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Home/ Questions/Q 850519
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T07:16:37+00:00 2026-05-15T07:16:37+00:00

I need to cache about 100 different selections for animating. The following is sample

  • 0

I need to cache about 100 different selections for animating. The following is sample code. Is there a syntax problem in the second sample? If this isn’t the way to cache selections, it’s certainly the most popular on the interwebs. So, what am I missing?

note: p in the $.path.bezier(p) below is a correctly declared object passed to jQuery.path.bezier (awesome animation library, by the way)

This works

    $(document).ready(function() {
        animate1();
        animate2();
    })
    function animate1() {
        $('#image1').animate({ path: new $.path.bezier(p) }, 3000);
        setTimeout("animate1()", 3000);
    }
    function animate2() {
        $('#image2').animate({ path: new $.path.bezier(p) }, 3000);
        setTimeout("animate2()", 3000);
    }

This doesn’t work

    var $one = $('#image1'); //problem with syntax here??
    var $two = $('#image2');
    $(document).ready(function() {
        animate1();
        animate2();
    })
    function animate1() {
        $one.animate({ path: new $.path.bezier(p) }, 3000);
        setTimeout("animate1()", 3000);
    }
    function animate2() {
        $two.animate({ path: new $.path.bezier(p) }, 3000);
        setTimeout("animate2()", 3000);
    }
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-15T07:16:38+00:00Added an answer on May 15, 2026 at 7:16 am

    If the images are not loaded when you call them, jQuery will return an empty object. Move your assignment inside your document.ready function:

    $(document).ready(function() {
        var $one = $('#image1');
        var $two = $('#image2');
        animate1();
        animate2();
    });
    // ... etc.
    

    If you need to cache them for later use outside of your initialization script then add them to a storage object:

    var my_storage_object = {};
    $(document).ready(function() {
        var $one, $two;
        my_storage_object.$one = $one = $('#image1');
        my_storage_object.$two = $two = $('#image2');
        animate1();
        animate2();
    });
    // ... etc.
    

    Then later on, outside of document.ready you can call:

    my_storage_object.$one //still has a reference to the jQuery object.
    
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