Make a single property optional in TypeScript

Question

In TypeScript, 2.2...

Let\'s say I have a Person type:

interface Person {
  name: string;
  hometown: string;
  nickname: string;
}

Answer 1

Ok well what you are really describing is two different "Types" of people (i.e. Person types) .. A normal person and a nick named person.

interface Person {
    name: string;
    hometown: string;
}

interface NicknamedPerson extends Person {
    nickname: string;
}

Then in the case where you don't really want a nicknamed person but just a person you just implement the Person interface.

An alternative way to do this if you wanted to hang on to just one Person interface is having a different implementation for a non nicknamed person:

interface Person {
  name: string;
  hometown: string;
  nickname: string;
}
class NicknamedPerson implements Person {
    constructor(public name: string, public hometown: string, public nickname: string) {}
}

class RegularPerson implements Person {
    nickname: string;
    constructor(public name: string, public hometown: string) {
        this.nickname = name;
    }
}

makePerson(input): Person {
     if(input.nickname != null) {
       return new NicknamedPerson(input.name, input.hometown, input.nickname);
     } else {
       return new RegularPerson(input.name, input.hometown);
     }
}

This enables you to still assign a nickname (which is just the persons name in case of an absence of a nickname) and still uphold the Person interface's contract. It really has more to do with how you intend on using the interface. Does the code care about the person having a nickname? If not, then the first proposal is probably better.