jQuery Deferred not working

问题:

I am trying out a code as

function search(query) {
    var dfr = $.Deferred();
    $.ajax({
        url: "http://search.twitter.com/search.json",
        data: {
            q: query
        },
        dataType: 'jsonp',
        success: dfr.resolve
    });
    return dfr.promise();
}

Test = {
    start: function(){
        alert("Starting");
    }
};

function gotresults(data) {
    alert(data.max_id);
}

function showDiv() {
    $('<div />').html("Results received").appendTo('body');
}

$.when(search('ashishnjain'))
    .then(gotresults)
    .then(showDiv);

This works as expected. However when I write it as:

Test.start()
    .then(search('ashishnjain'))
    .then(gotresults)
    .then(showDiv);

it just alerts "Starting" and terminates.A working example can be found at http://jsfiddle.net/XQFyq/2/. What am I doing wrong?

回答1:

Test is not a deferred object, so it does not have a method .then(). .when() IS a deferred object hence why it works when you call .when().

Your $.ajax() call IS a deferred object, so if you return that as part of your 'Test.start() method, you can add .then() callbacks (see example here), the .then() callbacks will be called once the ajax call has been resolved, i.e. has returned its data, however this isn't really the correct use of the deferred object I don't think. The following is more how it is intended to be used I believe:

function searchTwitter(query){
    $.ajax({
            url: "http://search.twitter.com/search.json",
            data: {
                q: query
            },
            dataType: 'jsonp',
            success: function(data){return data;}
        })
        .then(gotresults)
        .then(showDiv)
        .fail(showFailDiv);
};

function gotresults(data) {
    alert(data.max_id);
}

function showDiv() {
    $('<div />').html("Results received").appendTo('body');
}

function showFailDiv() {
    $('<div />').html("Results <b>NOT</b> received").appendTo('body');
}

// Starting can be done with a click:

$("#searchTwitter").click(function(){
   searchTwitter($("#searchName").val()); 
});

// OR a static call:
searchTwitter("ashishnjain");

See it working here

If you want the returned data in for example showDiv() change it to showDiv(data).....

---

Here is another example of how you could create your own deferred object instead of relying on the deferred object of the .ajax() call. This is a little closer to your original example - if you want to see it fail for example, change the url to http://DONTsearch.twitter.com/search.json example here:

var dfr;

function search(query) {
    $.ajax({
        url: "http://search.twitter.com/search.json",
        data: {
            q: query
        },
        dataType: 'jsonp',
        success: function(data){dfr.resolve(data);},
        error:  function(){dfr.reject();}
    });
}

Test = {
    start: function(){
        dfr = $.Deferred();
        alert("Starting");
        return dfr.promise();        
    }
};


function gotresults(data) {
    alert(data.max_id);
}

function showDiv() {
    $('<div />').html("Results received").appendTo('body');
}

function showFailDiv() {
    $('<div />').html("Results <b>NOT</b> received").appendTo('body');
}

Test.start()
    .then(search('ashishnjain'))
    .then(gotresults)
    .then(showDiv)
    .fail(showFailDiv);

---

Update to answer the comments:

In your version 11, you are not telling the deferred object of a failure, so it will never call the .fail() callback. To rectify this, use the ajax interpretation if the .fail() (error:.......) to advise the deferred object of a failure error: drf.reject - this will run the .fail() callback.

As for the reason you are seeing ShowMoreCode() run straight away is, the .then() calls are callbacks, if you pass it a string representation of a function like: .then(ShowDiv) once its turn comes the callback will look for a function with that name. If you pass a call to a function .then(someMoreCode('Ashish')) it will run the function. Try it, change .then(showDiv) to .then(showDiv()) you will notice as soon as the code runs, it will show the code from showDiv().

If you change .then(ShowMoreCode('Ashish')) to .then(ShowMoreCode), you can access the returned data from the $.ajax() call. Like this:

function someMoreCode(name) {
    alert('Hello ' + name.query);
}

Take a look here: working and NOT working .fail()