What is the best way to make a property required in a TypeScript type?

Question

What is the best way to make a property required in a TypeScript type?

I have defined a data type called User:

type User = { 
  id: string;
  name?: string;
  email?: string;
}

Now, I want to create a new type named UserWithName that is similar to User but with the name property being non-optional:

type UserWithName = {
  id: string;
  name: string;
  email?: string;
}

Instead of duplicating the type definition like I did above, I am looking for a way to generate UserWithName from User using generic utility types.
Required almost solves this issue, but it makes all properties mandatory whereas I only need to make one property mandatory.

typescript

Answer 1

Answer №1

Upon inspecting the source code for the Mandatory category, it reveals this:

type Mandatory<T> = {
  [P in keyof T]-?: T[P]
}

This same method can be applied to create a versatile type that fulfills your requirements:

type User = {
  id: string
  name?: string
  email?: string
}

type WithMandatory<T, K extends keyof T> = T & { [P in K]-?: T[P] }

type UserWithName = WithMandatory<User, 'name'>

// error: name field is missing
const user: UserWithName = {
  id: '12345',
}

Explore the concept further on TypeScript Playground

Answer 2

Upon inspecting the source code for the Mandatory category, it reveals this:

type Mandatory<T> = {
  [P in keyof T]-?: T[P]
}

This same method can be applied to create a versatile type that fulfills your requirements:

type User = {
  id: string
  name?: string
  email?: string
}

type WithMandatory<T, K extends keyof T> = T & { [P in K]-?: T[P] }

type UserWithName = WithMandatory<User, 'name'>

// error: name field is missing
const user: UserWithName = {
  id: '12345',
}

Explore the concept further on TypeScript Playground

Answer 3

Answer №2

If you prefer not to create a helper type, you have the option of combining Required with Pick

type Person = { 
  id: number;
  name?: string;
  age?: number;
}

const person: Required<Pick<Person, 'name'>> & Person = {
  id: 1,
  name: 'Alice'
}

Answer 4

If you prefer not to create a helper type, you have the option of combining Required with Pick

type Person = { 
  id: number;
  name?: string;
  age?: number;
}

const person: Required<Pick<Person, 'name'>> & Person = {
  id: 1,
  name: 'Alice'
}

Answer 5

Answer №3

Consider utilizing interfaces as an alternative approach:

interface Person  { 
  id: string;
  name?: string;
  email?: string;
}


interface PersonWithName extends Person {
  name: string;
}

By extending the Person type with a new interface, PersonWithName, you make the name property mandatory. You can experiment with this concept further on the typescript playground - any instance of PersonWithName without the name field will raise an error.

Answer 6

Consider utilizing interfaces as an alternative approach:

interface Person  { 
  id: string;
  name?: string;
  email?: string;
}


interface PersonWithName extends Person {
  name: string;
}

By extending the Person type with a new interface, PersonWithName, you make the name property mandatory. You can experiment with this concept further on the typescript playground - any instance of PersonWithName without the name field will raise an error.

Answer 7

Answer №4

This method seems more straightforward and performs effectively. Considering the User type in which you wish to make the name property non-optional:

type User = { 
  id: string;
  name?: string;
  email?: string;
}

To establish a named interface:

interface UserWithName extends User {
  name: NonNullable<User['name']>
}
let personWithName1: UserWithName;

To define a named type:

type UserWithName = User & {
  name: NonNullable<User['name']>
}
let personWithName2: UserWithName;

To create an unidentified type:

let personWithName3: User & { name: NonNullable<User['name']> }

I appreciate this approach because it eliminates the need for defining a new Generic, while still allowing flexibility to modify the type of the name property within the User object. Additionally, it provides error notifications if any changes are made to the name itself of the name property.

If there's uncertainty about the process, I'm utilizing:

The built-in NonNullable utility type.
Lookup types to access the property type.
Intersection types for extending a type (alternative to the extends keyword that specifically applies to named interfaces).

Answer 8

This method seems more straightforward and performs effectively. Considering the User type in which you wish to make the name property non-optional:

type User = { 
  id: string;
  name?: string;
  email?: string;
}

To establish a named interface:

interface UserWithName extends User {
  name: NonNullable<User['name']>
}
let personWithName1: UserWithName;

To define a named type:

type UserWithName = User & {
  name: NonNullable<User['name']>
}
let personWithName2: UserWithName;

To create an unidentified type:

let personWithName3: User & { name: NonNullable<User['name']> }

I appreciate this approach because it eliminates the need for defining a new Generic, while still allowing flexibility to modify the type of the name property within the User object. Additionally, it provides error notifications if any changes are made to the name itself of the name property.

If there's uncertainty about the process, I'm utilizing:

The built-in NonNullable utility type.
Lookup types to access the property type.
Intersection types for extending a type (alternative to the extends keyword that specifically applies to named interfaces).

Answer 9

Answer №5

I built upon the original recommendation.

type Necessary<T, K extends keyof T> = T & { [P in K]-?: T[P] }

type EnhancedWithNecessary<T, K extends keyof T> = Pick<T, Exclude<keyof T, K>> & Necessary<T, K>

now you have the ability to mandate multiple properties simultaneously

interface bar {x?: number, y?: String}

EnhancedWithNecessary<bar, 'x' | 'y'>

each property is now mandatory

Answer 10

I built upon the original recommendation.

type Necessary<T, K extends keyof T> = T & { [P in K]-?: T[P] }

type EnhancedWithNecessary<T, K extends keyof T> = Pick<T, Exclude<keyof T, K>> & Necessary<T, K>

now you have the ability to mandate multiple properties simultaneously

interface bar {x?: number, y?: String}

EnhancedWithNecessary<bar, 'x' | 'y'>

each property is now mandatory

Answer 11

Answer №6

If you're in search of a solution that involves utility types, the following code snippet should do the trick:

type WithRequiredProp<Type, Key extends keyof Type> = Type & Required<Pick<Type, Key>>;

Answer 12

If you're in search of a solution that involves utility types, the following code snippet should do the trick:

type WithRequiredProp<Type, Key extends keyof Type> = Type & Required<Pick<Type, Key>>;

Answer 13

Answer №7

If you're looking to make certain keys required, try using the syntax

SetRequired<Obj, 'key1' | 'key2'>

which is demonstrated here in the library known as type-fest.

Answer 14

If you're looking to make certain keys required, try using the syntax

SetRequired<Obj, 'key1' | 'key2'>

which is demonstrated here in the library known as type-fest.

Answer 15

Answer №8

If you find yourself in a situation where you must require a nested property, utilizing an intersection can come in handy. This method is effective for both interfaces and types.

// Consider using this as a type
interface Person {
  id: string
  details: {
    age?: number
    location?: string
  }
}

// Ensuring details.age is always present
type PersonWithAge = Person & { details: { age: number } }

Answer 16

If you find yourself in a situation where you must require a nested property, utilizing an intersection can come in handy. This method is effective for both interfaces and types.

// Consider using this as a type
interface Person {
  id: string
  details: {
    age?: number
    location?: string
  }
}

// Ensuring details.age is always present
type PersonWithAge = Person & { details: { age: number } }

What is the best way to make a property required in a TypeScript type?

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

Answer №7

Answer №8

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