POST['value'] = 'woot!';
*/
function doSomething($message) {
echo $message;
}
// Call the function if it exists
if (function_exists($_POST['functionName']) {
call_user_func($_POST['functionName'], $_POST['value']);
}
// The above would output "Woot!"