Working with two jquery document.ready functions

Question

I have two document.ready functions as shown below:

    $(document).ready(function() {
        $('#slider1').s3Slider({
            timeOut: 3000 
        });
    });
    $(document).ready(function() {
     $(".modalbox").fancybox();
     $("#contact").submit(function() {
    return false;
});

The first is a banner jquery that is supposed to rotate forever. The second is a contact form modal box that appears on the user's click. Is it possible for the two to coexist?

Why don't you put both things inside a single document.ready? — Th0rndike, Sep 03 '13 at 15:04
it would work indeed, but the best approach is to wrap them in a single document.ready() — maverickosama92, Sep 03 '13 at 15:05
By the way, you are not closing well second `document.ready`. You are just closing the `submit` event. — Alvaro, Sep 03 '13 at 15:05
Please explain what problem you're encountering. SO is not the correct venue for hypotheticals. — Ruan Mendes, Sep 03 '13 at 15:05
@maverickosama92 It's definitely not best approach. It depends on the context. For example if you use few libraries with `.ready()` then you won't be modifying them now will you? — freakish, Sep 03 '13 at 15:06

score 0 · Answer 1 · answered Sep 03 '13 at 15:04

0

Of course. You only add more handlers to ready event and all of them will fire once the ready event fires.

answered Sep 03 '13 at 15:04

freakish

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score 0 · Answer 2 · answered Sep 03 '13 at 15:06

0

Totally possible. In your example it is just an overhead but you can perfectly have two subscribers to an event.

answered Sep 03 '13 at 15:06

Aegis

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score 0 · Answer 3 · answered Sep 03 '13 at 15:07

0

$(init);
function init(){
    banner();
    contactForm();
}
function banner(){ 
    $('#slider1').s3Slider({timeOut: 3000});
}
function contactForm(){
    $(".modalbox").fancybox();
    $("#contact").submit(...);
}

answered Sep 03 '13 at 15:07

itsmikem

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Working with two jquery document.ready functions

3 Answers3