[英]Typescript: Union Type as interface
I have a situation where I need a function to accept a variety of different types. 我遇到的情况是我需要一个函数来接受各种不同的类型。 They will be separated by type guards within the function. 它们将由函数中的类型保护分隔。 So I am using a Union Type like this: 所以我正在使用这样的联合类型:
function(param: TypeA | TypeB): void {
let callback: (param: TypeA | TypeB): void = () => {};
if (isTypeA(param)) {
callback = doSomethingWithTypeA;
} else if (isTypeB(param)) {
callback = doSomethingWithTypeB;
}
return callback(param);
}
Where the function doSomethingWithTypeA
only accepts typeA and so on. 函数doSomethingWithTypeA
仅接受typeA,依此类推。
As you can see, always writing out (TypeA | TypeB)
is very verbose, especially since it's more than two types in my actual code. 如您所见,总是写出(TypeA | TypeB)
非常冗长,特别是因为在我的实际代码中,它是两个以上的类型。
Is there some way to create an interface that is (TypeA | TypeB)
? 有什么方法可以创建(TypeA | TypeB)
接口?
Or is there some other way to achieve this? 还是有其他方法可以做到这一点?
使用类型别名 :
type YourType = TypeA | TypeB // | ... and so on
How about creating an interface: 如何创建界面:
interface MyType {
}
class TypeA implements MyType {
}
class TypeB implements MyType {
}
Then you can check it in the method: 然后可以在方法中检查它:
function(param: MyType): void {
let callback: (param: MyType): void = () => {};
if (param instanceof TypeA) {
callback = doSomethingWithTypeA;
} else if (param instanceof TypeB) {
callback = doSomethingWithTypeB;
}
return callback(param);
}
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.