Question

我正在开发一个需要迭代 Employee[] | Guest[] 类型数组的 Typescript 函数。问题是 TypeScript 发出错误 TS2349: This expression is not callable... has signatures but none of those are compatible with each other。我希望能够迭代数组，并且只需要在使用局部变量之前使用类型保护，但情况似乎并非如此。有没有一种方法可以使用 map 函数而不必创建自定义类型？该函数只需要迭代这两种类型，所以我希望将参数的类型定义保持为 Employee[] | Guest[]。下面是一个简单的示例，其中 people.map 出现错误。

interface Employee {
  id: string;
  firstName: string;
}

interface Guest {
  firstName: string;
}

function getFirstNames(people: Employee[] | Guest[]): string[] {
  return people.map((x: Employee | Guest) => {
    return x.firstName;
  });
}

const employeeFirstNames = getFirstNames([
  {id: '1', firstName: 'Steve'},
  {id: '2', firstName: 'James'},
] as Employee[]);

const guestFirstNames = getFirstNames([
  {firstName: 'Mary'},
] as Guest[]);

Answer 1

将数组输入为 Employee 和 Guest 的并集。

function getFirstNames(people: (Employee | Guest)[]): string[] {
  return people.map((x: Employee | Guest) => x.firstName);
}

Answer 2

我认为在这种情况下你可以使用重载，如果你想传递具有固定元素类型的数组

function getFirstNames(people: Employee[]): string[];
function getFirstNames(people: Guest[]): string[];
function getFirstNames(people: (Employee | Guest)[]): string[] {
  return people.map((x: Employee | Guest) => {
    return x.firstName;
  });
}

在这种情况下，你不会担心公共接口，所以像下面这样的混合类型会出错

const guestFirstNames = getFirstNames([
   { id: '1', firstName: 'Steve' },
  { firstName: 'Mary' }
]); // <-- No overload matches this call. Overload 1 of 2, '(people: Employee[]): string[]', gave the following error. Property 'id' is missing in type '{ firstName: string; }' but required in type 'Employee'. Overload 2 of 2, '(people: Guest[]): string[]', gave the following error. Type '{ id: string; firstName: string; }' is not assignable to type 'Guest'. Object literal may only specify known properties, and 'id' does not exist in type 'Guest'.

如何在打字稿中使用多种类型的数组调用map函数

2 个答案: