Leveraging Express Mergeparams in TypeScript

Question

Leveraging Express Mergeparams in TypeScript

I've run into an issue while working on my small project in Typescript. The problem arises when I attempt to nest my router, as Typescript doesn't seem to acknowledge the parent's parameter.

Within the "child" file, I have the following code:

const childRouter = express.Router({ mergeParams: true });

childRouter.get('/', (req, res) => {
    const userName = req.params.username; 
    // This results in the error, Property 'username' does not exist on type '{}'
});

Then, in the "parent" file, the code looks like this:

import childRouter from './child';

const parentRouter = express.Router();

parentRouter.use('/:username', childRouter);

I'm at a loss on how to resolve this issue. It seems like Typescript is not recognizing that I'm using the parent's parameter. Any suggestions on how to address this?

typescript express

Answer 1

Answer №1

Regrettably, the information cannot be deduced from the parent route itself.

To tackle this situation in a more organized manner, one can utilize generics:

import { RouteParameters } from 'express-serve-static-core';

type WithParentParams<T extends string> = RouteParameters<':userName'> & RouteParameters<T>
//                       Insert any additional parent parameters here -^^^^^^^^^

const route = '/' as const;

childRouter.get<typeof route, WithParentParams<typeof route>>(route, async (req, res) => {
  const { userName } = req.params;

This approach may seem complex, as it requires setting up the WithParentParams interface and repeating the route as the initial parameter in

childRouter.get<routeType, paramsType>

.

Nevertheless, it is preferable to aim for clarity rather than relying on type casting.

Answer 2

Regrettably, the information cannot be deduced from the parent route itself.

To tackle this situation in a more organized manner, one can utilize generics:

import { RouteParameters } from 'express-serve-static-core';

type WithParentParams<T extends string> = RouteParameters<':userName'> & RouteParameters<T>
//                       Insert any additional parent parameters here -^^^^^^^^^

const route = '/' as const;

childRouter.get<typeof route, WithParentParams<typeof route>>(route, async (req, res) => {
  const { userName } = req.params;

This approach may seem complex, as it requires setting up the WithParentParams interface and repeating the route as the initial parameter in

childRouter.get<routeType, paramsType>

.

Nevertheless, it is preferable to aim for clarity rather than relying on type casting.

Answer 3

Answer №2

Correction: Upon rereading the original post, I realized my mistake.

To correctly define the request.params type in Express, make sure you have installed the types properly.

interface ParamsDictionary {
    [key: string]: string;
}

If you want to explicitly set the type, you can also use the following method:

const username = req.params.username as string

Answer 4

Correction: Upon rereading the original post, I realized my mistake.

To correctly define the request.params type in Express, make sure you have installed the types properly.

interface ParamsDictionary {
    [key: string]: string;
}

If you want to explicitly set the type, you can also use the following method:

const username = req.params.username as string

Answer 5

Answer №3

My solution to this issue may not be perfect, but it is versatile and takes into account the fact that the parent's parameters are unknown.

interface ParentParams { username: string };

childRouter.get('/:childId', (req, res) => {
    const { userName, childId } = req.params as typeof req.params & ParentParams;
    ...

You can define the parent's parameters that the child is not aware of, and include any additional child parameters in the route if needed. This approach ensures that Express TypeScript type checking works for both existing child parameters and the parent's parameters.

Answer 6

My solution to this issue may not be perfect, but it is versatile and takes into account the fact that the parent's parameters are unknown.

interface ParentParams { username: string };

childRouter.get('/:childId', (req, res) => {
    const { userName, childId } = req.params as typeof req.params & ParentParams;
    ...

You can define the parent's parameters that the child is not aware of, and include any additional child parameters in the route if needed. This approach ensures that Express TypeScript type checking works for both existing child parameters and the parent's parameters.

Answer 7

Answer №4

Here's a quick and dirty workaround:

childRouter.get('/', (req, res) => {
  const { username } = req.params as any;
  …
});

Answer 8

Here's a quick and dirty workaround:

childRouter.get('/', (req, res) => {
  const { username } = req.params as any;
  …
});

Answer 9

Answer №5

I have yet to come across a perfect solution,
However, I have come up with a rather effective workaround:

(all in the name of type support...)

uniqueUserId.ts

const uniqueUserIdRouter = (userid: string) => {

  router.get("/", async (req, res, next) => {
    res.send(`User's unique identifier is: ${userid}`);
  });

  return router;
};

export default uniqueUserIdRouter;

Now, you can easily set up a new router in users with access to the user's unique identifier...

users.ts

router.use("/:userid", (req, res, next) => {
  const { userid } = req.params;
  uniqueUserIdRouter(userid)(req, res, next);
});

Answer 10

I have yet to come across a perfect solution,
However, I have come up with a rather effective workaround:

(all in the name of type support...)

uniqueUserId.ts

const uniqueUserIdRouter = (userid: string) => {

  router.get("/", async (req, res, next) => {
    res.send(`User's unique identifier is: ${userid}`);
  });

  return router;
};

export default uniqueUserIdRouter;

Now, you can easily set up a new router in users with access to the user's unique identifier...

users.ts

router.use("/:userid", (req, res, next) => {
  const { userid } = req.params;
  uniqueUserIdRouter(userid)(req, res, next);
});

Leveraging Express Mergeparams in TypeScript

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

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