jQuery 递归 AJAX 调用承诺

问题描述

我仍在尝试弄清楚如何在递归 AJAX 调用中使用 jQuery 延迟对象.我有这样的代码

I'm still trying to figure out how to use jQuery deferred object in recursive AJAX call. I have a code like this

function request(page, items){

    //building the AJAX return value for JSFiddle dummy AJAX endpoint
    var ret = {
        totalPage: 10,
        currentPage: page,
        items: []
    };
    for (var i = page; i < (page + 5); i++){
        ret.items.push(i);
    }

    //calling the AJAX
    $.ajax({
        url: '/echo/json/',
        method: 'POST',
        dataType: 'json',
        data: {
            delay: 1,
            json: JSON.stringify(ret)
        },
        success: function(data){
            if (data.currentPage <= data.totalPage){
                var filtered = data.items.filter(function(el){
                    return el % 2 == 1;
                });
                var newitems = items.concat(filtered);
                console.dir(newitems);
                request(data.currentPage + 1, newitems);
            } else {
                console.dir(items);
                //resolve all item
            }
        }
    });
}

function requestAll(){
    request(1, []);
    //should return a promise tha contains all items
}

这是 JSFiddle http://jsfiddle.net/petrabarus/BHswy/

Here's the JSFiddle http://jsfiddle.net/petrabarus/BHswy/

我知道如何在单个 AJAX 调用中使用 promise，但我不知道如何在递归 AJAX 调用中使用它.我想以类似于下面的方式调用 requestAll 函数

I know how to use promise in single AJAX call, but I have no idea how to use it in a recursive AJAX call. I want to call the requestAll function in a way similar like below

var promise = requestAll();
promise.done(function(items){
    console.dir(items);
});

我该怎么做?

推荐答案

如果您想使用 Promise，则不应使用 success 参数.相反，您想要返回一个promise，并且您想要使用then将promise的结果转换为不同的东西，甚至可能是另一个promise.

You should not use the success parameter if you want to work with promises. Instead, you want to return a promise, and you want to use then to transform the results of a promise into something different, possibly even another promise.

function request(page) {
    …
    // return the AJAX promise
    return $.ajax({
        url: '/echo/json/',
        method: 'POST',
        dataType: 'json',
        data: {
            delay: 1,
            json: JSON.stringify(ret)
        }
    });
}

function requestOddsFrom(page, items) {
    return request(page).then(function(data){
        if (data.currentPage > data.totalPage) {
            return items;
        } else {
            var filtered = data.items.filter(function(el){ return el%2 == 1; });
            return requestOddsFrom(data.currentPage + 1, items.concat(filtered));
        }
    });
}

function requestAll(){
    return requestOddsFrom(1, []);
}

requestAll().then(function(items) {
    console.dir(items);
});

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