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
我找到了一个简单的方法，使用shift()
function waitReq(id)
{
jQuery.ajax(
{
type: 'POST',
url: ajaxurl,
data:
{
"page": id
},
success: function(resp)
{
...........
// check array length if not "0" continue to use next array value
if(ids.length)
{
waitReq(ids.shift()); // 2
)
},
error: function(resp)
{
....................
if(ids.length)
{
waitReq(ids.shift());
)
}
});
}
var ids = [1, 2, 3, 4, 5];
// shift() = delete first array value (then print)
waitReq(ids.shift()); // print 1
其他回答
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>
我找到了一个简单的方法，使用shift()
function waitReq(id)
{
jQuery.ajax(
{
type: 'POST',
url: ajaxurl,
data:
{
"page": id
},
success: function(resp)
{
...........
// check array length if not "0" continue to use next array value
if(ids.length)
{
waitReq(ids.shift()); // 2
)
},
error: function(resp)
{
....................
if(ids.length)
{
waitReq(ids.shift());
)
}
});
}
var ids = [1, 2, 3, 4, 5];
// shift() = delete first array value (then print)
waitReq(ids.shift()); // print 1
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}>`;
$("#image_container").append(img);
}
});
更新工作的单个或多个url: https://jsfiddle.net/euypj5w9/
我找到了一个很好的答案，这正是我在寻找的:)
jQuery ajaxQueue
//This handles the queues
(function($) {
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);
}
});
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>
