So far, we've looked at object types only in the context of 'object literals', defined using {}
with type aliases.
But TypeScript has many tools available that let you be more expressive with object types. You can model inheritance, create new object types from existing ones, and use dynamic keys.
Extending Objects
Let's start our investigation by looking at how to build object types from other object types in TypeScript.
Intersection Types
An intersection type lets us combine multiple object types into a single type. It uses the &
operator. You can think of it like the reverse of the |
operator. Instead of representing an "or" relationship between types, the &
operator signifies an "and" relationship.
Using the intersection operator &
combines multiple separate types into a single type.
Consider these types for Album
and SalesData
:
type Album = {
title: string;
artist: string;
releaseYear: number;
};
type SalesData = {
unitsSold: number;
revenue: number;
};
On their own, each type represents a distinct set of properties. While the SalesData
type on its own could be used to represent sales data for any product, using the &
operator to create an intersection type allows us to combine the two types into a single type that represents an album's sales data:
type AlbumSales = Album & SalesData;
The AlbumSales
type now requires objects to include all of the properties from both AlbumDetails
and SalesData
:
const wishYouWereHereSales: AlbumSales = {
title: "Wish You Were Here",
artist: "Pink Floyd",
releaseYear: 1975
unitsSold: 13000000,
revenue: 65000000,
};
If the contract of the AlbumSales
type isn't fulfilled when creating a new object, TypeScript will raise an error.
It's also possible to intersect more than two types:
type AlbumSales = Album & SalesData & { genre: string };
This is a useful method for creating new types from existing ones.
Intersection Types With Primitives
It's worth noting that intersection types can also be used with primitives, like string
and number
- though it often produces odd results.
For instance, let's try intersecting string
and number
:
type StringAndNumber = string & number;
What type do you think StringAndNumber
is? It's actually never
. This is because string
and number
have innate properties that can't be combined together.
This also happens when you intersect two object types with an incompatible property:
type User1 = {
age : number;
};
type User2 = {
age : string;
};
type User = User1 & User2 ;
In this case, the age
property resolves to never
because it's impossible for a single property to be both a number
and a string
.
Interfaces
So far, we've been only using the type
keyword to define object types. Experienced TypeScript programmers will likely be tearing their hair out thinking "Why aren't we talking about interfaces?!".
Interfaces are one of TypeScript's most famous features. They shipped with the very first versions of TypeScript and are considered a core part of the language.
Interfaces let you declare object types using a slightly different syntax to type
. Let's compare the syntax:
type Album = {
title: string;
artist: string;
releaseYear: number;
};
interface Album {
title: string;
artist: string;
releaseYear: number;
}
They're largely identical, except for the keyword and an equals sign. But it's a common mistake to think of them as interchangeable. They're not.
They have quite different capabilities, which we'll explore in this section.
interface extends
One of interface
's most powerful features is its ability to extend other interfaces. This allows you to create new interfaces that inherit properties from existing ones.
In this example, we have a base Album
interface that will be extended into StudioAlbum
and LiveAlbum
interfaces that allow us to provide more specific details about an album:
interface Album {
title: string;
artist: string;
releaseYear: number;
}
interface StudioAlbum extends Album {
studio: string;
producer: string;
}
interface LiveAlbum extends Album {
concertVenue: string;
concertDate: Date;
}
This structure allows us to create more specific album representations with a clear inheritance relationship:
const americanBeauty: StudioAlbum = {
title: "American Beauty",
artist: "Grateful Dead",
releaseYear: 1970,
studio: "Wally Heider Studios",
producer: "Grateful Dead and Stephen Barncard",
};
const oneFromTheVault: LiveAlbum = {
title: "One from the Vault",
artist: "Grateful Dead",
releaseYear: 1991,
concertVenue: "Great American Music Hall",
concertDate: new Date("1975-08-13"),
};
Just as adding additional &
operators add to an intersection, it's also possible for an interface to extend multiple other interfaces by separating them with commas:
interface BoxSet extends StudioAlbum, LiveAlbum {
numberOfDiscs: number;
}
Intersections vs interface extends
We've now covered two separate TypeScript syntaxes for extending object types: &
and interface extends
. So, which is better?
You should choose interface extends
for two reasons.
Better Errors When Merging Incompatible Types
We saw earlier that when you intersect two object types with an incompatible property, TypeScript will resolve the property to never
:
type User1 = {
age: number;
};
type User2 = {
age: string;
};
type User = User1 & User2;
When using interface extends
, TypeScript will raise an error when you try to extend an interface with an incompatible property:
interface User1 {
age : number;
}
interface User extends User1 {Interface 'User' incorrectly extends interface 'User1'.
