Question

想象一下，我有一个名为Person的类型，该类型适用于有工作或无工作的人。如果有工作，则需要提供工作详细信息，该详细信息会通过传递Job泛型而有所不同。

理想情况下，我认为这应该可行：type Person<Job = undefined> = { name: string, job: Job extends undefined? never : Job }，但不行。

我必须像这样进行编码：type People<Job = undefined> = Job extends undefined ? { name: string } : { name: string, job: Job }，这很冗长。

任何人都可以提供更好的解决方案吗？谢谢。请检查this playground或阅读以下内容：

type Teacher = { school: string }
type Engineer = { company: string }
type Job = Teacher | Engineer

// type People<Job = undefined> = Job extends undefined ? { name: string } : { name: string, job: Job }   // passed, but a lot verbose
type Person<Job = undefined> = { name: string, job: Job extends undefined? never : Job  }                 // error: personWithoutJob missing job 


const personWithoutJob: Person = { name: 'Ron' } 
const personWithJob: Person<Teacher> = { name: 'Angela', job: { school: 'a' } }

Answer 1

一种好的方法是交叉点类型。

赞：

type Person<Job = undefined> = { name: string } & (Job extends undefined ? {} : { job: Job });

updated playground

official intersection types documentation

打字稿：如何使用泛型指示可为空的属性

1 个答案: