Calling function inside jQuery document ready

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6 Answers

callinginsidefunction
90%

In firebug im calling functions inside the $(document).ready.., Meta Stack Overflow , Stack Overflow help chat ,Stack Overflow en español

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!');
   })();
});
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88%

A page can't be manipulated safely until the document is "ready." jQuery detects this state of readiness for you. Code included inside $( document ).ready() will only run once the page Document Object Model (DOM) is ready for JavaScript code to execute. Code included inside $( window ).on( "load", function() { ... }) will run once the entire page (images or iframes), not just the DOM, is ready.,The example below shows $( document ).ready() and $( window ).on( "load" ) in action. The code tries to load a website URL in an <iframe> and checks for both events:,You can also pass a named function to $( document ).ready() instead of passing an anonymous function.,Experienced developers sometimes use the shorthand $() for $( document ).ready(). If you are writing code that people who aren't experienced with jQuery may see, it's best to use the long form.

// A $( document ).ready() block.$( document ).ready(function() {    console.log( "ready!" );});
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handler Type: Function() A function to execute after the DOM is ready. , version added: 1.0.ready( handler ) handler Type: Function() A function to execute after the DOM is ready. ,Description: Specify a function to execute when the DOM is fully loaded.,jQuery offers several ways to attach a function that will run when the DOM is ready. All of the following syntaxes are equivalent:

$(document).ready(function() { // Handler for .ready() called.});
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65%

No, they don’t. When using jQuery we can define the functions we want to use after our document is loaded, inside of jQuery’s .ready() method. This way of defining functions ensures that our functions are scoped to the .ready() method and will not be available for access from outside of .ready(). If we define our functions outside of .ready() said functions will be available for use both inside and outside of .ready().,Do my functions have to be defined in jQuery’s .ready() method?,Defining a function outside of .ready():,Defining a function inside .ready():

Defining a function inside .ready():

$(document).ready(() => {
   function sayHello() { //defines sayHello inside the .ready() method
      console.log('hello');
   }
});

sayHello(); //this will error as sayHello is not available outside of the .ready() method
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75%

// jQuery document ready
$(document).ready(function() {

});
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40%

Im trying to debug my web app that uses jQuery.,In firebug im calling functions inside the $(document).ready..,The correct approach is to define the function outside document.ready and call it inside:,In firebug console I type validate() and it says its not a function

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!');
          }
       }
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