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

在@BBonifield回答的基础上,我写了一个实用函数,这样信号量逻辑就不会分布在所有ajax调用中。

untilAjax是一个实用函数,它在所有ajaxCalls完成后调用回调函数。

ajaxObjs是一个ajax设置对象的数组[http://api.jquery.com/jQuery.ajax/]。

Fn是回调函数

function untilAjax(ajaxObjs, fn) {
  if (!ajaxObjs || !fn) {
    return;
  }
  var ajaxCount = ajaxObjs.length,
    succ = null;

  for (var i = 0; i < ajaxObjs.length; i++) { //append logic to invoke callback function once all the ajax calls are completed, in success handler.
    succ = ajaxObjs[i]['success'];
    ajaxObjs[i]['success'] = function(data) { //modified success handler
      if (succ) {
        succ(data);
      }
      ajaxCount--;
      if (ajaxCount == 0) {
        fn(); //modify statement suitably if you want 'this' keyword to refer to another object
      }
    };
    $.ajax(ajaxObjs[i]); //make ajax call
    succ = null;
  };

示例:doSomething函数使用untilAjax。

function doSomething() {
  // variable declarations
  untilAjax([{
    url: 'url2',
    dataType: 'json',
    success: function(data) {
      //do something with success data
    }
  }, {
    url: 'url1',
    dataType: 'json',
    success: function(data) {
      //do something with success data
    }
  }, {
    url: 'url2',
    dataType: 'json',
    success: function(response) {
      //do something with success data
    }
  }], function() {
    // logic after all the calls are completed.
  });
}

其他回答

我找到了一个很好的答案,这正是我在寻找的:)

jQuery ajaxQueue

//This handles the queues    
(function($) {

  var ajaxQueue = $({});

  $.ajaxQueue = function(ajaxOpts) {

    var oldComplete = ajaxOpts.complete;

    ajaxQueue.queue(function(next) {

      ajaxOpts.complete = function() {
        if (oldComplete) oldComplete.apply(this, arguments);

        next();
      };

      $.ajax(ajaxOpts);
    });
  };

})(jQuery);

然后你可以像这样向队列添加一个ajax请求:

$.ajaxQueue({
        url: 'page.php',
        data: {id: 1},
        type: 'POST',
        success: function(data) {
            $('#status').html(data);
        }
    });

正如其他答案所提到的,你可以使用ajaxStop()等待直到所有ajax请求完成。

$(document).ajaxStop(function() {
     // This function will be triggered every time any ajax request is requested and completed
});

如果你想为特定的ajax()请求做这件事,你能做的最好的是在特定的ajax请求中使用complete()方法:

$.ajax({
    type: "POST",
    url: "someUrl",
    success: function(data) {
        // This function will be triggered when ajax returns a 200 status code (success)
    },
    complete: function() {
        // This function will be triggered always, when ajax request is completed, even it fails/returns other status code
    },
    error: function() {
        // This will be triggered when ajax request fail.
    }
});

但是,如果你只需要等待几个和特定的ajax请求被完成?使用美妙的javascript承诺等待,直到这些ajax你想等待完成。我做了一个简单易读的示例,向您展示promises如何与ajax一起工作。请看下一个例子。我使用setTimeout来阐明这个示例。

// Note: // resolve() is used to mark the promise as resolved // reject() is used to mark the promise as rejected $(document).ready(function() { $("button").on("click", function() { var ajax1 = new Promise((resolve, reject) => { $.ajax({ type: "GET", url: "https://miro.medium.com/max/1200/0*UEtwA2ask7vQYW06.png", xhrFields: { responseType: 'blob'}, success: function(data) { setTimeout(function() { $('#image1').attr("src", window.URL.createObjectURL(data)); resolve(" Promise ajax1 resolved"); }, 1000); }, error: function() { reject(" Promise ajax1 rejected"); }, }); }); var ajax2 = new Promise((resolve, reject) => { $.ajax({ type: "GET", url: "https://cdn1.iconfinder.com/data/icons/social-media-vol-1-1/24/_github-512.png", xhrFields: { responseType: 'blob' }, success: function(data) { setTimeout(function() { $('#image2').attr("src", window.URL.createObjectURL(data)); resolve(" Promise ajax2 resolved"); }, 1500); }, error: function() { reject(" Promise ajax2 rejected"); }, }); }); var ajax3 = new Promise((resolve, reject) => { $.ajax({ type: "GET", url: "https://miro.medium.com/max/632/1*LUfpOf7teWvPdIPTBmYciA.png", xhrFields: { responseType: 'blob' }, success: function(data) { setTimeout(function() { $('#image3').attr("src", window.URL.createObjectURL(data)); resolve(" Promise ajax3 resolved"); }, 2000); }, error: function() { reject(" Promise ajax3 rejected"); }, }); }); Promise.all([ajax1, ajax2, ajax3]).then(values => { console.log("We waited until ajax ended: " + values); console.log("My few ajax ended, lets do some things!!") }, reason => { console.log("Promises failed: " + reason); }); // Or if you want wait for them individually do it like this // ajax1.then(values => { // console.log("Promise 1 resolved: " + values) // }, reason => { // console.log("Promise 1 failed: " + reason) // }); }); }); img { max-width: 200px; max-height: 100px; } <script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script> <button>Make AJAX request</button> <div id="newContent"> <img id="image1" src=""> <img id="image2" src=""> <img id="image3" src=""> </div>

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
});

当所有ajax加载完成时,我使用大小检查

function get_ajax(link, data, callback) { $.ajax({ url: link, type: "GET", data: data, dataType: "json", success: function (data, status, jqXHR) { callback(jqXHR.status, data) }, error: function (jqXHR, status, err) { callback(jqXHR.status, jqXHR); }, complete: function (jqXHR, status) { } }) } function run_list_ajax(callback){ var size=0; var max= 10; for (let index = 0; index < max; index++) { var link = 'http://api.jquery.com/ajaxStop/'; var data={i:index} get_ajax(link,data,function(status, data){ console.log(index) if(size>max-2){ callback('done') } size++ }) } } run_list_ajax(function(info){ console.log(info) }) <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.8.1/jquery.min.js"></script>