jQuery .prop() compatibility

Question

I'm trying to test if the .prop() method exists on the current jQuery included (for compatibility reason) via:

if(typeof $.prop === 'function')

I would expect that the condition above is true for jQuery >= 1.6 and false for jQuery < 1.6 as I can understand from the docs

Anyway, testing this on jsfiddle, leads to:

typeof $.prop === 'function' is:

true when jQuery >= 1.6
false when jQuery < 1.6 and jQuery > 1.3
true when jQuery <= 1.3

here is the very very simple script which provide the results above (just switch jQuery version to see what I've described).

When I try to use that .prop() with jQuery i.e. 1.3 I get the .prop is not a function error. The same problem occours also testing outside jsfiddle. Is it normal such a behavior? How could I truly test if .prop() is available?

Thanks

i did test it on FF4, Safari5 and Chrome12 – Dalen Jun 12 '11 at 17:36 — Dalen, Jun 12 '11 at 17:36
Why support 1.3. It's really old. – Raynos Jun 12 '11 at 17:36 — Raynos, Jun 12 '11 at 17:36

score 6 · Accepted Answer · answered Jun 12 '11 at 17:37

6

alert(typeof $.fn.prop === 'function')

You want to check for the .prop method on the jQuery prototype which lives on $.fn. This is false in 1.3.

Also I would avoid feature detection for jQuery versions and instead support a particular version (and up).

answered Jun 12 '11 at 17:37

Raynos

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score 6 · Answer 2 · answered Dec 08 '11 at 10:25

What I did, is build a compatible function for prop, for jQuery versions, which doesn't use prop:

(function($){
    if (typeof $.fn.prop !== 'function')
    $.fn.prop = function(name, value){
        if (typeof value === 'undefined') {
            return this.attr(name);
        } else {
            return this.attr(name, value);
        }
    };
})(jQuery);

You can test this code: http://jsfiddle.net/JtK2Q

score 4 · Answer 3 · answered Jun 12 '11 at 17:37

4

You're checking for the existence of a static method.

You need to check for the instance method by writing $.fn.prop ($.fn is the same as $.prototype).

answered Jun 12 '11 at 17:37

SLaks

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It's the same as `$.fn.init.prototype` – Raynos Jun 12 '11 at 17:39

jQuery .prop() compatibility

3 Answers3