如何查看变量,在ajax中运行其他文件脚本

Example: I have file ajax demo_ajax.php will return:

<p>this is data in file ajax</p>
<script>
   demo();
</script>

And file script demo_script.js :

$(document).ready(function(){
  function demo(){
  // my code here
  }
});

My question is how to in file demo_ajax.php can call function demo() in file demo_script.js, I will try and get error function undefined

Change

$(document).ready(function(){
  function demo(){
  // my code here
  }
});

To

$(document).ready(function(){

}); 
  function demo(){
  // my code here
  }

Dont't forgot to include your script file on parent page

Try this...

$(document).ready(function() {
  $(this).demo();
});
// You can replace 'this' with another selector such as 'p'
// Then inside the 'demo' function, $(this) will refer to your selector
$.fn.demo = function()
{

}

include your

demo_script.js

on the top of

demo_ajax.php

and then it will work..