如何通过JQuery ajaxSend事件覆盖成功函数

Question

我试图覆盖ajaxsend事件的成功函数,但它不起作用的是代码:

    $(document).ajaxSend(function(event,xhr,options){
        console.log('ajaxSend');
        var tempSuccess = options.success;
        options.success = function(data, textStatus, jqXHR){
            console.log('start');
            tempSuccess(data, textStatus, jqXHR);
            console.log('end');
        }; xhr.success = options.success;});

在AJAX上我确实在控制台中看到了'ajax',但是在成功的时候我看不到开始和结束的调试信息.

我做错了什么？

Answer 1

你想要完成的是无法完成的ajaxSend.问题是ajaxSend显然使用原始xhr和options对象的副本,因此修改不会产生任何影响.您可以使用以下代码轻松测试:

$(document).ajaxSend(function(event, xhr, options){
    delete options.success;
    console.log(options.success);   // undefined
});
$.ajax({
    url: "test.html",
    success: function() { console.log("this will be printed nevertheless"); }
});

所以你不能ajaxSend用来覆盖成功的回调.相反,你将不得不"破解"jQuery的AJAX功能:

// closure to prevent global access to this stuff
(function(){
    // creates a new callback function that also executes the original callback
    var SuccessCallback = function(origCallback){
        return function(data, textStatus, jqXHR) {
            console.log("start");
            if (typeof origCallback === "function") {
                origCallback(data, textStatus, jqXHR);
            }
            console.log("end");
        };
    };

    // store the original AJAX function in a variable before overwriting it
    var jqAjax = $.ajax;
    $.ajax = function(settings){
        // override the callback function, then execute the original AJAX function
        settings.success = new SuccessCallback(settings.success);
        jqAjax(settings);
    };
})();

现在您可以$.ajax像往常一样使用:

$.ajax({
    url: "test.html",
    success: function() {
        console.log("will be printed between 'start' and 'end'");
    }
});

据我所知,任何jQuery的AJAX函数(如$.get()或.load())都在内部使用$.ajax,所以这应该适用于通过jQuery完成的每个AJAX请求(我还没有测试过这个......).

像这样的东西也应该通过黑客攻击"纯粹的"JavaScript XMLHttpRequest.prototype.请注意,以下内容在IE中不起作用,而ActiveXObject不是使用XMLHttpRequest.

(function(){
    // overwrite the "send" method, but keep the original implementation in a variable
    var origSend = XMLHttpRequest.prototype.send;
    XMLHttpRequest.prototype.send = function(data){
        // check if onreadystatechange property is set (which is used for callbacks)
        if (typeof this.onreadystatechange === "function") {
            // overwrite callback function
            var origOnreadystatechange = this.onreadystatechange;
            this.onreadystatechange = function(){
                if (this.readyState === 4) {
                    console.log("start");
                }
                origOnreadystatechange();
                if (this.readyState === 4) {
                    console.log("end");
                }
            };
        }
        // execute the original "send" method
        origSend.call(this, data);
    };
})();

用法(就像通常的XMLHttpRequest一样):

var xhr = new XMLHttpRequest();
xhr.open("POST", "test.html", true);
xhr.onreadystatechange = function(){
    if (xhr.readyState === 4) {
        console.log("will be printed between 'start' and 'end'");
    }
};
xhr.send();

Answer 2

您可以使用闭包来完成您所需要的：

function closure(handler) {
    return function(ev, xhr, options) {
        console.log("start");
        handler(ev, xhr, options);
        console.log("stop");
    }
}

$(document).ajaxSend(closure(function(ev, xhr, options) {
    console.log("hello");
}));