Key remapping with as clause in Typescript - Mini Project: Build & Apply

Key remapping with as clause

📖 Scenario: Imagine you have a list of users with their full names and ages. You want to create a new list where the keys are changed to shorter, friendlier names.

🎯 Goal: You will create a new array of objects by remapping the keys of the original user objects using TypeScript's key remapping with as clause.

📋 What You'll Learn

Create an array called users with objects having keys fullName and age

Create a type alias ShortUser that remaps fullName to name and keeps age

Create a new array shortUsers by mapping users to objects of type ShortUser

Print the shortUsers array

💡 Why This Matters

🌍 Real World

Key remapping helps when you get data from an API or database with long or unclear keys and want to rename them for easier use in your code.

💼 Career

Understanding key remapping is useful for frontend and backend developers working with TypeScript to keep code clean and maintainable when handling data transformations.

1

Create the initial users array

Create an array called users with these exact objects: { fullName: "Alice Johnson", age: 30 }, { fullName: "Bob Smith", age: 25 }, and { fullName: "Charlie Brown", age: 35 }.

Typescript

# Create the array users with the given objects
# Your code here

Hint

Use const users = [ ... ] with the exact objects inside.

2

Create the ShortUser type with key remapping

Create a type alias called ShortUser that remaps the key fullName to name and keeps the age key as is, using TypeScript's key remapping with as clause.

Typescript

const users = [
  { fullName: "Alice Johnson", age: 30 },
  { fullName: "Bob Smith", age: 25 },
  { fullName: "Charlie Brown", age: 35 }
];

// Create the type alias ShortUser with key remapping
// Your code here

Hint

Use type ShortUser = { [K in keyof typeof users[0] as K extends "fullName" ? "name" : K]: typeof users[0][K] }.

3

Map users to shortUsers with remapped keys

Create a new array called shortUsers by mapping over users and returning objects with keys name (from fullName) and age. Use the ShortUser type for the new objects.

Typescript

const users = [
  { fullName: "Alice Johnson", age: 30 },
  { fullName: "Bob Smith", age: 25 },
  { fullName: "Charlie Brown", age: 35 }
];

type ShortUser = {
  [K in keyof typeof users[0] as K extends "fullName" ? "name" : K]: typeof users[0][K]
};

// Create shortUsers by mapping users to objects with keys name and age
// Your code here

Hint

Use const shortUsers: ShortUser[] = users.map(user => ({ name: user.fullName, age: user.age })).

4

Print the shortUsers array

Write a console.log statement to print the shortUsers array.

Typescript

const users = [
  { fullName: "Alice Johnson", age: 30 },
  { fullName: "Bob Smith", age: 25 },
  { fullName: "Charlie Brown", age: 35 }
];

type ShortUser = {
  [K in keyof typeof users[0] as K extends "fullName" ? "name" : K]: typeof users[0][K]
};

const shortUsers: ShortUser[] = users.map(user => ({
  name: user.fullName,
  age: user.age
}));

// Print the shortUsers array
# Your code here

Hint

Use console.log(shortUsers) to show the new array.

Practice

1.

What does the as clause do in a TypeScript mapped type?

easy

A. It adds new keys without changing existing ones.

B. It changes the values of the original type.

C. It removes keys from the original type.

D. It renames the keys of the original type to new keys.

Solution

1. Step 1: Understand mapped types

   Mapped types create new types by transforming keys or values of an existing type.

2. Step 2: Role of as clause

   The as clause specifically allows renaming keys during this transformation.

3. Final Answer:

   It renames the keys of the original type to new keys. -> Option D

4. Quick Check:

   as clause = rename keys [OK]

Hint: Remember: as changes keys, not values [OK]

Common Mistakes:

• Thinking as changes values instead of keys
• Confusing as with adding new keys
• Assuming as removes keys

2.

