我如何让一个函数等待,直到所有的jQuery Ajax请求在另一个函数内完成?

简而言之,在执行下一个Ajax请求之前,我需要等待所有Ajax请求都完成。但如何?


当前回答

为了扩展Alex的回答,我举了一个带有可变论点和承诺的例子。我想通过ajax加载图像,并在它们全部加载后显示在页面上。

为了做到这一点,我使用了以下方法:

let urlCreator = window.URL || window.webkitURL;

// Helper function for making ajax requests
let fetch = function(url) {
    return $.ajax({
        type: "get",
        xhrFields: {
            responseType: "blob"
        },
        url: url,
    });
};

// Map the array of urls to an array of ajax requests
let urls = ["https://placekitten.com/200/250", "https://placekitten.com/300/250"];
let files = urls.map(url => fetch(url));

// Use the spread operator to wait for all requests
$.when(...files).then(function() {
    // If we have multiple urls, then loop through
    if(urls.length > 1) {
        // Create image urls and tags for each result
        Array.from(arguments).forEach(data => {
            let imageUrl = urlCreator.createObjectURL(data[0]);
            let img = `<img src=${imageUrl}>`;
            $("#image_container").append(img);
        });
    }
    else {
        // Create image source and tag for result
        let imageUrl = urlCreator.createObjectURL(arguments[0]);
        let img = `<img src=${imageUrl}>`;
        $("#image_container").append(img);
    }
});

更新工作的单个或多个url: https://jsfiddle.net/euypj5w9/

其他回答

注意:上面的答案使用的功能在编写这个答案时并不存在。我建议使用jQuery.when()而不是这些方法,但我把答案留给历史用途。

-

您可能可以使用一个简单的计数信号量,尽管如何实现它取决于您的代码。一个简单的例子是……

var semaphore  = 0,     // counting semaphore for ajax requests
    all_queued = false; // bool indicator to account for instances where the first request might finish before the second even starts

semaphore++;
$.get('ajax/test1.html', function(data) {
    semaphore--;
    if (all_queued && semaphore === 0) {
        // process your custom stuff here
    }
});

semaphore++;
$.get('ajax/test2.html', function(data) {
    semaphore--;
    if (all_queued && semaphore === 0) {
        // process your custom stuff here
    }
});

semaphore++;
$.get('ajax/test3.html', function(data) {
    semaphore--;
    if (all_queued && semaphore === 0) {
        // process your custom stuff here
    }
});

semaphore++;
$.get('ajax/test4.html', function(data) {
    semaphore--;
    if (all_queued && semaphore === 0) {
        // process your custom stuff here
    }
});

// now that all ajax requests are queued up, switch the bool to indicate it
all_queued = true;

如果你想让它像{async: false}一样操作,但又不想锁定浏览器,你可以用jQuery队列来完成同样的事情。

var $queue = $("<div/>");
$queue.queue(function(){
    $.get('ajax/test1.html', function(data) {
        $queue.dequeue();
    });
}).queue(function(){
    $.get('ajax/test2.html', function(data) {
        $queue.dequeue();
    });
}).queue(function(){
    $.get('ajax/test3.html', function(data) {
        $queue.dequeue();
    });
}).queue(function(){
    $.get('ajax/test4.html', function(data) {
        $queue.dequeue();
    });
});

你也可以使用async.js。

我认为它比美元好。因为你可以合并所有不支持承诺的异步调用,比如超时,SqlLite调用等,而不仅仅是ajax请求。

如果从头开始,我强烈建议使用$.when()。

尽管这个问题有超过一百万个答案,但我仍然没有找到任何对我的情况有用的答案。假设您必须处理现有的代码库,已经进行了一些ajax调用,并且不想引入承诺的复杂性和/或重做整个事情。

我们可以很容易地利用jQuery的.data, .on和.trigger函数,这些函数一直是jQuery的一部分。

Codepen

我的解决方案的优点是:

显然回调依赖于什么 函数triggerNowOrOnLoaded并不关心数据是否已经加载或我们仍在等待它 将其插入现有代码非常容易

$(function() { // wait for posts to be loaded triggerNowOrOnLoaded("posts", function() { var $body = $("body"); var posts = $body.data("posts"); $body.append("<div>Posts: " + posts.length + "</div>"); }); // some ajax requests $.getJSON("https://jsonplaceholder.typicode.com/posts", function(data) { $("body").data("posts", data).trigger("posts"); }); // doesn't matter if the `triggerNowOrOnLoaded` is called after or before the actual requests $.getJSON("https://jsonplaceholder.typicode.com/users", function(data) { $("body").data("users", data).trigger("users"); }); // wait for both types triggerNowOrOnLoaded(["posts", "users"], function() { var $body = $("body"); var posts = $body.data("posts"); var users = $body.data("users"); $body.append("<div>Posts: " + posts.length + " and Users: " + users.length + "</div>"); }); // works even if everything has already loaded! setTimeout(function() { // triggers immediately since users have been already loaded triggerNowOrOnLoaded("users", function() { var $body = $("body"); var users = $body.data("users"); $body.append("<div>Delayed Users: " + users.length + "</div>"); }); }, 2000); // 2 seconds }); // helper function function triggerNowOrOnLoaded(types, callback) { types = $.isArray(types) ? types : [types]; var $body = $("body"); var waitForTypes = []; $.each(types, function(i, type) { if (typeof $body.data(type) === 'undefined') { waitForTypes.push(type); } }); var isDataReady = waitForTypes.length === 0; if (isDataReady) { callback(); return; } // wait for the last type and run this function again for the rest of the types var waitFor = waitForTypes.pop(); $body.on(waitFor, function() { // remove event handler - we only want the stuff triggered once $body.off(waitFor); triggerNowOrOnLoaded(waitForTypes, callback); }); } <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <body>Hi!</body>

为了扩展Alex的回答,我举了一个带有可变论点和承诺的例子。我想通过ajax加载图像,并在它们全部加载后显示在页面上。

为了做到这一点,我使用了以下方法:

let urlCreator = window.URL || window.webkitURL;

// Helper function for making ajax requests
let fetch = function(url) {
    return $.ajax({
        type: "get",
        xhrFields: {
            responseType: "blob"
        },
        url: url,
    });
};

// Map the array of urls to an array of ajax requests
let urls = ["https://placekitten.com/200/250", "https://placekitten.com/300/250"];
let files = urls.map(url => fetch(url));

// Use the spread operator to wait for all requests
$.when(...files).then(function() {
    // If we have multiple urls, then loop through
    if(urls.length > 1) {
        // Create image urls and tags for each result
        Array.from(arguments).forEach(data => {
            let imageUrl = urlCreator.createObjectURL(data[0]);
            let img = `<img src=${imageUrl}>`;
            $("#image_container").append(img);
        });
    }
    else {
        // Create image source and tag for result
        let imageUrl = urlCreator.createObjectURL(arguments[0]);
        let img = `<img src=${imageUrl}>`;
        $("#image_container").append(img);
    }
});

更新工作的单个或多个url: https://jsfiddle.net/euypj5w9/

jQuery允许您指定是否希望ajax请求是异步的。您可以简单地使ajax请求同步,然后其余的代码直到它们返回才执行。

例如:

jQuery.ajax({ 
    async: false,
    //code
});