Chaining multiple ajax calls with jquery deferred objects

Question

While looking for a solution similar to as described here: How to chain ajax calls using jquery I am looking for a solution using jquery v1.52.

I have a set of ajax requests to be made. But each ajax request is to be sent only after completing the previous ajax call. I am trying to achieve this with jquery 1.5.2 but just unable to. Here is what I modified from the above mentioned example. It does not work. Can anyone help me get this working? Expected output is http://jsfiddle.net/k8aUj/3/

P.S: I cannot upgrade to version beyond 1.5.2

@Vishal, that answer doesn' t solve the problem to start an ajax call after the success of the previous one. — Fabrizio Calderan
– Fabrizio Calderan, Commented Sep 7, 2012 at 12:21
Ah! didn't saw the success part..Ignore the previous comment, I'll try to work out something. — Vishal
– Vishal, Commented Sep 7, 2012 at 12:25

Felix Kling · Accepted Answer · 2013-02-09 02:33:38Z

4

Yo! Solved! http://jsfiddle.net/sandhyasriraj/AaHZv/

var x = null;
var i = 0;
x= $.Deferred();
var countries=["US","CA","MX","bx","fs","ZX"];
function log(msg) {
    var $out=$("<div />");
    $out.html(msg);
    $("#console").append($out);
}


callX = function(j) {
    return $.ajax({
            type: "GET",
            url: "/echo/json/",
            data: {country:countries[i]},
            dataType: "JSON",
            success: function(){
                log("Successful request for [" + countries[j] + "]");
                i++;      
                x.resolve();
                xy();
               }
        });

}
x.resolve();
xy = function()
{
    debugger;
    if(i > 5)
        return;

     $.when(x).then(function() {
        x = $.Deferred();
        log("Making request for [" + countries[i] + "]");
        callX(i);
    });
}
xy();

is this what you want?

edited Feb 9, 2013 at 2:33

Felix Kling

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answered Sep 7, 2012 at 13:31

Sandy

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