How to make all AJAX calls sequential?

Jader Dias picture Jader Dias · Jul 20, 2009 · Viewed 37.8k times · Source

I use jQuery. And I don't want parallel AJAX calls on my application, each call must wait the previous before starting. How to implement it? There is any helper?

UPDATE If there is any synchronous version of the XMLHttpRequest or jQuery.post I would like to know. But sequential != synchronous, and I would like an asynchronous and sequential solution.

Answer

Todd Chaffee picture Todd Chaffee · Sep 26, 2012

There's a much better way to do this than using synchronous ajax calls. Jquery ajax returns a deferred so you can just use pipe chaining to make sure that each ajax call finishes before the next runs. Here's a working example with a more in depth example you can play with on jsfiddle.

// How to force async functions to execute sequentially 
// by using deferred pipe chaining.

// The master deferred.
var dfd = $.Deferred(),  // Master deferred
    dfdNext = dfd; // Next deferred in the chain
    x = 0, // Loop index
    values = [], 

    // Simulates $.ajax, but with predictable behaviour.
    // You only need to understand that higher 'value' param 
    // will finish earlier.
    simulateAjax = function (value) {
        var dfdAjax = $.Deferred();

        setTimeout(
            function () {
                dfdAjax.resolve(value);
            },
            1000 - (value * 100)
        );

        return dfdAjax.promise();
    },

    // This would be a user function that makes an ajax request.
    // In normal code you'd be using $.ajax instead of simulateAjax.
    requestAjax = function (value) {
        return simulateAjax(value);
    };

// Start the pipe chain.  You should be able to do 
// this anywhere in the program, even
// at the end,and it should still give the same results.
dfd.resolve();

// Deferred pipe chaining.
// What you want to note here is that an new 
// ajax call will not start until the previous
// ajax call is completely finished.
for (x = 1; x <= 4; x++) {

    values.push(x);

    dfdNext = dfdNext.pipe(function () {
        var value = values.shift();
        return requestAjax(value).
            done(function(response) {
                // Process the response here.

            });

    });

}

Some people have commented they have no clue what the code does. In order to understand it, you first need to understand javascript promises. I am pretty sure promises are soon to be a native javascript language feature, so that should give you a good incentive to learn.