TypeScript require with type checking

Question

I am using TypeScript v1.4.1 and would like to require an external module (in this case "chai") and have it be type checked.

However, I am running into some sort of naming conflict with this code:

/// <reference path="../typings/node/node.d.ts" />
/// <reference path="../typings/chai/chai.d.ts" />
/// <reference path="../typings/mocha/mocha.d.ts" />

var chai = require("chai");

var expect = chai.expect;
var assert = chai.assert;

describe("TEST", () =>
{
   it("true should be true", (done)=>
   {
      expect(true).to.be.true;
      done();
   });
});

With this definition file:

declare module chai {
   ...
}
declare module "chai" {
   export = chai;
}

Compiling gives these errors:

test/test.ts(5,5): error TS2300: Duplicate identifier 'chai'.
typings/chai/chai.d.ts(6,16): error TS2300: Duplicate identifier 'chai'.

It seems my only option is rename my chai variable name in test.ts. That seems clunky AND won't type check the use of the renamed chai variable.

Any advice?

curpa · Accepted Answer · 2015-02-16 19:45:02Z

17

Use the import keyword with require instead of var

import chai = require('chai');

And compile with --module commonjs if you're not already

Or, if for some reason you don't want the test code to be an external module, adding a type annotation will preserve type checking.

var c: typeof chai = require("chai");

edited Feb 16, 2015 at 19:45

answered Feb 16, 2015 at 19:04

curpa

2,4101 gold badge16 silver badges12 bronze badges

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

4 Comments

EToreo Over a year ago

If I do that I can't use --out because this file becomes an external module according to TypeScript: a top level import or export statement makes the file an external module.

curpa Over a year ago

I updated my answer to include another option, though you'll still have to rename the variable. It might be helpful for you to include some more info about how you're bundling the code and running the tests because it's not clear to me why you would want to use --out.

EToreo Over a year ago

Thank you. We found that :typeof option and it works. Let me pick your brain a little though... It is our understanding that if we do no use the --out compile flag, then all our files will be compiled as external modules. Is that true?

curpa Over a year ago

Yes that's true. If you're runtime (e.g. node, requirejs) includes a module loader, that seems like the desired behavior.

kshetline · Accepted Answer · 2020-05-04 07:46:33Z

0

Since TypeScript 3.9 Beta was released it's possible to use require with typing

Example:

const {someValue} = require('fs')

"TypeScript now automatically detects the types of imports you’re using to keep your file’s style clean and consistent."

ref. https://devblogs.microsoft.com/typescript/announcing-typescript-3-9-beta/

edited May 4, 2020 at 7:46

kshetline

14k6 gold badges49 silver badges96 bronze badges

answered Apr 5, 2020 at 17:55

Przemek Struciński

5,2381 gold badge32 silver badges20 bronze badges

1 Comment

Anatoliy Arkhipov Over a year ago

What does it mean, to "to keep your file’s style clean and consistent"? I'm not sure if it has actually started to type-check require, because nothing changed for me after updating to 3.9.9.

Collectives™ on Stack Overflow

TypeScript require with type checking

2 Answers 2

4 Comments

1 Comment

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

4 Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Linked

Related