确定一个对象属性是否是可观察的

我使用的是 帅哥2.0版

如果循环遍历对象的所有属性，那么如何测试每个属性是否是 ko.observable？以下是我目前为止的尝试:

    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);
}

但到目前为止，一切都是假的。

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最佳答案

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.

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

ko.isObservable(vm[key])
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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

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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.


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