Types of property 'age' are incompatible.
Type 'string' is not assignable to type 'number'.2430Interface 'User' incorrectly extends interface 'User1'.
Types of property 'age' are incompatible.
Type 'string' is not assignable to type 'number'. age : string;
}
This is very different because it actually sources an error. With intersections, TypeScript will only raise an error when you try to access the age
property, not when you define it.
So, interface extends
is better for catching errors when building out your types.
Better TypeScript Performance
When you're working in TypeScript, the performance of your types should be at the back of your mind. In large projects, how you define your types can have a big impact on how fast your IDE feels, and how long it takes for tsc
to check your code.
interface extends
is much better for TypeScript performance than intersections. With intersections, the intersection is recomputed every time it's used. This can be slow, especially when you're working with complex types.
Interfaces are faster. TypeScript can cache the resulting type of an interface based on its name. So if you use interface extends
, TypeScript only has to compute the type once, and then it can reused it every time you use the interface.
Conclusion
interface extends
is better for catching errors and for TypeScript performance. This doesn't mean you need to define all your object types using interface
- we'll get to that later. But if you need to make one object type extend another, you should use interface extends
where possible.
Types vs Interfaces
Now we know how good interface extends
is for extending object types, a natural question arises. Should we use interface
for all our types by default?
Let's look at a few comparison points between types and interfaces.
Types Can be Anything
Type aliases are a lot more flexible than interfaces. A type
can represent anything – union types, object types, intersection types, and more.
type Union = string | number;
When we declare a type alias, we're just giving a name (or alias) to an existing type.
On the other hand, an interface
can only represent object types (and functions, which we'll look at much later).
Declaration Merging
Interfaces in TypeScript have an odd property. When multiple interfaces with the same name in the same scope are created, TypeScript automatically merges them. This is known as declaration merging.
Here's an example of an Album
interface with properties for the title
and artist
:
interface Album {
title: string;
artist: string;
}
But let's imagine that, in the same file, you accidentally declare another Album
interface with properties for the releaseYear
and genres
:
interface Album {
title: string;
artist: string;
}
interface Album {
releaseYear: number;
genres: string[];
}
TypeScript automatically merges these two declarations into a single interface that includes all of the properties from both declarations:
// Under the hood:
interface Album {
title: string;
artist: string;
releaseYear: number;
genres: string[];
}
This is very different from type
, which would give you an error if you tried to declare the same type twice:
type Album = {Duplicate identifier 'Album'.2300Duplicate identifier 'Album'. title : string;
artist : string;
};
type Album = {Duplicate identifier 'Album'.2300Duplicate identifier 'Album'. releaseYear : number;
genres : string[];
};
Coming from a JavaScript point of view, this behavior of interfaces feels pretty weird. I have lost hours of my life to having two interfaces with the same name in the same 2,000+ line file. It's there for a good reason - that we'll explore in a later chapter - but it's a bit of a gotcha.
Declaration merging, and its somewhat unexpected behavior, makes me a little wary of using interfaces.
Conclusion
So, should you use type
or interface
for declaring simple object types?
I tend to default to type
unless I need to use interface extends
. This is because type
is more flexible and doesn't declaration merge unexpectedly.
But, it's a close call. I wouldn't blame you for going the opposite way. Many folks coming from a more object-oriented background will prefer interface
because it's more familiar to them from other languages.
Exercises
Exercise 1: Create an Intersection Type
Here we have a User
type and a Product
type, both with some common properties like id
and createdAt
:
type User = {
id: string;
createdAt: Date;
name: string;
email: string;
};
type Product = {
id: string;
createdAt: Date;
name: string;
price: number;
};
Your task is to create a new BaseEntity
type that includes the id
and createdAt
properties. Then, use the &
operator to create User
and Product
types that intersect with BaseEntity
.
Exercise 1: Create an Intersection Type
Exercise 2: Extending Interfaces
After the previous exercise, you'll have a BaseEntity
type along with User
and Product
types that intersect with it.
This time, your task is to refactor the types to be interfaces, and use the extends
keyword to extend the BaseEntity
type. For extra credit, try creating and extending multiple smaller interfaces.
Exercise 2: Extending Interfaces
Solution 1: Create an Intersection Type
To solve this challenge, we'll create a new BaseEntity
type with the common properties:
type BaseEntity = {
id: string;
createdAt: Date;
};
Once the BaseEntity
type is created, we can intersect it with the User
and Product
types:
type User = {
id: string;
createdAt: Date;
name: string;
email: string;
} & BaseEntity;
type Product = {
id: string;
createdAt: Date;
name: string;
price: number;
} & BaseEntity;
Then, we can remove the common properties from User
and Product
:
type User = {
name: string;
email: string;
} & BaseEntity;
type Product = {
name: string;
price: number;
} & BaseEntity;
Now User
and Product
have exactly the same behavior that they did before, but with less duplicated code.
