php - return the value of an ajax post back to javascript

Question

I'm trying to test the successful submission of data from javascript to a php file by having that php file return the results of the javascript post back to javascript. I'm getting a successful response in the ajax post, but the data is an empty string. How do I find out what data was posted? Here's my code:

JAVASCRIPT:

var benefitsArray = ["someData","someOtherData"];

$('#drop-submit').on('click',function(){
if (benefitsArray.length > 0){

    var formData = { "benefits" : benefitsArray };
    debugger;
    $.ajax({
        url : "dd-receiver.php",
        type: "POST",
        data : formData,
        success: function(data, textStatus, jqXHR)
        {
            console.log(data); //result is "";
            debugger;
            //data - response from server
        },
        error: function (jqXHR, textStatus, errorThrown)
        {
            console.log('failure');
        }
    });
}
});

PHP:

<?php 
echo $_POST["benefits"]
?>

UPDATE:

I got a response by, in the php code, doing:

echo json_encode($_POST['benefits']);

but the problem is that in the javascript, if I log the data, the result is

"["someData","someOtherData"]" (a string)

and not

["someData","someOtherData"] (an array)

how do I get it to return an array and not a string?

Does adding return false; at the end of the click function help? Without it, you're probably submitting the form and none of the rest of the function executes. — Barmar
– Barmar, Commented Jan 7, 2014 at 20:16
BTW, echoing an array will just echo the word "Array", not the contents. — Barmar
– Barmar, Commented Jan 7, 2014 at 20:16
In the Javascript, if you add dataType: "json" as an option to $.ajax, jQuery will parse the JSON automatically and provide data as an object. — snwflk
– snwflk, Commented Jan 7, 2014 at 20:36

user229044 · Accepted Answer · 2014-01-07 20:37:41Z

2

You're not parsing the JSON being sent to you.

You can make jQuery do this for you by adding dataType: 'JSON' to your $.ajax options...

 $.ajax({
   dataType: 'JSON',
   url : "dd-receiver.php",
   type: "POST",
   data : formData,
   success: function(data, textStatus, jqXHR) ...

Or manually with JSON.parse:

   success: function(data, textStatus, jqXHR) {
     benefits = JSON.parse(data);
     ...
   }

answered Jan 7, 2014 at 20:37

user229044♦

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