Removing item from array without iterating

Question

I've looked through some documentation, but I feel like there's an easier method to remove one item from an array without using an iteration loop.

http://jsfiddle.net/G97bt/1/

Updated jsFiddle: http://jsfiddle.net/G97bt/6/ using not()

<button id="1">1</button>
<button id="2">2</button>
<button id="3">3</button>

<div id="1a">Test</div>
<div id="2a">Test</div>
<div id="3a">Test</div>

var $myList = $("#1a, #2a, #3a");

$("#1").on("click", function() {
    $myList.fadeOut(); // I want to fade DIVs #2a + #3a, not ALL
    $("#1a").fadeIn("slow");    
});
$("#2").on("click", function() {
    $myList.fadeOut(); // I want to fade DIVs #1a + #3a, not ALL
    $("#2a").fadeIn("slow");    
});
$("#3").on("click", function() {
    $myList.fadeOut(); // I want to fade DIVs #1a + #2a, not ALL    
    $("#3a").fadeIn("slow");    
});

This is how I'm envisioning it would function:

$myList.remove['#1a'].fadeOut();

javscript .splice() ? developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/… — mituw16
– mituw16, Commented Mar 31, 2014 at 13:40
Ugh. I even read through using splice() but it didn't click. Thank you, and sorry for a silly question :) — Sophisticake
– Sophisticake, Commented Mar 31, 2014 at 13:42

xdazz · Accepted Answer · 2014-03-31 13:57:39Z

10

Like below, by using .not to remove elements from the set of matched elements:

$myList.not('#1a').fadeOut();

And note $myList is not an array but a jQuery object(even it behaves like an array).

You could also rewrite your code like below:

var $myList = $("#1a, #2a, #3a");

$("#1,#2,#3").on("click", function() {
    $myList.not('#'+this.id+'a').fadeOut();
    $('#'+this.id+'a').fadeIn("slow");    
});

The working demo.

edited Mar 31, 2014 at 13:57

answered Mar 31, 2014 at 13:41

xdazz

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2 Comments

Sophisticake Over a year ago

stackoverflow.com/questions/22762173/…

xdazz Over a year ago

@Sophisticake Because .not() is designed to do such job. And splice() is a method of array, not jQuery object. splice() returns the removed element by position, etc...

Anoop Joshi P · Accepted Answer · 2014-03-31 13:42:24Z

0

You can use like this

$("button").click(function(){
  $("[id$=a]").hide();

  $("#"+this.id+"a").show();
});

answered Mar 31, 2014 at 13:42

Anoop Joshi P

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Comments

Elias Soares · Accepted Answer · 2014-03-31 13:43:16Z

0

Use

$arrayName.splice(position, number of elements)

answered Mar 31, 2014 at 13:43

Elias Soares

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