创建一个类型为 object 的函数

Question

I have some methods in TypeScript that are passed functions like getJson . In the application, this is an actual getJson function that does what it implies, but for testing purposes, getJson is mocked with a function that is the same shape as the original getJson .

I'm creating a type for these dependencies that currently looks like this:

import { getJson } from 'get-json';

interface Dependencies {
  getJson: typeof getJson;
  ... others ...
}

This works great, but I have quite a few dependencies and may need to add more, and there's a lot of redundancy with typing the function name and then typeof -function- again.

Is there any way to create the type that is an object with keys that match the function names whose values are the functions?

Answer 1

You can use typeof on the whole object, and typescript will do the work for you:

const function1 = (x: string) => undefined;
const function2 = (x: number, y: boolean) => 5;
const function3 = () => ({x: true, y: 5});

const myObject = {
  function1, 
  function2, 
  function3
}

type MyObjectType = typeof myObject;

Now MyObjectType is

type MyObjectType = {
    function1: (x: string) => undefined;
    function2: (x: number, y: boolean) => number;
    function3: () => {
        x: boolean;
        y: number;
    };
}

And you can apply that type to any other object.

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