Key remapping with as clause in Typescript - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to rename the key 'name' to 'fullName' in the object.

Typescript

const person = { name: 'Alice', age: 30 };
const { name: [1] } = person;
console.log(fullName);

Drag options to blanks, or click blank then click option'

AfullName

Bname

CpersonName

DuserName

Attempts:

3 left

💡 Hint

Common Mistakes

Using the same key name without renaming, which keeps the original variable name.

Trying to rename without the colon syntax.

Using name: fullName renames the key name to fullName when destructuring.

2fill in blank

medium

Complete the code to rename the key 'age' to 'years' when destructuring.

Typescript

const user = { age: 25, city: 'NY' };
const { age: [1] } = user;
console.log(years);

Drag options to blanks, or click blank then click option'

AageYears

BuserAge

Cyears

Dage

Attempts:

3 left

💡 Hint

Common Mistakes

Using the original key name instead of the new variable name.

Forgetting the colon between old and new key names.

The syntax age: years renames the age property to years in the destructured variable.

3fill in blank

hard

Fix the error in the destructuring assignment to rename 'title' to 'bookTitle'.

Typescript

const book = { title: '1984', author: 'Orwell' };
const { [1] } = book;
console.log(bookTitle);

Drag options to blanks, or click blank then click option'

Atitle: bookTitle

Btitle

CbookTitle

Dtitle as bookTitle

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'as' instead of colon for renaming.

Not renaming and just using the original key name.

The correct syntax to rename a key during destructuring is oldKey: newKey. So title: bookTitle is correct.

4fill in blank

hard

Fill both blanks to rename 'firstName' to 'name' and 'birthYear' to 'year' in the object.

Typescript

const person = { firstName: 'John', birthYear: 1990 };
const { [1], [2] } = person;
console.log(name, year);

Drag options to blanks, or click blank then click option'

AfirstName: name

BbirthYear: year

CfirstName

DbirthYear

Attempts:

3 left

💡 Hint

Common Mistakes

Not renaming keys and using original names.

Mixing up the order of keys.

Use firstName: name and birthYear: year to rename keys during destructuring.

5fill in blank

hard

Fill all three blanks to rename keys 'a' to 'alpha', 'b' to 'beta', and 'c' to 'gamma' in the object.

Typescript

const obj = { a: 1, b: 2, c: 3 };
const { [1], [2], [3] } = obj;
console.log(alpha, beta, gamma);

Drag options to blanks, or click blank then click option'

Aa: alpha

Bb: beta

Cc: gamma

Da

Attempts:

3 left

💡 Hint

Common Mistakes

Using original keys without renaming.

Using invalid syntax like 'as' for renaming.

Each key is renamed using oldKey: newKey syntax during destructuring.

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