Solution 2: Extending Interfaces
Instead of using the type
keyword, the BaseEntity
, User
, and Product
, can be declared as interfaces. Remember, interfaces do not use an equals sign like type
does:
interface BaseEntity {
id: string;
createdAt: Date;
}
interface User {
name: string;
email: string;
}
interface Product {
name: string;
price: number;
}
Once the interfaces are created, we can use the extends
keyword to extend the BaseEntity
interface:
interface User extends BaseEntity {
name: string;
email: string;
}
interface Product extends BaseEntity {
name: string;
price: number;
}
For the extra credit, we can take this further by creating WithId
and WithCreatedAt
interfaces that represent objects with an id
and createdAt
property. Then, we can have User
and Product
extend from these interfaces by adding commas:
interface WithId {
id: string;
}
interface WithCreatedAt {
createdAt: Date;
}
interface User extends WithId, WithCreatedAt {
name: string;
email: string;
}
interface Product extends WithId, WithCreatedAt {
name: string;
price: number;
}
We've now refactored our intersections to use interface extends
- our TypeScript compiler will thank us.
Dynamic Object Keys
When using objects, it's common that we won't always know the exact keys that will be used.
In JavaScript, we can start with an empty object and add keys and values to it dynamically:
// JavaScript Example
const albumAwards = {};
albumAwards.Grammy = true;
albumAwards.MercuryPrize = false;
albumAwards.Billboard = true;
However, when we try to add keys dynamically to an object in TypeScript, we'll get errors:
// TypeScript Example
const albumAwards = {};
albumAwards .Grammy = true;Property 'Grammy' does not exist on type '{}'.2339Property 'Grammy' does not exist on type '{}'.albumAwards .MercuryPrize = false;Property 'MercuryPrize' does not exist on type '{}'.2339Property 'MercuryPrize' does not exist on type '{}'.albumAwards .Billboard = true;Property 'Billboard' does not exist on type '{}'.2339Property 'Billboard' does not exist on type '{}'.
This can feel unhelpful. You might think that TypeScript, based on its ability to narrow our code, should be able to figure out that we're adding keys to an object.
In this case, TypeScript prefers to be conservative. It's not going to let you add keys to an object that it doesn't know about. This is because TypeScript is trying to prevent you from making a mistake.
We need to tell TypeScript that we want to be able to dynamically add keys. Let's look at some ways to do this.
Index Signatures for Dynamic Keys
Let's take another look at the code above.
const albumAwards = {};
albumAwards .Grammy = true;Property 'Grammy' does not exist on type '{}'.2339Property 'Grammy' does not exist on type '{}'.
The technical term for what we're doing here is 'indexing'. We're indexing into albumAwards
with a string key, Grammy
, and assigning it a value.
To support this behavior, we want to tell TypeScript that whenever we try to index into albumAwards
with a string, we should expect a boolean value.
To do that, we can use an 'index signature'.
Here's how we would specify an index signature for the albumAwards
object.
const albumAwards: {
[index: string]: boolean;
} = {};
albumAwards.Grammy = true;
albumAwards.MercuryPrize = false;
albumAwards.Billboard = true;
The [index: string]: boolean
syntax is an index signature. It tells TypeScript that albumAwards
can have any string key, and the value will always be a boolean.
We can choose any name for the index
. It's just a description.
const albumAwards: {
[iCanBeAnything: string]: boolean;
} = {};
The same syntax can also be used with types and interfaces:
interface AlbumAwards {
[index: string]: boolean;
}
const beyonceAwards: AlbumAwards = {
Grammy: true,
Billboard: true,
};
Index signatures are one way to handle dynamic keys. But there's a utility type that some argue is even better.
Using a Record Type for Dynamic Keys
The Record
utility type is another option for supporting dynamic keys.
Here's how we would use Record
for the albumAwards
object, where the key will be a string and the value will be a boolean:
const albumAwards: Record<string, boolean> = {};
albumAwards.Grammy = true;
The first type argument is the key, and the second type argument is the value. This is a more concise way to achieve a similar result as an index signature.
