How to get return value to use it another function as a parameter sync in Jquery

Question

I have an ajax function that returns some data. What I need is, to use it another function as a parameter. Actually I did something like below , but it doesnt work. I need the return value of ajaxDataRenderer. How can I do that ?

    $(document).ready(function () {

        var ajaxDataRenderer = function (url) {
            var ret = null;
            $.ajax({
                async: false,
                url: url,
                dataType: "json",
                success: function (data) {
                    ret = data;
                }
            });
            return ret;
        };

        // The url for our json data
        var jsonurl = "Service/test.aspx";

        var plot2 = $.jqplot('chart2', ajaxDataRenderer(jsonurl));

    });

Thanks in advance,

Answer 1

How about using real asynchronous AJAX:

var ajaxDataRenderer = function (url, successHandler) {
    $.ajax({
        url: url,
        dataType: "json",
        success: successHandler
   });
};

$(document).ready(function () {
    // The url for our json data
    var jsonurl = "Service/test.aspx";
    ajaxDataRenderer(jsonurl, function(data) {
        var plot2 = $.jqplot('chart2', data);
    });
});

Answer 2

You cannot do this since the ajax request is asynchronous. This question has been asked many times before: How to return value from $.getJSON

Answer 3

You could call the 'jqplot' function inside the success callback -

success: function (data) {
    ret = data;
    var plot2 = $.jqplot('chart2', data);
}