typescript - 如何在 TypeScript 中引用属于联合类型的类型？

Question

我在 TypeScript 中为 Action 定义了联合类型:

type Action = {
  type: 'reset',
} | {
  type: 'add',
  payload: number
} | {
  type: 'minus',
  payload: number
}

在某些函数中，我想引用联合类型的一部分，比如:

function handleAdd(state: State, action: {
  type: 'add',
  payload: number
}): State => {
  // handle the add
}

有没有办法简化action的输入所以我不需要重复

的完整定义

{
  type: 'add',
  payload: number
}

这里？

更新:

我不想单独预定义每个 Action 类型。是否有其他解决方案，例如 Action[type='add']？

Answer 1

这可以通过 distributive conditional types 来实现:

type ExtractType<T extends Action['type'], A = Action> =
    A extends { type: T } ? A : never;

type Add = ExtractType<'add'>; // { type: 'add'; payload: number; }

Playground

Distributive conditional types are automatically distributed over union types during instantiation. For example, an instantiation of T extends U ? X : Y with the type argument A | B | C for T is resolved as (A extends U ? X : Y) | (B extends U ? X : Y) | (C extends U ? X : Y)