How do I add TypeScript types to a JavaScript module without switching to TypeScript?

How do I add TypeScript types to a JavaScript module without switching to TypeScript?

By : user2172969
Date : October 22 2020, 08:10 AM
wish of those help What you will want to create are Definition Files. These files are used to describe your JavaScript.
Let's say for simplicity you have a function called add, it will take two numbers or an array of numbers.
code :
exports.add = function add(a, b) {
  if(args.length == 1) return a.reduce((sum, val) => sum + val, 0)
  return a + b; 
 * Sums up two numbers.
 * @param {number} a The first number to add.
 * @param {number} b The second number to add.
 * @returns {number} The resulting sum of the two numbers.
export declare function add(a: number, b: number): number
 * Sums up a list of numbers.
 * @param {number[]} numbers An array of numbers to add.
 * @returns {number} The resulting sum of all the numbers.
export declare function add(numbers: number[]): number
  "types": "types/index.d.ts"

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How to reference @types/module in typescript?

How to reference @types/module in typescript?

By : 侯博维
Date : March 29 2020, 07:55 AM
it fixes the issue That is a TypeScript 2.0+ plus. It will not work with versions earlier than that.
As of now 2.0 is still in beta but there is a release candidate available as the typescript@rc version. To install i run:
code :
npm install -g typescript@rc
npm install -g typescript@next
typescript - representing module pattern in types

typescript - representing module pattern in types

By : y91
Date : March 29 2020, 07:55 AM
will help you Sadly, there is no idiomatic way to extend function in your case. The only fallback is to cast menu function to any.
code :
var menu: ContextMenu = function (x:number, y:number) {
    // ....
} as any
function menu () {}
namespace menu {
  export function remove() {}
menu.remove() // compiles
TypeScript can't find module flatpickr, even though it includes typescript types

TypeScript can't find module flatpickr, even though it includes typescript types

By : Mindaugas Steponavič
Date : March 29 2020, 07:55 AM
To fix this issue Your tsconfig.json uses "module": "es6". According to the compiler options definition, this causes the compiler to lookup modules in a classic manner. The classic lookup works for some scenarios but does not support node modules defined via packages.json, which is the case for flatpickr.
What you need to change for this to work for you is either set the tsconfig.json to "moduleResolution": "node" or perhaps use "module": "commonjs" which will switch both the emitted modules and resolution mechanism to node.js standards.
TypeScript compilerOptions.types with a module

TypeScript compilerOptions.types with a module

By : Kangda
Date : March 29 2020, 07:55 AM
Any of those help The pattern you are describing is often used to define UMD packages. Their contents can be imported or — if there is no module bundler — accessed from the global scope.
TypeScript handbook explains what their definitions should look like. Your case is defined in the module-function.d.ts template, which can be reduced to:
code :
declare function depB(): string;

export as namespace depB;
export = depB;
How to allow partial TypeScript types when mocking functions - Jest, TypeScript

How to allow partial TypeScript types when mocking functions - Jest, TypeScript

By : yash
Date : October 16 2020, 06:10 PM
This might help you I have a function which I would like to mock for testing purposes in TypeScript. In my tests, all I care about are the json and the status. However, when using Jest's jest.spyOn the type of my mocked function is set to return a http Response type. This is awkward as it means I have to manually go and implement a bunch of functions and properties that are irrelevant and arbitrary. , You can use as...
code :
export function mockApi(json: object, status: number): void {
  ).mockImplementation(() =>
      json: () => Promise.resolve(json),
    } as http.Response), // <-- here
type X {
  a: number
  b: number

const x = { a: 2 } as X // OK
const y = { a: 3, c: 2 } as X // NOT OK, because c does not exist in X
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