如何根据同一类型中提到的键在打字稿中声明类型？

Question

I want to declare a type in typescript ApiResponse and mention 3 keys in it which are isError, error, and content. What I want is that type should be declared as such that either isError and error exist or content exists.

type ApiResponse<T> = {
    isError?: boolean;
    error?: ErrorContent;
    content: this.isError ? undefined : User; // this is something I want . how should I do it.
}

I want this so that when I call a function which wants a parameter of User type doesn't give an error that the parameter is undefined

Answer 1

We can't define type based on dynamic value here,

1. one we need to use generic to get User type
2. Instead of isError boolean we should use status kind of enumeration ( success, error )

so that we represent invalid state properly. Try like below,

type ErrorContent = {};
type User = {};

interface SuccessResponse<T> {
  status: "success";
  content: T; // this is something I want . how should I do it.
}

interface ErrorResponse {
  status: "error";
  error: ErrorContent;
}

type ApiResponse<T> = SuccessResponse<T> | ErrorResponse;

const success: ApiResponse<User> = {
  status: "success",
  content: {}
};

const failure: ApiResponse<User> = {
  status: "error",
  error: {},
};

Answer 2

It is impossible to define type differently by variable. Simply you can define the type using | operator.

type ApiResponse<T> = {
  isError?: boolean;
  error?: ErrorContent;
  content: User | undefined;
}