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Home/ Questions/Q 517299
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T07:49:11+00:00 2026-05-13T07:49:11+00:00

I’m a webdesigner that’s trying to get the hang of JavaScript and jQuery, and

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I’m a webdesigner that’s trying to get the hang of JavaScript and jQuery, and I want to learn how to write shorter, more concise code – to avoid being ridiculed by the developers at work 😉

I have this snippet:

// toggle divs when checkbox is checked
$('.product-optional-checkbox1').click(function () {
    $('.product-optional-toggle1').toggle('fast');  
});

$('.product-optional-checkbox2').click(function () {
    $('.product-optional-toggle2').toggle('fast');  
});

$('.product-optional-checkbox3').click(function () {
    $('.product-optional-toggle3').toggle('fast');  
});

// hide divs
$('.product-optional-toggle1').hide(); 
$('.product-optional-toggle2').hide(); 
$('.product-optional-toggle3').hide();

…that I want to reduce using a for-loop, like this:

for( var i = 1; i < 4; ++i ) {
    $('.product-optional-checkbox' + i).click(function () {
        $(this).parent('div').find('div').toggle('fast');  
    });
    $('.product-optional-toggle' + i).css({ display: 'none'}); 
};

It works fine in FF, however in IE7 it toggles twice. Anyone know who to solve a problem like this?

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  1. Editorial Team
    Editorial Team
    2026-05-13T07:49:11+00:00Added an answer on May 13, 2026 at 7:49 am

    It’s hard to say without seeing the HTML structure but maybe going to the parent then descendants is finding multiple divs?

    In your example, you could replace:

    $(this).parent('div').find('div').toggle('fast');
    

    With code more similar to the original examples:

    $('.product-optional-toggle' + i).toggle('fast');
    

    However, it would be much better to ditch all the numbers and just use a class of .product-optional-checkbox. This way you can add a click function to all elements of that class in one go and avoid the loop:

    $('.product-optional-checkbox').click(function () {
        // do stuff using $(this)
    });
    
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