CODEWITHSUNDEEP
javascriptjqueryslice
Katai
Katai

Reputation: 2937

First .slice() and then do things with all other elements

I was quite sure that I already did this in some earlier version of jQuery, but http://api.jquery.com/category/traversing/ seems to suggest otherwise.

What I'm looking for is similar kind of the opposite of .addBack() - a traversing function that uses "all other" elements (not .not()!)

Preusdo Example:

$('.some-class li').slice(33,55).hide().allOthers().show()

Edit: This is not actually a hide() / show() based problem, this is just a simple example to clarify what I meant.

First, I'ld like to manipulate a set of elements selected with .slice(), and then manipulate all elements that were not selected by .slice().

Is there a handy traversing function I've missed that does just that? I know how to solve it in general, but a ".allOthers()" method that I might have missed would certainly be more handy and clearer.

Upvotes: 2

Views: 67

Answers (3)

Ruan Mendes
Ruan Mendes

Reputation: 92274

In your case you can just call show before calling slice

$('.some-class li').show().slice(33,55).hide();

It's true that there is no method to get all others, the closest is to get back they previous collection as you mentioned, http://api.jquery.com/addback/

You could implement a plugin, since I'm on my mobile, I'll just write some straight code

// o(n*m), could be improved
function allOthers(jqObj) {
   var current = [].concat(jqObj);
   var prev = jqObj.addBack();

   return prev.filter(function(obj){
      return !current.includes(obj);
   });

}

Upvotes: 2

Katai
Katai

Reputation: 2937

After testing @JuanMendes suggestion, I played around with it a bit and found quite a compact way to implement this kind of functionality, due to jQuery's prevObject:

$.fn.others = function() {
  return this.prevObject.not( this );
}

I didn't test it too much with other methods, so it might needs some further changes - but it seems to work fine with .slice() at least.

https://jsfiddle.net/1L3db7k4/

Upvotes: 1

gurvinder372
gurvinder372

Reputation: 68393

First show all of them and then hide from 33 to 55, here is the demo

$('.some-class li').show().slice(33,55).hide();

Upvotes: 1

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