Record
can also support a union type as keys, but an index signature can't:
const albumAwards1 : Record <"Grammy" | "MercuryPrize" | "Billboard", boolean> = {
Grammy : true,
MercuryPrize : false,
Billboard : true,
};
const albumAwards2 : {
[index : "Grammy" | "MercuryPrize" | "Billboard"]: boolean;An index signature parameter type cannot be a literal type or generic type. Consider using a mapped object type instead.1337An index signature parameter type cannot be a literal type or generic type. Consider using a mapped object type instead.} = {
Grammy : true,
MercuryPrize : false,
Billboard : true,
};
Index signatures can't use literal types, but Record
can. We'll look at why this is when we explore mapped types in a later chapter.
The Record
type helper is a repeatable pattern that's easy to read and understand, and is a bit more flexible than an index signature. It's my go-to for dynamic keys.
Combining Known and Dynamic Keys
In many cases there will be a base set of keys we know we want to include, but we also want to allow for additional keys to be added dynamically.
For example, say we are working with a base set of awards we know were nominations, but we don't know what other awards are in play. We can use the Record
type to define a base set of awards and then use an intersection to extend it with an index signature for additional awards:
type BaseAwards = "Grammy" | "MercuryPrize" | "Billboard";
type ExtendedAlbumAwards = Record<BaseAwards, boolean> & {
[award: string]: boolean;
};
const extendedNominations: ExtendedAlbumAwards = {
Grammy: true,
MercuryPrize: false,
Billboard: true, // Additional awards can be dynamically added
"American Music Awards": true,
};
This technique would also work when using an interface and the extends
keyword:
interface BaseAwards {
Grammy: boolean;
MercuryPrize: boolean;
Billboard: boolean;
}
interface ExtendedAlbumAwards extends BaseAwards {
[award: string]: boolean;
}
This version is preferable because, in general, interface extends
is preferable to intersections.
Being able to support both default and dynamic keys in our data structures allows a lot of flexibility to adapt to changing requirements in your applications.
PropertyKey
A useful type to know about when working with dynamic keys is PropertyKey
.
The PropertyKey
type is a global type that represents the set of all possible keys that can be used on an object, including string, number, and symbol. You can find its type definition inside of TypeScript's ES5 type definitions file:
// inside lib.es5.d.ts
declare type PropertyKey = string | number | symbol;
Because PropertyKey
works with all possible keys, it's great for working with dynamic keys where you aren't sure what the type of the key will be.
For example, when using an index signature you could set the key type to PropertyKey
in order to allow for any valid key type:
type Album = {
[key: PropertyKey]: string;
};
object
Similar to string
, number
, and boolean
, object
is a global type in TypeScript.
It represents more types than you might expect. Instead of representing only objects like {}
or new Object()
, object
represents any non-primitive type. This includes arrays, functions, and objects.
So a function like this:
function acceptAllNonPrimitives(obj: object) {}
Would accept any non-primitive value:
acceptAllNonPrimitives({});
acceptAllNonPrimitives([]);
acceptAllNonPrimitives(() => {});
But error on primitives:
acceptAllNonPrimitives (1 );Argument of type 'number' is not assignable to parameter of type 'object'.2345Argument of type 'number' is not assignable to parameter of type 'object'.acceptAllNonPrimitives ("hello" );Argument of type 'string' is not assignable to parameter of type 'object'.2345Argument of type 'string' is not assignable to parameter of type 'object'.acceptAllNonPrimitives (true );Argument of type 'boolean' is not assignable to parameter of type 'object'.2345Argument of type 'boolean' is not assignable to parameter of type 'object'.
This means that the object
type is rarely useful by itself. Using Record
is usually a better choice. For instance, if you want to accept any object type, you can use Record<string, unknown>
.
Exercises
Exercise 1: Use an Index Signature for Dynamic Keys
Here we have an object called scores
, and we are trying to assign several different properties to it:
const scores = {};
scores .math = 95;Property 'math' does not exist on type '{}'.2339Property 'math' does not exist on type '{}'.scores .english = 90;Property 'english' does not exist on type '{}'.2339Property 'english' does not exist on type '{}'.scores .science = 85;Property 'science' does not exist on type '{}'.2339Property 'science' does not exist on type '{}'.
Your task is to give scores
a type annotation to support the dynamic subject keys. There are three ways: an inline index signature, a type, an interface, or a Record
.
Exercise 1: Use an Index Signature for Dynamic Keys
Exercise 2: Default Properties with Dynamic Keys
Here, we're trying to model a situation where we want some required keys - math
, english
, and science
- on our scores object.
