你要完成什么不能ajaxSend
来完成。问题是,ajaxSend
显然与原始的xhr
和options
对象的副本一起使用,所以这些修改不会产生任何影响。您可以轻松地用下面的代码测试:
$(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"); }
});
所以你不能使用ajaxSend
覆盖成功回调。相反,你将不得不“黑客” 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;
$.ajax = function(settings){
// override the callback function, then execute the original AJAX function
settings.success = new SuccessCallback(settings.success);
jqAjax(settings);
};
})();
现在,你可以简单地使用$.ajax
像往常一样:
$.ajax({
url: "test.html",
success: function() {
console.log("will be printed between 'start' and 'end'");
}
});
据我所知,任何的jQuery的AJAX功能(如$.get()
或)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,尽管...)。
类似的东西也应该用“纯”的JavaScript由黑客
XMLHttpRequest.prototype
工作。请注意,以下内容在IE中不起作用,IE使用
ActiveXObject
而不是
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);
};
})();
使用(就像一个平常的XMLHttpRequest):
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();
相关:http://stackoverflow.com/questions/293668/jquery-ajax-prevent-callback-from-running – pimvdb 2012-08-17 10:31:32
感谢,但它不回答我的问题,在浏览器获取HTTP响应后,ajaxComplete触发,我询问有关在HTTP请求之前触发的ajaxSend – ciochPep 2012-08-17 10:39:38
您是对的。我只是指出回答说必须运行回调,而且你不能阻止回调。当然,这不完全是你的问题;我也对解决方案感兴趣。 – pimvdb 2012-08-17 10:42:20