Jquery code error: Undefined is not a function

Question

Simple problem that I cannot find the answer to. I have made a jsfiddle of the problem.

I have this Javascript code:

$(document).ready(function(){
    $('#display-feedback-form').off('click').on('click', function(){
        $('#feedback_form').show();
    });
});

Here is the fiddle with an example: JSFiddle

Why am I getting this error? Jquery is loaded correctly on the page.

esqew · Accepted Answer · 2014-07-23 13:43:46Z

7

Sure, you're loading jQuery correctly, but you're loading jQuery 1.6.4.

$.off() and $.on() weren't added until jQuery 1.7.

answered Jul 23, 2014 at 13:43

esqew

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Comments

Rodney G · Accepted Answer · 2014-07-23 13:43:54Z

3

.on() and .off() weren't introduced until jQuery 1.7. You are loading 1.6.4.

answered Jul 23, 2014 at 13:43

Rodney G

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Comments

Dhaval Modi · Accepted Answer · 2014-07-23 13:47:13Z

1

Try This

$(document).ready(function(){
    $('#display-feedback-form').click(function(){
        $('#feedback_form').show();
    });
});

JSFiddle

answered Jul 23, 2014 at 13:47

Dhaval Modi

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Comments

alexP · Accepted Answer · 2014-07-23 13:52:54Z

1

You can use .die() and .live() in jQuery 1.6.4. Until 1.7 there is no .off() and .on() function.

$(document).ready(function(){
    $('#display-feedback-form').die('click').live('click', function(){
        $('#feedback_form').show();
    });
});

DEMO

edited Jul 23, 2014 at 13:52

answered Jul 23, 2014 at 13:47

alexP

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