How to call second jQuery.ajax instance on success of first and update page

Question

I have some jQuery that is triggered on click of a link with the class \'changetag\'. I\'m using $.ajax() to update the database via changetag.php.

Answer 1

Ajax calls are (by default) asynchronous. That means that this code:

$("#li_"+I).toggleClass("off on");
element.toggleClass("off on");

return false;

could be executed before the ajax call preceding it is finished. This is a common problem for programmers who are new to ajax and asynchronous code execution. Anything you want to be executed after the ajax call is done must be put into a callback, such as your success handler:

$.ajax({
    type: "POST",
    url: "_js/changetag.php",
    data: info,
    success: function(){
        $("#li_"+I).toggleClass("off on");
        element.toggleClass("off on");
    }
});

Likewise, you could put the second ajax call in there as well:

$.ajax({
    type: "POST",
    url: "_js/changetag.php",
    data: info,
    success: function(){
        $("#li_"+I).toggleClass("off on");
        element.toggleClass("off on");

        $.ajax({
            url: "_js/loaddeals_v2.php",
            success: function(results){
                $('#listresults').empty();
                $('#listresults').append(results);
            }
        });
    }
});

With jQuery 1.5's Deferred Object, you can make this slicker.

function firstAjax() {
    return $.ajax({
        type: "POST",
        url: "_js/changetag.php",
        data: info,
        success: function(){
            $("#li_"+I).toggleClass("off on");
            element.toggleClass("off on");
        }
    });
}

// you can simplify this second call and just use $.get()
function secondAjax() {
    return $.get("_js/loaddata.php", function(results){
        $('#listresults').html(results);
    });
}

// do the actual ajax calls
firstAjax().success(secondAjax);

This is nice because it lets you un-nest callbacks - you can write code that executes asynchronously, but is written like synchronously-executed code.

Answer 2

Deprecation Notice: The jqXHR.success(), jqXHR.error(), and jqXHR.complete() callbacks are removed as of jQuery 3.0. You can use jqXHR.done(), jqXHR.fail(), and jqXHR.always() instead. https://api.jquery.com/jQuery.ajax/#jqXHR