jQuery init() function.

Question

Is there a way to force widgets to output jquery init() functions in one '$(document).ready()'?
It should look like this:

$(document).ready(function($) {
     $('.style1').plugin1Init();
     $('.style2').plugin2Init();
     $('.style3').plugin3Init();
});

I'm asking because I'm trying to speed up my wordpress theme. I have several widgets which outputs jquery code in $(document).ready() functions.

Three '$(document).ready()' functions reduces page loading speed almost three times according to Jquery speed test

Example code on a page when I'm using 3 widgets:

$(document).ready(function($) {
     $('.style1').plugin1Init();
});

$(document).ready(function($) {
     $('.style2').plugin2Init();
});

$(document).ready(function($) {
     $('.style3').plugin3Init();
});

Having multiple document ready blocks is not a problem at all. There is only a single event which will call all registered functions in a loop. If your page loads slowly it's due to something else. — ThiefMaster
– ThiefMaster, Commented Jun 15, 2012 at 17:46

AriePutranto · Accepted Answer · 2012-06-16 00:45:25Z

1

You can call $(document).ready(function($) { once

See:

$(document).ready(function($) {
    $('.style1').plugin1Init();
    $('.style2').plugin2Init();
    $('.style3').plugin3Init();
});

answered Jun 16, 2012 at 0:45

AriePutranto

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