How pass a interface's property as parameter in typescript?-CodePudding

Sorry if this question is too stupid :D
I have an interface:

interface APIRoutes {
  SIGNUP  : string,
  LOGIN   : string,
}

And a const, like routes config.

const APIs: APIRoutes = {
  SIGNUP  : Config.baseURL   "/signup",
  LOGIN   : Config.baseURL   "/login",
}

I want to declare a function, it will take one of those route as parameter:

const callAPI = async (APIRoute: string, body: any, headers: any) => {
    //...
}

The problem is, I don't know how to make this function only accept one of the APIRoutes properties.
I mean something like this:

callAPI(APIs.SIGNUP, ...)                 // => OK
callAPI(APIs.STACKOVERFLOWISAWESOME, ...) // => ERROR

How can I do this? Not only interface, whatever can solve this problem is great.

Thanks

CodePudding user response：

You need keyof operator. It takes an object type and produces a union of its keys.

interface APIRoutes {
  SIGNUP  : string,
  LOGIN   : string,
}

const APIs: APIRoutes = {
  SIGNUP  : "/signup",
  LOGIN   : "/login",
}

const callAPI = async (APIRoute: keyof APIRoutes) => {
    
}

console.log(callAPI('asdasd'));
console.log(callAPI('SIGNUP'));
console.log(callAPI('LOGIN'));

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