CODEWITHSUNDEEP
javascript
Paritosh
Paritosh

Reputation: 4503

javascript order in which code runs

I have the following code on my webpage:

   <script type="text/javascript">        
        debugger;  // 1
        var dataClassObj = new DataClassList();
        dataClassObj.init();
        var viewModel = dataClassObj.getModel();
        debugger; // 2
   </script>

in a js file I have:

var DataClassList = function () {
};

DataClassList.prototype = function () {
var viewModel;

// private memebers
var init = function () {
        var self = this;
        $.ajax({
            url: "xxx",
            type: 'GET',
            data: "{}",
            contentType: "application/json; charset=utf-8",
            dataType: "json",
            success: function(data) {
                    // code
                });

                // code
                self.viewModel = viewModel;
                ko.applyBindings(viewModel);
                debugger;  // 3
            },
            error: function(xhr, status, error) {
                $('#lblError').show();
            }
        });
    },
    getModel = function () {
        debugger;  // 4
        return viewModel;
    };

// public members
return {
    init: init,
    getModel: getModel
};

}();

When I run this using Chrome's developer tool, hitting the debugger points. I thought it would run in 1->3->4->2, but it seeme always hit the debugger statements in this order 1->4->2->3 i'm confused as to why it does that, I thought Javascript was synchronous so it would hit 1 and then make the call to init which would trigger 3 and then call 4 and finally 2.

Upvotes: 0

Views: 46

Answers (2)

Bradley Trager
Bradley Trager

Reputation: 3590

JavaScript is syncronous with the exception of ajax calls, setTimeout() and setInterval().

Upvotes: 0

Lucas Holt
Lucas Holt

Reputation: 3826

$.ajax is asynchronous. It will do that call after the other code has run (usually).

Upvotes: 2

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