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

Question

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 ?

Answer 1

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!');
   })();
});