Determine if an object property is ko.observable

Question

I'm using KnockoutJS version 2.0.0

If I'm looping through all properties of an object, how can I test whether each property is a ko.observable? Here's what I've tried so far:

    var vm = {
        prop: ko.observable(''),
        arr: ko.observableArray([]),
        func: ko.computed(function(){
            return this.prop + " computed";
        }, vm)
    };

    for (var key in vm) {
        console.log(key, 
            vm[key].constructor === ko.observable, 
            vm[key] instanceof ko.observable);
    }

But so far everything is false.

RP Niemeyer · Accepted Answer · 2013-07-31T15:36:51.543

163

Knockout includes a function called ko.isObservable(). You can call it like ko.isObservable(vm[key]).

Update from comment:

Here is a function to determine if something is a computed observable:

ko.isComputed = function (instance) {
    if ((instance === null) || (instance === undefined) || (instance.__ko_proto__ === undefined)) return false;
    if (instance.__ko_proto__ === ko.dependentObservable) return true;
    return ko.isComputed(instance.__ko_proto__); // Walk the prototype chain
};

UPDATE: If you are using KO 2.1+ - then you can use ko.isComputed directly.

edited Jul 31 '13 at 15:36

answered Mar 08 '12 at 22:13

RP Niemeyer

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Thank you. Do you by chance know how to tell if an observable is computed? I can determine if an observable is an observable array via `$.isArray(vm[key]())`, but do you know how to differentiate observables from a `ko.computed`?? – Adam Rackis Mar 08 '12 at 22:19
7

KO 2.1 that should be out in the next few weeks will include a `ko.isComputed` function. The code would be the equivalent to what I added to the answer above. – RP Niemeyer Mar 09 '12 at 02:36

score 29 · Answer 2 · answered Mar 08 '12 at 22:21

29

Knockout has the following function which I think is what you are looking for:

ko.isObservable(vm[key])

answered Mar 08 '12 at 22:21

Mark Robinson

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score 3 · Answer 3 · answered Nov 21 '17 at 20:01

3

To tack on to RP Niemeyer's answer, if you're simply looking to determine if something is "subscribable" (which is most often the case). Then ko.isSubscribable is also available.

answered Nov 21 '17 at 20:01

pim

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score 0 · Answer 4 · edited Apr 09 '15 at 21:11

0

I'm using

ko.utils.unwrapObservable(vm.key)

Update: As of version 2.3.0, ko.unwrap was added as substitute for ko.utils.unwrapObservable

edited Apr 09 '15 at 21:11

Homer

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answered Jul 29 '14 at 09:27

Ivan Rodriguez

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That unwraps an observable. It doesn't test whether a property is an observable. – Adam Rackis Jul 29 '14 at 14:48
1

You're right. But usually if you need to know to get the actual observable value ;) – Ivan Rodriguez Jul 30 '14 at 07:57
1

It provides a valid way to access a property that may or may not be observable however it doesn't answer the question per se. – Ozil Jun 14 '17 at 10:41

Determine if an object property is ko.observable

4 Answers4

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