TypeScript:可用于许多函数的独立函数类型

Question

我想为函数创建一个可重用的类型，这样我就可以说：

type MyFunc = ...
const func1: MyFunc = ...
const func2: MyFunc = ...

以下不起作用，但基本上我想要：

type Book = {
id: string;
title: string;
}
type createBook = (book: Partial<Book>) => Book;
const someBookFactory: createBook = ({ title }) => ({ id: /* autogenerated */, title });
const someOtherBookFactory: createBook = ({ id, title }) => ({ id, title });

这在TypeScript中可能吗？

Answer 1

type createBook = (book: Partial<Book>) => Book;
function getId(): string {
return Math.random().toString();
}
// Partial means they can be undefined so you need to provide default values
const someBookFactory: createBook = ({ title = "untitled" }) => ({ id: getId(), title });
const someOtherBookFactory: createBook = ({ id = getId(), title = "untitled" }) => ({ id, title });

最重要的是，Book是一份包含2个必填字段的合同，如果您没有将它们全部传递出去，那么在返回之前的某个时候，您就违反了合同。所有具有静态类型的语言都是这样。

// You can also have both fallbacks handled inside
const anotherBookFactory: createBook = ({ id, title }) => {
return {
id: id ?? getId(),
title: title ?? "untitled"
};
};