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Home/ Questions/Q 4014344
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Editorial Team
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Editorial Team
Asked: May 20, 20262026-05-20T09:29:08+00:00 2026-05-20T09:29:08+00:00

I’m trying to pass an array to jquery’s .ajax() function’s data parameter. The first

  • 0

I’m trying to pass an array to jquery’s .ajax() function’s data parameter. The first approach is that I made my 2-dimensional array like this:

var arr = new Array();
for(i in someArray){
      arr[i] = new Array();
  arr[i].lon = "x";
  arr[i].lat = "y";
}

Then I try to pass this as data in $.ajax():

$.ajax({
    data: { vals : arr },
    async: false,
    type: "POST",
    url: "namedb.php",
    dataType: "script",
    success: function(data){
      result = data;
      alert(result);
        }
});

test.php just returns all the values of $_POST. So alert() here returns:

Array
(
)

But if I changed the code to:

var arr = new Array();
for(i in someArray){
      arr[i] = new Array();
  arr[i] = { lon: "x", lat: "y" };
}

the alert() returns what I expected:

Array
(
    [vals] => Array
        (
            [0] => Array
                (
                    [lat] => "y",
                    [lon] => "x"
                )
            ...
        )
)

I know that both methods initializes the variables/attributes of each element of arr (or am I wrong?). But why do the 2 approach behaves differently? (Sorry I could’ve shorten my question, but I guess I need to explain how I found it).

EDIT: I had added the initialization (arr[i] = new Array();). I must have erased it during the editing of the question. But still the same problem.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-20T09:29:09+00:00Added an answer on May 20, 2026 at 9:29 am

    That is because you did not initialize the array elements.

    var arr = new Array();
    for(i in someArray){
      arr[i] = {}; // initialize it!
      arr[i].lon = "x";
      arr[i].lat = "y";
    }
    

    Actually, I prefer to write

    var arr = new Array();
    for(i in someArray){
      arr[i] = { lon: "x", lat: "y" };
    }
    

    Because I don’t have to type arr[i] 3 times.


    Using for in loops over an array is wrong, you should use for loops or jQuery’s $.each

    var arr = [];
    for(var i = 0; i < someArray.length; i ++) {
      arr[i] = { lon: "x", lat: "y" };
    }
    

    Or

    var arr = [];
    $.each(someArray, function(i, value) {
      arr[i] = { lon: "x", lat: "y" };
    });
    

    Using Array.prototype.push for adding things to an array is easier, because you don’t need to know the next index of an array, also compare with arr[arr.length]=something.

    var arr = [];
    $.each(someArray, function(i, value) {
      arr.push({ lon: "x", lat: "y" });
    });
    

    And now what you’re doing is collect data from an array and translate it into another array. jQuery already has a function that does this: $.map.

    var arr = $.map(someArray, function(value, i) {
      return { lon: "x", lat: "y" };
    });
    

    Note that when using $.each, the arguments are i, value, when you use $.map, they are switched.

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