But we also want to add dynamic properties. In this case, athletics
, french
, and spanish
:
interface Scores {}
// @ts-expect-error science should be provided Unused '@ts-expect-error' directive.2578Unused '@ts-expect-error' directive.const scores : Scores = {
math : 95,
english : 90,
};
scores .athletics = 100;Property 'athletics' does not exist on type 'Scores'.2339Property 'athletics' does not exist on type 'Scores'.scores .french = 75;Property 'french' does not exist on type 'Scores'.2339Property 'french' does not exist on type 'Scores'.scores .spanish = 70;Property 'spanish' does not exist on type 'Scores'.2339Property 'spanish' does not exist on type 'Scores'.
The definition of scores should be erroring, because science
is missing - but it's not, because our definition of Scores
is currently an empty object.
Your task is to update the Scores
interface to specify default keys for math
, english
, and science
while allowing for any other subject to be added. Once you've updated the type correctly, the red squiggly line below @ts-expect-error
will go away because science
will be required but missing. See if you can use interface extends
to achieve this.
Exercise 2: Default Properties with Dynamic Keys
Exercise 3: Restricting Object Keys With Records
Here we have a configurations
object, typed as Configurations
which is currently unknown.
The object holds keys for development
, production
, and staging
, and each respective key is associated with configuration details such as apiBaseUrl
and timeout
.
There is also a notAllowed
key, which is decorated with a @ts-expect-error
comment. But currently, this is not erroring in TypeScript as expected.
type Environment = "development" | "production" | "staging";
type Configurations = unknown;
const configurations : Configurations = {
development : {
apiBaseUrl : "http://localhost:8080",
timeout : 5000,
},
production : {
apiBaseUrl : "https://api.example.com",
timeout : 10000,
},
staging : {
apiBaseUrl : "https://staging.example.com",
timeout : 8000,
},
// @ts-expect-error Unused '@ts-expect-error' directive.2578Unused '@ts-expect-error' directive. notAllowed : {
apiBaseUrl : "https://staging.example.com",
timeout : 8000,
},
};
Update the Configurations
type so that only the keys from Environment
are allowed on the configurations
object. Once you've updated the type correctly, the red squiggly line below @ts-expect-error
will go away because notAllowed
will be disallowed properly.
Exercise 3: Restricting Object Keys With Records
Exercise 4: Dynamic Key Support
Consider this hasKey
function that accepts an object and a key, then calls object.hasOwnProperty
on that object:
const hasKey = (obj: object, key: string) => {
return obj.hasOwnProperty(key);
};
There are several test cases for this function:
The first test case checks that it works on string keys, which doesn't present any issues. As anticipated, hasKey(obj, "foo")
would return true and hasKey(obj, "bar")
would return false:
it("Should work on string keys", () => {
const obj = {
foo: "bar",
};
expect(hasKey(obj, "foo")).toBe(true);
expect(hasKey(obj, "bar")).toBe(false);
});
A test case that checks for numeric keys does have issues because the function is expecting a string key:
// @errors: 2345
import { it, expect } from "vitest";
const hasKey = (obj: object, key: string) => {
return obj.hasOwnProperty(key);
};
// ---cut---
it("Should work on number keys", () => {
const obj = {
1: "bar",
};
expect(hasKey(obj, 1)).toBe(true);
expect(hasKey(obj, 2)).toBe(false);
});
Because an object can also have a symbol as a key, there is also a test for that case. It currently has type errors for fooSymbol
and barSymbol
when calling hasKey
:
it ("Should work on symbol keys", () => {
const fooSymbol = Symbol ("foo");
const barSymbol = Symbol ("bar");
const obj = {
[fooSymbol ]: "bar",
};
expect (hasKey (obj , fooSymbol )).toBe (true);Argument of type 'typeof fooSymbol' is not assignable to parameter of type 'string'.2345Argument of type 'typeof fooSymbol' is not assignable to parameter of type 'string'. expect (hasKey (obj , barSymbol )).toBe (false);Argument of type 'typeof barSymbol' is not assignable to parameter of type 'string'.2345Argument of type 'typeof barSymbol' is not assignable to parameter of type 'string'.});
Your task is to update the hasKey
function so that all of these tests pass. Try to be as concise as possible!
Exercise 4: Dynamic Key Support
Solution 1: Use an Index Signature for Dynamic Keys
Here are the three solutions:
You can use an inline index signature:
const scores: {
[key: string]: number;
} = {};
Or an interface:
interface Scores {
[key: string]: number;
}
Or a type:
type Scores = {
[key: string]: number;
};
Or finally, a record:
const scores: Record<string, number> = {};
Solution 2: Default Properties with Dynamic Keys
Here's how to add an index signature to the Scores
interface to support dynamic keys along with the required keys:
interface Scores {
[subject: string]: number;
math: number;
english: number;
science: number;
}
Creating a RequiredScores
interface and extending it looks like this:
interface RequiredScores {
math: number;
english: number;
science: number;
}
interface Scores extends RequiredScores {
[key: string]: number;
}
These two are functionally equivalent, except for the fact that you get access to the RequiredScores
interface if you need to use that seprately.