Which of the following is the correct syntax to rename keys in a mapped type using as?

type Original = { a: number; b: string };
type Renamed = { [K in keyof Original as ???]: Original[K] };

easy

A. as K

B. `new_${K}`

C. K as string

D. K + 'Key'

Solution

1. Step 1: Understand template literal types in key remapping

   To rename keys dynamically, template literals like `new_${K}` are used inside the as clause.

2. Step 2: Check syntax correctness

   Only `new_${K}` is valid syntax for renaming keys using template literals in TypeScript.

3. Final Answer:

   `new_${K}` -> Option B

4. Quick Check:

   Use template literals inside as for renaming [OK]

Hint: Use backticks and ${} for key renaming [OK]

Common Mistakes:

• Using plus (+) operator instead of template literals
• Writing as K inside brackets incorrectly
• Trying to cast keys with as string

3.

What is the resulting type of this code?

type User = { id: number; name: string };
type PrefixedUser = { [K in keyof User as `user_${K}`]: User[K] };

medium

A. { userId: number; userName: string }

B. { id: number; name: string }

C. { user_id: number; user_name: string }

D. Error: Invalid key remapping

Solution

1. Step 1: Analyze key remapping with template literals

   The keys id and name are renamed to userId and userName respectively using `user_${K}`.

2. Step 2: Confirm values remain unchanged

   The values for each key remain the same as in User.

3. Final Answer:

   { userId: number; userName: string } -> Option A

4. Quick Check:

   Keys renamed with prefix 'user_' [OK]

Hint: Template literals rename keys, values stay same [OK]

Common Mistakes:

• Assuming keys stay the same
• Confusing camelCase with snake_case
• Expecting a syntax error

4.

Identify the error in this mapped type:

type Data = { x: number; y: string };
type NewData = { [K in keyof Data as K.toUpperCase()]: Data[K] };

medium

A. Cannot use toUpperCase() on type keys directly.

B. Missing return type for mapped type.

C. Keys must be numbers, not strings.

D. Values cannot be accessed with Data[K].

Solution

1. Step 1: Check key remapping expression

   The expression K.toUpperCase() is a runtime string method and cannot be used in type-level key remapping.

2. Step 2: Understand type-level string manipulation

   TypeScript requires template literal types or conditional types for key transformations, not runtime methods.

3. Final Answer:

   Cannot use toUpperCase() on type keys directly. -> Option A

4. Quick Check:

   Runtime methods disallowed in type keys [OK]

Hint: Use template literals, not runtime string methods [OK]

Common Mistakes:

• Trying to call string methods on keys in types
• Confusing runtime and compile-time code
• Ignoring TypeScript's type system limits

5.

Given this type, how would you create a new type that renames only keys starting with is by prefixing them with flag_, keeping others unchanged?

type Flags = { isActive: boolean; isAdmin: boolean; count: number };

hard

A. type NewFlags = { [K in keyof Flags]: Flags[K] };

B. type NewFlags = { [K in keyof Flags as `flag_${K}`]: Flags[K] };

C. type NewFlags = { [K in keyof Flags as K extends `is${string}` ? `flag_${K}` : K]: Flags[K] };

D. type NewFlags = { [K in keyof Flags as K.toLowerCase()]: Flags[K] };

Solution

1. Step 1: Use conditional type in as clause

   The condition K extends `is${string}` checks if the key starts with 'is'.

2. Step 2: Rename keys conditionally

   If true, rename key to `flag_${K}`, else keep original key K.

3. Step 3: Keep values unchanged

   Values remain the same as in Flags.

4. Final Answer:

   type NewFlags = { [K in keyof Flags as K extends `is${string}` ? `flag_${K}` : K]: Flags[K] }; -> Option C

5. Quick Check:

   Conditional key renaming with template literals [OK]

Hint: Use conditional as with template literals [OK]

Common Mistakes:

• Renaming all keys without condition
• Using runtime string methods in types
• Not preserving keys that don't match condition