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

debugging, firebug, jquery

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

Problem

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 ?

Original source