Pick type in Typescript - Mini Project: Build & Apply

Using the Pick Type in TypeScript

📖 Scenario: You are working on a user management system. You have a full user profile object, but sometimes you only need a smaller part of that profile, like just the user's name and email.

🎯 Goal: Learn how to use the Pick type in TypeScript to create a new type with only selected properties from an existing type.

📋 What You'll Learn

Create a full user profile type with specific properties

Create a new type using Pick to select only some properties

Create a variable of the new picked type with exact properties

Print the picked user information to the console

💡 Why This Matters

🌍 Real World

In real projects, you often have big data objects but only need a small part for certain tasks, like showing user contact info or sending emails.

💼 Career

Understanding <code>Pick</code> helps you write cleaner, safer TypeScript code by reusing types and avoiding repetition.

1

Create a full user profile type

Create a TypeScript type called UserProfile with these exact properties: id as number, name as string, email as string, and age as number.

Typescript

// Your code here

Need a hint?

Use the type keyword to define a new type with the given properties and their types.

2

Create a picked user type with Pick

Create a new type called UserContact using Pick to select only the name and email properties from UserProfile.

Typescript

type UserProfile = {
  id: number;
  name: string;
  email: string;
  age: number;
};

// Your code here

Need a hint?

Use Pick<OriginalType, "prop1" | "prop2"> to create a new type with only the chosen properties.

3

Create a variable of the picked type

Create a constant variable called contactInfo of type UserContact with name set to "Alice" and email set to "[e-mail]".

Typescript

type UserProfile = {
  id: number;
  name: string;
  email: string;
  age: number;
};

type UserContact = Pick;

// Your code here

Need a hint?

Define a constant with the picked type and assign the exact property values as strings.

4

Print the picked user information

Write a console.log statement to print the contactInfo variable.

Typescript

type UserProfile = {
  id: number;
  name: string;
  email: string;
  age: number;
};

type UserContact = Pick;

const contactInfo: UserContact = {
  name: "Alice",
  email: "[e-mail]"
};

// Your code here

Need a hint?

Use console.log(contactInfo) to show the picked user information in the console.