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 ?

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

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

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