I have a question on jQuery $(document).ready
Let\'s say we have a HTML page which includes 2 JavaScript files
- Will both these ready event function get fired ?
Yes, they will both get fired.
- what will the order in which they get fired, since the document will be ready at the same time for both of them?
In the way they appear (top to bottom), because the ready event will be fired once, and all the event listeners will get notified one after another.
- Is this approach recommended OR we should ideally have only 1 $(document).ready ?
It is OK to do it like that. If you can have them in the same block code it would be easier to manage, but that's all there is to it. Update: Apparently I forgot to mention, you will increase the size of your JavaScript code if you do this in multiple files.
- Is the order of execution same across all the browsers (IE,FF,etc)?
Yes, because jQuery takes the cross-browser normalization at hand.
See here: jQuery - is it bad to have multiple $(document).ready(function() {}); and here: Tutorials:Multiple $(document).ready()
readyList
with Deferred objects..js
file and be sure that it provides and binds its own ready handler).You can count on both handlers being executed in order of their script inclusion and globalVar
being 2
after the second script reference, in any current browser.
If you want full control I strongly recommend only one $(document).ready();
If you load partial portions of HTML through ajax and the ajax response includes a $(document).ready();-script and you want to fire $(document).ready(); from script1.js, script2.js and so on in the ajax callback.. You have to duplicate PLENTY of code....
Good Luck!
/ $(window).ready(); ;)