调用ajax返回函数外部？(Calling ajax returned outside the function? [duplicate])

调用ajax返回函数外部？(Calling ajax returned outside the function? [duplicate])

这个问题在这里已有答案：

如何从异步调用返回响应？ 32个答案 jQuery：ajax调用成功后返回数据[重复] 5个答案

我试图调用在其函数之外返回的ajax，下面是我尝试过的代码但是没有用。

var names; $.ajax({ type: 'POST', url: 'ajax/get_upcoming.php' }).success(function (data) { names = data[1]; }); alert(names);

This question already has an answer here:

How do I return the response from an asynchronous call? 35 answers jQuery: Return data after ajax call success [duplicate] 5 answers

I am trying to call the ajax returned outside its function, below is the code I tried but didn't work.

var names; $.ajax({ type: 'POST', url: 'ajax/get_upcoming.php' }).success(function (data) { names = data[1]; }); alert(names);

最满意答案

因为成功函数是回调函数，并在ajax请求获得响应时调用，所以改为尝试这个

var names; $.ajax({ type: 'POST', url: 'ajax/get_upcoming.php', success: performOperation }); function performOperation(data) { // write your code here names = data[1]; }

Because success function is callback function and called when ajax request get the response , so instead try this

var names; $.ajax({ type: 'POST', url: 'ajax/get_upcoming.php', success: performOperation }); function performOperation(data) { // write your code here names = data[1]; }