如何处理 Typescript 中的错误返回类型？

Question

I have a function that can return either an object of type Person or an Error. I want to conditionally read the values of keys in the returned object if it's not an Error.

The last line of the code snippet below gives an error:

Property 'firstName' does not exist on type 'Person | Error'

The error can be seen at this link .

 interface Person { firstName: string; lastName: string; } function greeter(person: Person): Person | Error { if (person.firstName === 'Malcolm') { return new Error('This firstName is not allowed'); } return { firstName: person.firstName, lastName: person.lastName, }; } const x = greeter({ firstName: "Malcolm", lastName: "Reynolds", }); const { firstName, lastName } = x;

I have to return the Error for some scenarios. Can't skip it.

Thanks in advance!

Answer 1

Firstly I would suggest you to wrap greeter() inside try ... catch block to handle the error.

Still if you want to do it this way, you can do the following.

const val = greeter({
    firstName: "Malcolm",
    lastName:  "Reynolds",
})

if (typeof val == Person) { 
const x = val as Person;
const {firstName, lastName} = x;
}

Here you can check the type of result and perform operation accordingly.