Let´s use the Partial utility type to make all of the properties in a type optional,

interface Employee {
  id: number;
  name: string;
  salary: number;
}

const emp: Partial<Employee> = {};

emp.name = 'James';

We used the Partial utility type to construct a new type with all of the properties of the provided type set to optional.

interface Employee {
  id: number;
  name: string;
  salary: number;
}

// ?? type T = {
//     id?: number | undefined;
//     name?: string | undefined;
//     salary?: number | undefined;
// }
type T = Partial<Employee>;

The same approach can be used to get a type that consists of an object’s properties where all of them are marked as optional.

const obj = {
  id: 1,
  name: 'James',
  salary: 100,
};

// ?? type T = {
//     id?: number | undefined;
//     name?: string | undefined;
//     salary?: number | undefined;
// }
type T = Partial<typeof obj>;

Notice that we had to use the typeof type operator, because Partial expects a type.