Solution 3: Restricting Object Keys
A Failed First Attempt at Using Record
We know that the values of the Configurations
object will be apiBaseUrl
, which is a string, and timeout
, which is a number.
It may be tempting to use a Record to set the key as a string and the value an object with the properties apiBaseUrl
and timeout
:
type Configurations = Record<
string,
{
apiBaseUrl: string;
timeout: number;
}
>;
However, having the key as string
still allows for the notAllowed
key to be added to the object. We need to make the keys dependent on the Environment
type.
The Correct Approach
Instead, we can specify the key
as Environment
inside the Record:
type Configurations = Record<
Environment,
{
apiBaseUrl: string;
timeout: number;
}
>;
Now TypeScript will throw an error when the object includes a key that doesn't exist in Environment
, like notAllowed
.
Solution 4: Dynamic Key Support
The obvious answer is to change the key
's type to string | number | symbol
:
const hasKey = (obj: object, key: string | number | symbol) => {
return obj.hasOwnProperty(key);
};
However, there's a much more succinct solution.
Hovering over hasOwnProperty
shows us the type definition:
(method) Object.hasOwnProperty(v: PropertyKey): boolean
Recall that the PropertyKey
type represents every possible value a key can have. This means we can use it as the type for the key parameter:
const hasKey = (obj: object, key: PropertyKey) => {
return obj.hasOwnProperty(key);
};
Beautiful.
Reducing Duplication with Utility Types
When working with object types in TypeScript, you'll often find yourself in situations where your object types share common properties. This can lead to a lot of duplicated code.
We've seen how using interface extends
can help us model inheritance, but TypeScript also gives us tools to directly manipulate object types. With its built-in utility types, we can remove properties from types, make them optional, and more.
Partial
The Partial utility type lets you create a new object type from an existing one, except all of its properties are optional.
Consider an Album interface that contains detailed information about an album:
interface Album {
id: number;
title: string;
artist: string;
releaseYear: number;
genre: string;
}
When we want to update an album's information, we might not have all the information at once. For example, it can be difficult to decide what genre to assign to an album before it's released.
Using the Partial
utility type and passing in Album
, we can create a type that allows us to update any subset of an album's properties:
type PartialAlbum = Partial<Album>;
Now we have a PartialAlbum
type where id
, title
, artist
, releaseYear
, and genre
are all optional.
This means we can create a function which only receives a subset of the album's properties:
const updateAlbum = (album: PartialAlbum) => {
// ...
};
updateAlbum({ title: "Geogaddi", artist: "Boards of Canada" });
Required
On the opposite side of Partial
is the Required
type, which makes sure all of the properties of a given object type are required.
This Album
interface has the releaseYear
and genre
properties marked as optional:
interface Album {
title: string;
artist: string;
releaseYear?: number;
genre?: string;
}
We can use the Required
utility type to create a new RequiredAlbum
type:
type RequiredAlbum = Required<Album>;
With RequiredAlbum
, all of the original Album
properties become required, and omitting any of them would result in an error:
const doubleCup: RequiredAlbum = {
title: "Double Cup",
artist: "DJ Rashad",
releaseYear: 2013,
genre: "Juke",
};
Required with Nested Properties
An important thing to note is that both Required
and Partial
only work one level deep. For example, if the Album
's genre
contained nested properties, Required<Album>
would not make the children required:
type Album = {
title : string;
artist : string;
releaseYear ?: number;
genre ?: {
parentGenre ?: string;
subGenre ?: string;
};
};
type RequiredAlbum = Required <Album >;
If you find yourself in a situation where you need a deeply Required type, check out the type-fest library by Sindre Sorhus.
Pick
The Pick utility type allows you to create a new object type by picking certain properties from an existing object.
For example, say we want to create a new type that only includes the title
and artist
properties from the Album
type:
type AlbumData = Pick<Album, "title" | "artist">;
This results in AlbumData
being a type that only includes the title
and artist
properties.
This is extremely useful when you want to have one object that relies on the shape of another object. We'll explore this more in the chapter on deriving types from other types.
Omit
The Omit helper type is kind of like the opposite of Pick. It allows you to create a new type by excluding a subset of properties from an existing type.
