TypeScript: How to create a type or interface that is an object that has one of n keys or is a string?

Question

I need to create an interface that is either a string or an object with one of three keys.

Basically I have a function that depending on the error returns something:

export const determineError = (error: ServerAlerts): AlertError => {
  if (typeof error !== "string") {
    if (error.hasOwnProperty("non_field_errors")) {
      return error.non_field_errors[0];
    } else if (error.hasOwnProperty("detail")) {
      return error.detail;
    } else if (error.hasOwnProperty("email")) {
      return error.email[0];
    } else {
      return UNKNOWN_ERROR;
    }
  } else {
    return error;
  }
};

Here are the types:

export type AlertError =
  | "Unable to log in with provided credentials."
  | "E-mail is not verified."
  | "Password reset e-mail has been sent."
  | "Verification e-mail sent."
  | "A user is already registered with this e-mail address."
  | "Facebook Log In is cancelled."
  | string;

export interface ServerAlerts {
  non_field_errors: [string];
  detail: string;
  email: [string];
}

But the way I designed ServerAlerts here does not work for me, since ServerAlerts can also be a string and if it has one of its keys, it only has one.

How would you design such a type or interface?

EDIT: I tried making the keys optional by giving them a question mark, but then my linter complains in the respective key's error return statement in determineError .

Answer 1

If I'm understanding you correctly, just declare the parameter as being either ServerAlerts or string :

export const determineError = (error: ServerAlerts|string): AlertError => {
// -----------------------------------------------^^^^^^^

In a comment you've said that all three of the ServerAlerts properties are optional, so you need to mark them as such with ? :

interface ServerAlerts {
  non_field_errors?: [string];
  detail?: string;
  email?: [string];
}

However, that means that anything typed object will also work, because all the fields are optional. So if you do both of those things, you get:

determineError("foo");                       // Works
determineError({ non_field_errors: ["x"] }); // Works
determineError({ detail: "x" });             // Works
determineError({ email: ["x"] });            // Works
determineError({});                          // Works (because all fields are optional)
let nonLiteralServerAlerts: object;
nonLiteralServerAlerts = { foo: ["x"] };
determineError(nonLiteralServerAlerts);      // Works (because all fields are optional)
determineError({ foo: ["x"] });              // Fails (correctly)

Playground example

Which suggests you might just use object in the parameter signature. If you want to require one of the three fields (which I believe would do away with that UNKNOWN_ERROR branch), you'd define three interfaces and make ServerAlerts a union of them:

interface ServerAlertsNonFieldErrors {
  non_field_errors: [string];
}

interface ServerAlertsDetail {
  detail: string;
}

interface ServerAlertsEmail {
  email: [string];
}

type ServerAlerts = ServerAlertsNonFieldErrors | ServerAlertsDetail | ServerAlertsEmail;

Then you'd use type assertions when returning the specific field:

if (error.hasOwnProperty("non_field_errors")) {
  return (error as ServerAlertsNonFieldErrors).non_field_errors[0];
// ------------^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

If you do that, then you get:

determineError("foo");                       // Works
determineError({ non_field_errors: ["x"] }); // Works
determineError({ detail: "x" });             // Works
determineError({ email: ["x"] });            // Works
determineError({});                          // Fails (correctly)
let nonLiteralServerAlerts: object;
nonLiteralServerAlerts = { foo: ["x"] };
determineError(nonLiteralServerAlerts);      // Fails (correctly)
determineError({ foo: ["x"] });              // Fails (correctly)

Playground Example