如何根据数组的值定义类型？

Question

如果我有一个看起来像数组的类型：

type names = ['Mike', 'Jeff', 'Ben'];

我可以很容易地定义另一种类型，它具有以下项目的值names：

type UserName = names[number]

对于函数：

function hello(name: UserName) {
  console.log(`Hello, ${name}!`)
}

我只能传递一个Mike，Jeff，Ben发挥作用hello。如果我提供其他值，例如John，则无法编译。

如果我没有type names而是const 数组 names怎么办？

const names = ['Mike', 'Jeff', 'Ben'];

type UserName = ???;

function hello(name: UserName) {
  console.log(`Hello, ${name}!`)
}

hello('Mike');

是否可以定义这样的类型UserName？

Answer 1

In TypeScript 3.4, which should be released in March 2019 it will be possible to tell the compiler to infer the type of a tuple of string literals as a tuple of string literals, instead of as string[], by using the as const syntax. It should look like this:

const names = ['Mike', 'Jeff', 'Ben'] as const; // TS3.4 syntax
type Names = typeof names; // type Names = readonly ['Mike', 'Jeff', 'Ben'] 
type UserName = Names[number]; // 'Mike' | 'Jeff' | 'Ben'

Until then (in TypeScript 3.0 through 3.3) you can get this effect by using a helper function which gives the compiler hints to infer a narrower type:

type Narrowable = string | number | boolean | undefined | null | void | {};
const tuple = <T extends Narrowable[]>(...t: T)=> t;
const names = tuple('Mike', 'Jeff', 'Ben');

type Names = typeof names; // type Names = ['Mike', 'Jeff', 'Ben'] 
type UserName = Names[number]; // 'Mike' | 'Jeff' | 'Ben'

(Note that in both cases you can skip the intermediate Names type and just define type UserName = (typeof names)[number] if you prefer)

Okay, hope that helps. Good luck!