For example, we could use Omit to create the same AlbumData
type we created with Pick, but this time by excluding the id
, releaseYear
and genre
properties:
type AlbumData = Omit<Album, "id" | "releaseYear" | "genre">;
A common use case is to create a type without id
, for situations where the id
has not yet been assigned:
type AlbumData = Omit<Album, "id">;
This means that as Album
gains more properties, they will flow down to AlbumData
too.
On the surface, using Omit is straightforward, but there is a small quirk to be aware of.
Omit is Looser than Pick
When using Omit, you are able to exclude properties that don't exist on an object type.
For example, creating an AlbumWithoutProducer
type with our Album
type would not result in an error, even though producer
doesn't exist on Album
:
type Album = {
id: string;
title: string;
artist: string;
releaseYear: number;
genre: string;
};
type AlbumWithoutProducer = Omit<Album, "producer">;
If we tried to create an AlbumWithOnlyProducer
type using Pick, we would get an error because producer
doesn't exist on Album
:
type AlbumWithOnlyProducer = Pick <Album , "producer" >;Type '"producer"' does not satisfy the constraint 'keyof Album'.2344Type '"producer"' does not satisfy the constraint 'keyof Album'.
Why do these two utility types behave differently?
When the TypeScript team was originally implementing Omit, they were faced with a decision to create a strict or loose version of Omit. The strict version would only permit the omission of valid keys (id
, title
, artist
, releaseYear
, genre
), whereas the loose version wouldn't have this constraint.
At the time, it was a more popular idea in the community to implement a loose version, so that's the one they went with. Given that global types in TypeScript are globally available and don't require an import statement, the looser version is seen as a safer choice, as it is more compatible and less likely to cause unforeseen errors.
While it is possible to create a strict version of Omit, the loose version should be sufficient for most cases. Just keep an eye out, since it may error in ways you don't expect.
We'll implement a strict version of Omit later in this book.
For more insights into the decisions behind Omit, refer to the TypeScript team's original discussion and pull request adding Omit
, and their final note on the topic.
Omit And Pick Don't Work Well With Union Types
Omit
and Pick
have some odd behaviour when used with union types. Let's look at an example to see what I mean.
Consider a scenario where we have three interface types for Album
, CollectorEdition
, and DigitalRelease
:
type Album = {
id: string;
title: string;
genre: string;
};
type CollectorEdition = {
id: string;
title: string;
limitedEditionFeatures: string[];
};
type DigitalRelease = {
id: string;
title: string;
digitalFormat: string;
};
These types share two common properties - id
and title
- but each also has unique attributes. The Album
type includes genre
, the CollectorEdition
includes limitedEditionFeatures
, and DigitalRelease
has digitalFormat
:
After creating a MusicProduct
type that is a union of these three types, say we want to create a MusicProductWithoutId
type, mirroring the structure of MusicProduct
but excluding the id
field:
type MusicProduct = Album | CollectorEdition | DigitalRelease;
type MusicProductWithoutId = Omit<MusicProduct, "id">;
You might assume that MusicProductWithoutId
would be a union of the three types minus the id
field. However, what we get instead is a simplified object type containing only title
– the other properties that were shared across all types, without id
.
// Expected:
type MusicProductWithoutId =
| Omit<Album, "id">
| Omit<CollectorEdition, "id">
| Omit<DigitalRelease, "id">;
// Actual:
type MusicProductWithoutId = {
title: string;
};
This is particularly annoying given that Partial
and Required
work as expected with union types:
type PartialMusicProduct = Partial<MusicProduct>;
// Hovering over PartialMusicProduct shows:
type PartialMusicProduct =
| Partial<Album>
| Partial<CollectorEdition>
| Partial<DigitalRelease>;
This stems from how Omit
processes union types. Rather than iterating over each union member, it amalgamates them into a single structure it can understand.
The technical reason for this is that Omit
and Pick
are not distributive. This means that when you use them with a union type, they don't operate individually on each union member.
The DistributiveOmit
and DistributivePick
Types
In order to address this, we can create a DistributiveOmit
type. It's defined similarly to Omit but operates individually on each union member. Note the inclusion of PropertyKey
in the type definition to allow for any valid key type:
type DistributiveOmit<T, K extends PropertyKey> = T extends any
? Omit<T, K>
: never;
When we apply DistributiveOmit
to our MusicProduct
type, we get the anticipated result: a union of Album
, CollectorEdition
, and DigitalRelease
with the id
field omitted:
type MusicProductWithoutId = DistributiveOmit<MusicProduct, "id">;
// Hovering over MusicProductWithoutId shows:
type MusicProductWithoutId =
| Omit<Album, "id">
| Omit<CollectorEdition, "id">
| Omit<DigitalRelease, "id">;
Structurally, this is the same as:
type MusicProductWithoutId =
| {
title: string;
genre: string;
}
| {
title: string;
limitedEditionFeatures: string[];
}
| {
title: string;
digitalFormat: string;
};
In situations where you need to use Omit with union types, using a distributive version will give you a much more predictable result.
