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Home/ Questions/Q 6786213
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T17:13:07+00:00 2026-05-26T17:13:07+00:00

I have a function: var myAnimation = function(){ $(.next_action).css({‘bottom’:-100%,left:0}).animate({‘bottom’:0},1000); $(‘.active’).animate({‘top’:-100%},1000); } that takes two

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I have a function:

var myAnimation = function(){
    $(".next_action").css({'bottom':"-100%","left":"0"}).animate({'bottom':"0"},1000);
    $('.active').animate({'top':"-100%"},1000);
}

that takes two objects an animates them, producing a sliding effect.

In order to get multiple direction support, I encapsulated the positions in arguments:

 var myAnimation = function(inE,outE){
        $(".next_action").css({inE:"-100%","left":"0"}).animate({inE:"0"},1000);
        $('.active').animate({outE:"-100%"},1000);
    }

Which I can now call thusly:

myAnimation('bottom','top');

As far as I am concerned, these functions should be extactly equivelent, but the first works, the second does not. No animation takes place at all, the new object simple jumps into position.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-26T17:13:08+00:00Added an answer on May 26, 2026 at 5:13 pm

    {inE:"-100%","left":"0"} creates an object with a key inE and a key left. The only time you need to enclose a key with quotes in JavaScript is if it is a reserved word (if, else etc), or contains control characters (-, :, {).

    Instead you’ll need something like this;

    var myAnimation = function(inE, outE) {
        var animObjA = {
            left: 0
        };
        var animObjB = {};
        var animObjC = {};
    
        animObjA[inE] = "-100%";
        animObjB[inE] = "0";
        animObjC[outE] = "-100%";
    
        $(".next_action").css(animObjA).animate(animObjB, 1000);
        $('.active').animate(animObjC, 1000);
    }
    

    Which uses square bracket notation to set the properties of the objects

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