How to make interfaces optional in typescript

First of all let us look at some random example,

interface personMale{
      gender:"male";
      salary:number;
}
interface personFemale{
      gender:"female";
      weight:number
}
type person={
      name:string;
      age:number;


}&(personMale|personFemale)


const person1:person={
      name:"Sayem",
      age:27,
      gender:"male",
      salary:0
}
const person2:person={
      name:"Setara",
      age:24,
      gender:"female",
      weight:55
}

Here we should use "male" and "salary" together and "female" and " "weight" together. If we want to use "male" and "weight" or "female" and "salary" together, it will through error.

In Typescript, we can define optional properties in an interface by adding a question mark (?) to the property name. This tells Typescript that this property may or may not exist on the object.

As mentioned earlier, the basic way to make a property optional is by appending a question mark(?) to the property name. Here is an simple example,

interface User {
  id: number;
  name?: string;
  email?: string;
}


let user: User = { id: 1 };

In the above example, name and email are optional. If we create an object named ‘user’ with only ‘id’ property Typescript will not complain.

Using utility type:
Typescript provides several utility types to manipulate types, another being Partial, which makes all properties in a type T optional. Here’s how we can use it,

interface User {
  id: number;
  name: string;
  email: string;
}


type OptionalUser = Partial<User>;


let user: OptionalUser = { id: 1 };

In the above example, OptionalUser is a new type where all properties of User are optional. Hence, we can assign an object with only the id property to the user.