jquerydebuggingfirebug

How to call a function inside $(document).ready


Im trying to debug my web app that uses jQuery.

In firebug im calling functions inside the $(document).ready..

 function val() { console.log('validated outside doc.ready'); }
 $(document).ready(function()
 {

    console.log('document ready...');

    function validate() { console.log('validated!'); }
 }

In firebug console I type validate() and it says its not a function

If i type val() it works fine.

How do i call validate from the console ?


Solution

  • You are not calling a function like that, you just define the function.

    The correct approach is to define the function outside document.ready and call it inside:

    // We define the function
    function validate(){
      console.log('validated!');
    }
    
    $(document).ready(function(){
      // we call the function
      validate();
    });
    

    Another option is to self invoke the function like that:

    $(document).ready(function(){
       // we define and invoke a function
       (function(){
         console.log('validated!');
       })();
    });