Pick type in Typescript

Introduction

The Pick type helps you create a new type by choosing only some properties from an existing type. It keeps your code simple and focused.

When you want to use only a few properties from a big object type.

When you want to send part of an object to a function or API.

When you want to create a smaller version of a type for forms or UI.

When you want to avoid repeating property types manually.

When you want to make your code safer by limiting properties.

Syntax

Typescript

Pick<Type, Keys>

Type is the original object type.

Keys is a list of property names you want to keep, written as a union of string literals.

Examples

This creates a new type with only name and age from Person.

Typescript

type Person = {
  name: string;
  age: number;
  city: string;
};

type PersonNameAndAge = Pick<Person, 'name' | 'age'>;

This creates a type with only the model property from Car.

Typescript

type Car = {
  make: string;
  model: string;
  year: number;
};

type CarModelOnly = Pick<Car, 'model'>;

Sample Program

This program shows how to use Pick to create a smaller type with only id and username from User. Then it creates an object of that smaller type and prints it.

Typescript

type User = {
  id: number;
  username: string;
  email: string;
  isAdmin: boolean;
};

// We want a type with only id and username

type UserPreview = Pick<User, 'id' | 'username'>;

const user: User = {
  id: 1,
  username: 'alice',
  email: '[e-mail]',
  isAdmin: false
};

const preview: UserPreview = {
  id: user.id,
  username: user.username
};

console.log(preview);

Important Notes

You can only pick properties that exist in the original type.

Pick helps keep your types DRY (Don't Repeat Yourself).

Summary

Pick creates a new type by selecting specific properties from another type.

It helps you work with smaller, focused types.

Use it to make your code safer and easier to maintain.

Practice

1. What does the Pick type do in TypeScript?

easy

A. Removes properties from a type.

B. Creates a new type by selecting specific properties from an existing type.

C. Combines two types into one.

D. Makes all properties in a type optional.

Solution

1. Step 1: Understand the purpose of Pick

   The Pick type is used to create a new type by choosing certain properties from an existing type.

2. Step 2: Compare with other options

   Removing properties is done by Omit, combining types by intersection, and making properties optional by Partial.

3. Final Answer:

   Creates a new type by selecting specific properties from an existing type. -> Option B

4. Quick Check:

   Pick selects properties = A [OK]

Hint: Pick selects properties to create smaller types [OK]

Common Mistakes:

• Confusing Pick with Omit
• Thinking Pick makes properties optional
• Mixing Pick with type intersections

2. Which of the following is the correct syntax to create a type PersonName with only name and age properties from Person using Pick?

easy

A. type PersonName = Pick;

B. type PersonName = Pick;

C. type PersonName = Pick;

D. type PersonName = Pick;

Solution

1. Step 1: Recall Pick syntax

   Pick takes two parameters: the original type and a union of property names as string literals.

2. Step 2: Identify correct union syntax

   The union of keys is written as 'name' | 'age' inside quotes and separated by |.

3. Final Answer:

   type PersonName = Pick; -> Option A

4. Quick Check:

   Pick syntax uses union of strings = B [OK]

Hint: Use quotes and | to list keys in Pick [OK]

Common Mistakes:

• Omitting quotes around keys
• Using array syntax instead of union
• Passing keys as object instead of union

3. Given the type interface User { id: number; name: string; email: string; } and type UserPreview = Pick;, what is the type of userPreview in this code?

const userPreview: UserPreview = { id: 1, name: 'Alice' };

medium

A. { name: string; email: string; }

B. { id: number; name: string; email: string; }

C. { id: number; }

D. { id: number; name: string; }

Solution

1. Step 1: Understand Pick selects properties

   UserPreview picks only 'id' and 'name' from User, so it has only those two properties.

2. Step 2: Match the type of userPreview

   The object assigned has 'id' and 'name' only, matching the Pick type exactly.

3. Final Answer:

   { id: number; name: string; } -> Option D

4. Quick Check:

   Pick selects only chosen keys = A [OK]

Hint: Pick type has only selected keys, no extras [OK]

Common Mistakes:

• Assuming all original properties remain
• Confusing Pick with Partial
• Including properties not picked

4. Identify the error in this TypeScript code using Pick:

interface Product { id: number; name: string; price: number; }
type ProductInfo = Pick;

medium

A. The interface Product is missing a 'cost' property.

B. Pick cannot be used with more than one key.

C. 'cost' is not a key of Product, causing a type error.

D. The syntax for Pick is incorrect; keys should be in an array.

Solution

1. Step 1: Check keys used in Pick

   Pick requires keys to exist in the original type. 'cost' is not a key in Product.

2. Step 2: Understand TypeScript error

   Using a non-existent key causes a type error because Pick cannot select keys that don't exist.

3. Final Answer:

   'cost' is not a key of Product, causing a type error. -> Option C

4. Quick Check:

   Pick keys must exist in type = C [OK]

Hint: Pick keys must match existing properties exactly [OK]

Common Mistakes:

• Using keys not in the original type
• Thinking Pick keys can be arrays
• Assuming Pick works with any string keys

5. You have a type interface Employee { id: number; name: string; department: string; salary: number; }. You want to create a type EmployeeSummary that includes only id, name, and a new property departmentCode which is a string. How can you do this using Pick and adding the new property?

hard

A. type EmployeeSummary = Pick & { departmentCode: string };

B. type EmployeeSummary = Pick;

C. type EmployeeSummary = Omit & { departmentCode: string };

D. type EmployeeSummary = Partial<Employee> & { departmentCode: string };

Solution

1. Step 1: Use Pick to select existing keys

   Pick 'id' and 'name' from Employee to get those properties only.

2. Step 2: Add new property with intersection

   Use & to add a new property 'departmentCode' with type string to the picked type.

3. Final Answer:

   type EmployeeSummary = Pick & { departmentCode: string }; -> Option A

4. Quick Check:

   Pick keys + add new property with & = D [OK]

Hint: Combine Pick with & to add new properties [OK]

Common Mistakes:

• Trying to Pick a non-existent property
• Using Omit instead of Pick incorrectly
• Using Partial which makes all properties optional