How to handle ExpressJS optional route parameters with TypeScript?

Question

I have the following code:

`index.ts`

import express from "express";
import { Uuid } from "./Uuid";
import { UuidGen } from "./Uuid_gen";

const app = express();
const port = process.env.port || 3000;

app.get("/Uuidgen/1/:operatingAgency/:locId?", (request, result) =>
    result.send(UuidGen.generateType1Uuid(request.params.operatingAgency, request.params.locId).toString())
);

app.listen(port, () => console.log(`UUID Generator listening on port ${port}`));

Now, when the file is saved with a .ts extension, the TypeScript compiler complains:

Now, when the 'Quick Fix' is applied, this is the resulting code:

app.get("/Uuidgen/1/:operatingAgency/:locId?", (request, result) =>
    result.send(UuidGen.generateType1Uuid(request.params.operatingAgency, request.params["locId?"]).toString())

This compiles fine, but the locId parameter is never passed to the callback function and is empty. What can I do? A temporary solution is to change the extension to .js and ensure "checkJs": false in tsconfig.json, but ideally I'd like my entire project in TypeScript.

I have @types/express listed under devDependencies in package.json.

There's an open PR that addresses this problem. Until it gets merged, you may want to use a query string instead. — Fad
– Fad, Commented Jun 14, 2021 at 14:01
Speak of the devil, it was reviewed not two days ago. Thanks; I tried using <string>req.query.locId instead, but this does not work either. Am I missing something? — SRSR333
– SRSR333, Commented Jun 14, 2021 at 14:14

lngovn · Accepted Answer · 2021-06-19 02:21:06Z

1

The express library has the Request interface with generic type for Params, Query, Body. This would work for this case. For example:

import express, { Request } from 'express';

const app = express();
const port = process.env.port || 3000;

interface RequestParams {
    locId?: number;
    operatingAgency: string;
}

app.get('/Uuidgen/1/:operatingAgency/:locId?', (request: Request<RequestParams>, result) =>
  result.send(UuidGen.generateType1Uuid(request.params.operatingAgency, request.params.locId).toString()),
);

app.listen(port, () => console.log(`UUID Generator listening on port ${port}`));

answered Jun 19, 2021 at 2:21

lngovn

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