How to easily make every property in a type alias or interface to optional properties in TypeScript?

Originally posted here!

To easily make every property on a type alias or interface to optional properties, you can use the Partial utility type and pass the type alias or interface as the first type argument to it using the angled brackets symbol (<>).

TL;DR

// a simple type
type Person = {
  name: string;
  age: number;
};

// make all the properties in the
// `Person` type to be optional
type OptionalPerson = Partial<Person>;

// contents of the `OptionalPerson` type
/*
{
  name?: string | undefined;
  age?: number | undefined;
}
*/

For example, let's say we have a type called Person with 2 properties called name having the type of string and age having the type of number like this,

// a simple type
type Person = {
  name: string;
  age: number;
};

Now to make all the properties in the Person type to be optional we can use the Partial utility type and pass the Person interface as the first type argument to it using the angled brackets symbol (<>).

It can be done like this,

// a simple type
type Person = {
  name: string;
  age: number;
};

// make all the properties in the
// `Person` type to be optional
type OptionalPerson = Partial<Person>;

// contents of the `OptionalPerson` type
/*
{
  name?: string | undefined;
  age?: number | undefined;
}
*/

Now if you hover over the OptionalPerson type you can see that all the properties are now optional which is what we want to happen.

See the above code live in codesandbox.

That's all 😃!

Feel free to share if you found this useful 😃.

---