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variable scope in AJAX calls

开发者 https://www.devze.com 2023-01-24 15:29 出处:网络
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 wh开发者_C百科ich i

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 wh开发者_C百科ich 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);
  }
});

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

How is this possible?


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

Here's a few from a google:

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

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

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