javascript - 使用 jQuery 并行执行多个 promise 后等待完成

Question

我想使用 jquery 的延迟/ promise 实现对多个异步 ajax 请求进行排队:

function doSomething() {
    console.log('doSomething')};

function makeMultiAjaxRequests1() {
    console.log('makeMultiAjaxRequests1')};

function makeMultiAjaxRequests2() {
    console.log('makeMultiAjaxRequests2')};

var step1 = function () { 
    var promise = new $.Deferred().promise();   
    makeMultiAjaxRequests1(); 
    return promise; }

var step2 = function () {
    var promise = new $.Deferred().promise();
    makeMultiAjaxRequests2();
    return promise; } 

step1()
   .then(step2())
   .done(doSomething());

$.when(step1(), 
       step2())
   .done(function () {
    doSomething();
});

这是 fiddle link .所以我的问题是:

在 step1 和 step2 并行执行的模式中，代码没有到达最后一个处理函数。为什么？

Answer 1

你需要解析step1和step2中的deferred对象

试试这个

function doSomething() {
    console.log('doSomething')};

function makeMultiAjaxRequests1(deferred) {
    console.log('makeMultiAjaxRequests1')
    deferred.resolve()};

function makeMultiAjaxRequests2(deferred) {
    console.log('makeMultiAjaxRequests2')
    deferred.resolve()};

var step1 = function () { 
    var deferred = new $.Deferred();   
    makeMultiAjaxRequests1(deferred); 
    return deferred; }

var step2 = function () {
    var deferred = new $.Deferred();
    makeMultiAjaxRequests2(deferred);
    return deferred; } 

step1().then(step2).done(doSomething);

$.when(step1(), step2()).done(function () {
    doSomething();
});