Building a typeguard in Typescript that handles errors

type VerifiedContext = Required<ApolloContext>;
function authenticateUser(context: ApolloContext): context is VerifiedContext {
  if (!context.user) {
    throw new AuthenticationError("Login required for this operation");
  }

  return true;
}

Here is an example of using the authentication middleware:

async fetchUser(_, { id }, context) {
    authenticateUser(context);

    const user = context.user;
    // Expected behavior from Typescript to determine that user is defined and not null
  },

Is there a way to assist Typescript in correctly inferring the type? For example, to indicate that if the code runs after the middleware, the user is considered "verified"?

The main objective is to avoid the need for conditional checks when implementing the authentication middleware.

Answer №1

If you're looking for something similar to a type assertion at block-scope level, unfortunately, TypeScript does not currently support this feature (as of version 3.3). However, you can visit the GitHub issue linked and show your support or share your use case to encourage its implementation. At the moment, there is no direct way to narrow the type of a variable without using an if check.

So what can you do as an alternative? In situations like this, I often replace the type guard with a function that returns the narrowed object. Instead of using x is T, you can simply return T:

function authMiddleware(context: ApolloContext): LoggedContext {
  if (!context.user) {
    throw new AuthenticationError("Resolver requires login");
  }
  // The following assertion is a way to narrow the type
  return context as LoggedContext;
}

When utilizing this function, make sure to use the returned value instead of the original parameter for all subsequent references:

const aContext = authMiddleware(context);
// Use aContext instead of context going forward:

const user = aContext.user; // string

Hopefully, this approach gives you some ideas on how to proceed. Best of luck!

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