函数参数中的 TypeScript 类型区分联合答案

【问题标题】：TypeScript type discriminated unions in function arguments函数参数中的 TypeScript 类型区分联合
【发布时间】：2022-01-25 02:41:11
【问题描述】：

函数的第二个参数的类型取决于第一个参数的字符串值。我想得到这样的东西：

    async action (name: 'create', args: { table: string, object: StorageObject }): Promise<StorageObject>;
    async action (name: 'createOrUpdate', args: { table: string, query: StorageQuery, object: StorageObject }): Promise<Array<StorageObject>>;
    async action (name: 'read', args: { table: string, query: StorageQuery }): Promise<Array<StorageObject>>;
    async action (name: 'update', args: { table: string, query: StorageQuery, object: StorageObject }): Promise<Array<StorageObject>>;
    async action (name: 'delete', args: { table: string, query: StorageQuery }): Promise<Array<StorageObject>> {
        ...
    }

目前，我有：TS2394: This overload signature is not compatible with its implementation signature.

【问题讨论】：

使用重载时需要 n + 1 个签名 - n 个重载加上 1 个实现签名：typescriptlang.org/docs/handbook/2/…
这能回答你的问题吗？ Typescript Method overloading for different type of parameters but same response

标签： typescript types

【解决方案1】：

当您进行重载时，实现签名需要与每个重载兼容。顺便说一句，最后一个签名不会暴露在.d.ts

  async action(name: 'create', args: { table: string, object: StorageObject }): Promise<StorageObject>;
  async action(name: 'createOrUpdate', args: { table: string, query: StorageQuery, object: StorageObject }): Promise<Array<StorageObject>>;
  async action(name: 'read', args: { table: string, query: StorageQuery }): Promise<Array<StorageObject>>;
  async action(name: 'update', args: { table: string, query: StorageQuery, object: StorageObject }): Promise<Array<StorageObject>>;
  async action(name: 'delete', args: { table: string, query: StorageQuery }): Promise<Array<StorageObject>>;

  // The actual implementation that needs to cover every case above 
  async action(name: 'create' | 'delete' | 'update' | 'read' | 'createOrUpdate', args: { table: string, query: StorageQuery, object: StorageObject }): Promise<Array<StorageObject> | StorageObject> {
    if (name === 'create' || name === 'delete') {
      return Promise.resolve(new StorageObject())
    }

    return Promise.resolve([new StorageObject()])
  }

Playground

【讨论】：