2017-08-14 25 views
0

似乎流中的類類型始終引用該類的實例,並且使用typeof來引用實際的類本身。所以,如果我想要一個變量來引用基類的子類(不是實例),我可以這樣做:

class MyBaseClass {} 
class MySubClass extends MyBaseClass {} 
let a: $Subtype<MyBaseClass> = MySubClass; // fails 
let b: $Subtype<MyBaseClass> = new MySubClass(); // works, but I don't want this. 
let c: $Subtype<typeof MyBaseClass> = MySubClass; // works! Ok, we're good 

不過,我似乎無法與類型參數做到這一點!例如,以下內容:

type GenericSubclass<T> = $Subtype<typeof T>; 
// fails with `^ identifier `T`. Could not resolve name` 

如果我嘗試以下打字稿招(見Generic and typeof T in the parameters),它也將失敗:

type ValidSubclass<T> = { new(): T }; 
const c: ValidSubclass<BaseClass> = MySubClass; 
// fails with: property `new`. Property not found in statics of MySubClass 

請注意,我試圖new__proto__constructor

什麼給?有沒有解決方法?

回答

1
typeof MyBaseClass 

Class<MyBaseClass> 

所以你可以做

type GenericSubclass<T> = $Subtype<Class<T>>;