jquery自定义延迟函数 [英] jquery custom deferred functions

问题描述

我有三个函数正在尝试运行，前两个正在做一些异步的东西，需要第三个使用的数据。我希望只有当1和2都完成时才会触发第三个函数。这是一般结构，但最终功能是在1和2结束前开始。

I have three functions i'm trying to run, the first two are doing some async stuff that need data for the third to use. I want the third function to fire only when 1 and 2 are both done. this is the general structure but the final function is firing before 1 and 2 finish.

function run() {
    var data1 = {};
    var data2 = {};

    $.when(first(), second()).done(constructData());

    function first() {
        var d = new $.Deferred();

        //do a bunch of stuff async
        data1 = {};

        d.resolve();
    }
    function second() {


        var d = new $.Deferred();

        //do a bunch of stuff async
        data2 = {};
        d.resolve();
    }
    function constructData() {
        //do stuff with data1 and data2
    }

}

答案是不立即调用构造数据

Answer was to not call construct data immediately

 $.when(first(), second()).done(constructData);

推荐答案

您应该返回promise对象。你在这一行中也有错误：

You should return promise object. You also have an error in this line:

$.when(first(), second()).done(constructData());

它应该是

$.when(first(), second()).done(constructData); // don't call constructData immediately

所有这一切可能是：

function run() {
    var data1 = {};
    var data2 = {};

    $.when(first(), second()).done(constructData);

    function first() {
        return $.Deferred(function() { // <-- see returning Deferred object
            var self = this;

            setTimeout(function() {   // <-- example of some async operation
                data1 = {func: 'first', data: true};
                self.resolve();       // <-- call resolve method once async is done
            }, 2000);
        });
    }
    function second() {
        return $.Deferred(function() {
            var self = this;
            setTimeout(function() {
                data2 = {func: 'second', data: true};
                self.resolve();
            }, 3000);
        });
    }
    function constructData() {
        //do stuff with data1 and data2
        console.log(data1, data2);
    }
}

http://jsfiddle.net/FwXZC/

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