在AJAX调用变量的作用域 [英] variable scope in AJAX calls

问题描述

一个问题，我一直在问自己是怎么可能的JavaScript有仍处于一个AJAX请求回调函数的引用，当变量是在isssues AJAX调用的函数声明。下面的例子

one question I always ask myself is how is it possible that javascript has still a reference in a callback function of a AJAX request when the variable was declared in the function which isssues the AJAX call. Here an example

var loadMask = {name:"test"};

form.submit({
  url: 'request.php',
  timeout : 180000,
  success: function(the_form, action_object)
  {    
    console.log(loadMask);
  }
});

尽管loadMask是成功的功能它仍然是可见的外部声明（和定义）的事实中。

despite the fact that loadMask was declared outside of the success function it is still visible (and defined) inside.

这怎么可能？

推荐答案

这是有可能使用一种叫做关闭。有许多资源，这样的：

This is possible using something called closures. There are many resources for this:

下面是来自谷歌的几个：

Here's a few from a google:

http://www.webreference.com/programming/javascript/rg36/

http://jibbering.com/faq/notes/closures/

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