Does Typescript support "subset types"?

Question

Let's say I have an interface:

interface IUser {   email: string;   id: number;   phone: string; }; 

Then I have a function that expects a subset (or complete match) of that type. Maybe it will pass an entire object, made it will just pass in {email: "[email protected]"}. I want the type checker to allow for both.

Example:

function updateUser(user: IUser) {   // Update a "subset" of user attributes:   $http.put("/users/update", user); } 

Does Typescript support this sort of behavior yet? I could find it very useful, particularly with paradigms like Redux.

To clarify, the goal is:

1. Avoid re-writing an interface and manually setting all attributes to optional.
2. Avoid assignment of unexpected attributes (such as spelling mistakes).
3. Avoid imperative logic such as if statements, which forfeit benefits of compile time type checking.

UPDATE: Typescript has announced support for mapped types which should solve this problem once published.

Answer 1

It's worth noting that Partial<T>, as suggested in the accepted answer, makes all fields optional, which is not necessarily what you need.

If you want to make some fields required (e.g. id and email), you need to combine it with Pick:

type UserWithOptionalPhone = Pick<IUser, 'id' | 'email'> & Partial<IUser> 

---

Some explanation:

What Pick does is that it lets you specify a subset of the interface succinctly (without creating a whole new interface repeating the field types, as suggested by other answers), and then lets you use those, and only those fields.

function hello1(user: Pick<IUser, 'id' | 'email'>) { }  hello1({email: '@', id: 1}); //OK  hello1({email: '@'}); //Not OK, id missing  hello1({email: '@', id: 1, phone: '123'}); //Not OK, phone not allowed  

Now, this is not exactly what we need, as we want to allow, but not require phone. To do that, we "merge" the partial and the "picked" version of our type by creating an intersection type, which then will have id and email as required fields, and everything else as optional – exactly how we wanted it.

function hello2(user: Pick<IUser, 'id' | 'email'> & Partial<IUser>) { }  hello2({email: '@', id: 1}); //OK  hello2({email: '@', id: 1, phone: '123'}); //OK  hello2({email: '@'}); //Not OK, id missing 

Answer 2

Typescript now supports partial types.

The correct way to create a partial type is:

type PartialUser = Partial<IUser>;