如何在 Javascript 中使用 Q 顺序运行承诺? [英] How to sequentially run promises with Q in Javascript?

问题描述

我很难按顺序执行承诺.

I am having a hard time running promises sequentially.

var getDelayedString = function(string) {
    var deferred = Q.defer();

    setTimeout(function() {
        document.write(string+" ");
        deferred.resolve();
    }, 500);

    return deferred.promise;
};

var onceUponATime = function() {
    var strings = ["Once", "upon", "a", "time"];

    var promiseFuncs = [];

    strings.forEach(function(str) {
        promiseFuncs.push(getDelayedString(str));
    });

    //return promiseFuncs.reduce(Q.when, Q());
    return promiseFuncs.reduce(function (soFar, f) {
        return soFar.then(f);
    }, Q());    
};

getDelayedString("Hello")
.then(function() {
    return getDelayedString("world!")
})
.then(function() {
    return onceUponATime();
})
.then(function() {
    return getDelayedString("there was a guy and then he fell.")
})
.then(function() {
    return getDelayedString("The End!")
})

onceUponATime() 应该按顺序输出 ["Once", "upon", "a", "time"] 但由于某些原因它们会立即输出.

onceUponATime() should sequentially output ["Once", "upon", "a", "time"] but instead they are being output immediately for some reason.

jsFiddle 在这里:http://jsfiddle.net/6Du42/2/

jsFiddle here: http://jsfiddle.net/6Du42/2/

知道我做错了什么吗?

推荐答案

但由于某种原因，它们被立即输出.

but instead they are being output immediately for some reason.

你已经在这里给他们打电话了:

You are calling them already here:

promiseFuncs.push(getDelayedString(str));
//                                ^^^^^

你需要推送 function(){ return getDelayedString(str);}.顺便说一句，与其在 each 循环中使用推送到数组，不如使用 map.实际上你并不真正需要它，但可以直接在 strings 数组上reduce:

You would need to push function(){ return getDelayedString(str); }. Btw, instead of using pushing to an array in an each loop you rather should use map. And actually you don't really need that anyway but can reduce over the strings array directly:

function onceUponATime() {
    var strings = ["Once", "upon", "a", "time"];

    return strings.reduce(function (soFar, s) {
        return soFar.then(function() {
            return getDelayedString(s);
        });
    }, Q());    
}

哦，还有 不要使用 document.write.

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