AJAX：$ _GET不返回任何值

Question

Been trying to figure this out for hours...

I am attempting to use AJAX by grabbing values from a jQuery slider I have contained in an <input> tag. AJAX does not fail (see code), and when I console.log the variable I am trying to send through Data:, it prints out correctly.

note: The jQuery and php are on the same page, so no url: parameter, or whatever you will call it is used.

Here is the <script> portion:

   jQuery(function($){
        $("input[id='protein-slider']").ready(function(){
            var p_range;
            var request;
            $("input[id='protein-slider']").each(function () {
            p_range = $(this).val();
            });
            console.log(p_range) //Returns correct values

            request = $.ajax({
                type: "POST",
                data: p_range,
            });  
            request.done(function (response, textStatus, jqXHR){
                console.log("Hooray, it worked!"); //This prints, no error
            });
            request.fail(function (jqXHR, textStatus, errorThrown){
                console.error(
                    "The following error occurred: "+
                    textStatus, errorThrown
                );
            });
        });
    });

and the <?php ?> code in correlation with this function is:

                echo $_SERVER['QUERY_STRING']; //returns nothing
                $income_data = $_GET['p_range']; 
                echo $income_data; //returns nothing
                var_dump($income_data); //returns "NULL"

I've lost track of all of the other methods I have done and the other countless awnsers I have read on here. None seem to work and it's getting quite depressing...

Hopefully one of you guys can spot something that I am doing stupid...

Thanks!

Answer 1

Your ajax is for POST, change it to GET

request = $.ajax({
    type: "POST",
    data: p_range,
}); 

or you access your params in php as

$var = $_POST['p_range'];

Answer 2

Your data needs to be in key/value pair

data: {p_range: p_range},

and

$income_data = $_POST['p_range'];