How do you handle express controller method parameters when using typescript?

Question

I'm using express for routing and I don't need to use the Request object in my controller method. However, I can't find a simple way to force the exclusion of the Request parameter.

This first example passes typescript validation.

  public async findAll(req: Request, res: Response) {
    // search the db for all transactions
    const transactions = await Transaction.find();

    res.status(200).send({ transactions, incoming: req.body });
  }

This example compiles into valid javascript but doesn't pass typescript validation as I get the error 'req' is declared but its value is never read.

  public async findAll(req: Request, res: Response) {
    // search the db for all transactions
    const transactions = await Transaction.find();

    res.status(200).send({ transactions });
  }

Ideally I don't want to include the Request in my method parameter at all. Is this possible without an intricate workaround?

Estus Flask · Accepted Answer · 2019-01-05 18:37:48Z

2

'req' is declared but its value is never read error may be caused by TypeScript noUnusedParameters option. If it interferes with development process, it can be disabled in favour of similar TSLint rule which can be fine-tuned to cause a warning instead of an error.

Unused parameters can be conventionally underscored to avoid the check. This is the case for TypeScript:

public async findAll(_req: Request, res: Response) {...}

answered Jan 5, 2019 at 18:37

Estus Flask

226k79 gold badges478 silver badges620 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Pouria Over a year ago

Wow, I never knew you could use underscores to bypass the lint rules. Exactly what I'm looking for. Also, now that I know what to search for, here's what I needed: stackoverflow.com/questions/38835001/…

Pouria · Accepted Answer · 2019-01-05 18:07:35Z

1

I just found a solution with some more trial and error. Please post if you can find a more elegant solution.

To avoid the typescript compiler error message, simply make the function parameter an empty object ({}).

The code from the question will now look like

public async findAll({}, res: Response) {
  // search the db for all transactions
  const transactions = await Transaction.find();

  res.status(200).send({ transactions });
}

Hope this helps someone in the future as I had trouble finding an existing post.

answered Jan 5, 2019 at 18:07

Pouria

1271 silver badge12 bronze badges

1 Comment

Estus Flask Over a year ago

Didn't notice the post. It's not a good way to solve the problem, but an interesting one. This is not empty object but destructuring syntax. It will make the function a bit less efficient and would fail in case the first arg is not an object, but is acceptable in this case. The benefit is that doesn't require to invent unique names for params that won't be used.

Collectives™ on Stack Overflow

How do you handle express controller method parameters when using typescript?

2 Answers 2

1 Comment

1 Comment

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

1 Comment

Your Answer

Sign up or log in

Post as a guest

Linked

Related