How to call a function inside $(document).ready
问题 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 ? 回答1: You are not calling a function like that, you just define