javascript - jquery document.ready multiple declaration - Stack Overflow

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I realized that I can specify $(document).ready(function(){}); more than once.
Suppose like this

$(document).ready(function(){
    var abc = "1122";
    //do something..
});

$(document).ready(function(){
    var abc = "def";
    //do something..
});

1. Is this standard ? Those codes work on my FF (16.0.2). I just a little afraid that other browser may not.
2. What actually happen ? How jQuery handle those code ?

Thanks.

I realized that I can specify $(document).ready(function(){}); more than once.
Suppose like this

$(document).ready(function(){
    var abc = "1122";
    //do something..
});

$(document).ready(function(){
    var abc = "def";
    //do something..
});

1. Is this standard ? Those codes work on my FF (16.0.2). I just a little afraid that other browser may not.
2. What actually happen ? How jQuery handle those code ?

Thanks.

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Share Improve this question Follow asked Nov 29, 2012 at 10:17 Hendry H.Hendry H. 1,52033 gold badges1414 silver badges2828 bronze badges 1

• Try this out – chridam Commented Nov 29, 2012 at 10:20

Add a ment  | 

3 Answers 3

Sorted by: Reset to default 10

Yes, it's standard and remended and jQuery guarantees those functions will be executed in order of declaration (see this related answer for details regarding the implementation).

The abc variable are local to your functions and don't interfere. You cannot see the value of abc declared in one of the callbacks from the other one.

Yes, multiple document.ready function can be used in the same page. It will not going to give any error or exception in any browser. and the function will be executed for upper to lower order.

Even though it's standard, I'd rather use only one $(document.ready) ... to keep it DRY, otherwise you may end up with let's say 10 of these declarations on your page, which will make it look like unmaintanable spaghetti code (but JS allows you to do that too)

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