For completeness, the DistributivePick
type can be defined in a similar way:
type DistributivePick<T, K extends PropertyKey> = T extends any
? Pick<T, K>
: never;
Exercises
Exercise 1: Expecting Certain Properties
In this exercise, we have a fetchUser
function that uses fetch
to access an endpoint named APIUser
and it return a Promise<User>
:
interface User {
id : string;
name : string;
email : string;
role : string;
}
const fetchUser = async (): Promise <User > => {
const response = await fetch ("/api/user");
const user = await response .json ();
return user ;
};
const example = async () => {
const user = await fetchUser ();
type test = Expect <Equal < typeof user , { name : string ; email : string }> >;Type 'false' does not satisfy the constraint 'true'.2344Type 'false' does not satisfy the constraint 'true'.};
Since we're in an asynchronous function, we do want to use a Promise
, but there's a problem with this User
type.
In the example
function that calls fetchUser
, we're only expecting to receive the name
and email
fields. These fields are only part of what exists in the User
interface.
Your task is to update the typing so that only the name
and email
fields are expected to be returned from fetchUser
.
You can use the helper types we've looked at to accomplish this, but for extra practice try using just interfaces.
Exercise 1: Expecting Certain Properties
Exercise 2: Updating a Product
Here we have a function updateProduct
that takes two arguments: an id
, and a productInfo
object derived from the Product
type, excluding the id
field.
interface Product {
id: number;
name: string;
price: number;
description: string;
}
const updateProduct = (id: number, productInfo: Product) => {
// Do something with the productInfo
};
The twist here is that during a product update, we might not want to modify all of its properties at the same time. Because of this, not all properties have to be passed into the function.
This means we have several different test scenarios. For example, update just the name, just the price, or just the description. Combinations like updating the name and the price or the name and the description are also tested.
updateProduct (1, { Argument of type '{ name: string; }' is not assignable to parameter of type 'Product'.
Type '{ name: string; }' is missing the following properties from type 'Product': id, price, description2345Argument of type '{ name: string; }' is not assignable to parameter of type 'Product'.
Type '{ name: string; }' is missing the following properties from type 'Product': id, price, description name : "Book",
});
updateProduct (1, { Argument of type '{ price: number; }' is not assignable to parameter of type 'Product'.
Type '{ price: number; }' is missing the following properties from type 'Product': id, name, description2345Argument of type '{ price: number; }' is not assignable to parameter of type 'Product'.
Type '{ price: number; }' is missing the following properties from type 'Product': id, name, description price : 12.99,
});
Your challenge is to modify the productInfo
parameter to reflect these requirements. The id
should remain absent from productInfo
, but we also want all other properties in productInfo
to be optional.
Exercise 2: Updating a Product
Solution 1: Expecting Certain Properties
There are quite a few ways to solve this problem. Here are a few examples:
Using Pick
Using the Pick utility type, we can create a new type that only includes the name
and email
properties from the User
interface:
type PickedUser = Pick<User, "name" | "email">;
Then the fetchUser
function can be updated to return a Promise
of PickedUser
:
const fetchUser = async (): Promise<PickedUser> => {
...
Using Omit
The Omit utility type can also be used to create a new type that excludes the id
and role
properties from the User
interface:
type OmittedUser = Omit<User, "id" | "role">;
Then the fetchUser
function can be updated to return a Promise
of OmittedUser
:
const fetchUser = async (): Promise<OmittedUser> => {
...
Extending an Interface
We could create an interface NameAndEmail
that contains a name
and email
property, along with updating the User
interface to remove those properties in favor of extending them:
interface NameAndEmail {
name: string;
email: string;
}
interface User extends NameAndEmail {
id: string;
role: string;
}
Then the fetchUser
function could return a Promise
of NameAndEmail
:
const fetchUser = async (): Promise<NameAndEmail> => {
// ...
};
Omit
will mean that the object grows as the source object grows. Pick
and interface extends
will mean that the object will stay the same size. So depending on requirements, you can choose the best approach.
Solution 2: Updating a Product
Using a combination of Omit
and Partial
will allow us to create a type that excludes the id
field from Product
and makes all other properties optional.
In this case, wrapping Omit<Product, "id">
in Partial
will remove the id
while making all of the remaining properties optional:
const updateProduct = (
id: number,
productInfo: Partial<Omit<Product, "id">>,
) => {
// Do something with the productInfo
};