如何使TypeScript接口方法的参数直接遵循另一接口属性类型？

要通过IShippingAddressParams定义setShippingAddress的参数类型，同时避免传递包裹对象，只需让IShippingAddressParams直接对应IAddress的结构即可。以下两种方案都能满足需求：

方案一：接口继承 #

通过让参数接口继承IAddress，复用其结构定义参数类型：

interface IAddress {
  name: string;
  addressLine1: string;
  addressLine2: string;
  landmark: string;
  city: string;
  state: string;
  pincode: number;
}

// 继承IAddress，使参数接口与地址结构完全一致
interface IShippingAddressParams extends IAddress {}

interface IOrderBuilder {
  setCustomer({ customerId }: { customerId: string; }): Order;
  // ...其他方法
  
  // 用IShippingAddressParams定义参数类型
  setShippingAddress(address: IShippingAddressParams): Order;

  // ...其他方法
}

// 实现时直接传入IAddress类型参数，符合预期
class OrderBuilder implements IOrderBuilder {
  setCustomer({ customerId }: { customerId: string; }): Order {
    // 业务逻辑实现
    return {} as Order;
  }

  setShippingAddress(address: IAddress): Order {
    // 业务逻辑实现
    return {} as Order;
  }
}

方案二：类型别名 #

用类型别名直接关联IAddress，作为参数类型的定义载体：

interface IAddress {
  name: string;
  addressLine1: string;
  addressLine2: string;
  landmark: string;
  city: string;
  state: string;
  pincode: number;
}

// 类型别名直接指向IAddress，作为参数类型定义
type IShippingAddressParams = IAddress;

interface IOrderBuilder {
  setCustomer({ customerId }: { customerId: string; }): Order;
  // ...其他方法
  
  setShippingAddress(address: IShippingAddressParams): Order;

  // ...其他方法
}

// 实现方式完全符合需求
class OrderBuilder implements IOrderBuilder {
  setCustomer({ customerId }: { customerId: string; }): Order {
    return {} as Order;
  }

  setShippingAddress(address: IAddress): Order {
    return {} as Order;
  }
}

原理说明 #

TypeScript采用结构类型系统，只要两个类型的结构匹配就会被视为兼容。通过上述两种方式让IShippingAddressParams与IAddress结构对齐，既满足了用独立接口/类型定义参数的要求，又不需要额外包裹对象，实现时直接传入IAddress类型参数即可。

内容的提问来源于stack exchange，提问作者Prerak Jain