Mapped type with template literals in Typescript - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a mapped type in TypeScript?

A mapped type creates a new type by transforming each property of an existing type. It loops over keys and applies changes, like making properties optional or readonly.

beginner

What are template literal types in TypeScript?

Template literal types let you create new string types by combining or transforming other string types, similar to string templates but at the type level.

intermediate

How do mapped types and template literal types work together?

You can use template literal types inside mapped types to create new property names by combining or modifying existing keys, like adding prefixes or suffixes.

intermediate

Example: What does this mapped type do?

type PrefixedKeys<T> = { [K in keyof T as `get${Capitalize<string & K>}`]: () => T[K] }

It creates a new type where each property name is prefixed with 'get' and capitalized, and the value is a function returning the original property type.

beginner

Why use mapped types with template literals?

They help automate creating new types with modified keys, reducing repetitive code and improving type safety by generating consistent property names.

What does the 'as' keyword do in a mapped type with template literals?

A It filters keys from the original type

B It changes the value types

C It makes properties optional

D It renames the keys using the template literal expression

The 'as' keyword lets you rename keys in a mapped type, often using template literals to create new property names.

Which TypeScript feature allows creating new string types by combining other string types?

A Template literal types

B Union types

C Intersection types

D Enums

Template literal types let you build new string types by combining or transforming existing string types.

What does this mapped type produce?

type ReadonlyKeys<T> = { readonly [K in keyof T]: T[K] }

A A type with all properties readonly

B A type with all properties optional

C A type with keys renamed

D A type with function properties

This mapped type makes every property in T readonly by adding the readonly modifier.

In this example, what is the type of 'getName'?

type Person = { name: string };
type Getters = { [K in keyof Person as `get${Capitalize}`]: () => Person[K] };

A () => number

B () => string

C string

D number

'getName' is a function returning the original 'name' property type, which is string.

Why might you use mapped types with template literals in a real project?

A To improve runtime performance

B To write less code for loops

C To automatically create related property names and types, reducing manual errors

D To create CSS styles

Mapped types with template literals help generate consistent property names and types automatically, improving code safety and maintainability.

Explain how mapped types and template literal types can be combined to transform property names in TypeScript.

Think about how you can rename keys by adding text before or after the original key.

Describe a practical use case where mapped types with template literals improve your TypeScript code.

Consider how you might create functions for each property automatically.

Practice

1.

What does a mapped type with template literals in TypeScript allow you to do?

Example: { [K in keyof T as `get${Capitalize}`]: () => T[K] }

easy

A. Create new variables at runtime

B. Rename keys dynamically in object types

C. Change values of an object at runtime

D. Execute functions based on key names

Solution

1. Step 1: Understand mapped types

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

2. Step 2: Role of template literals

   Template literals let you rename keys dynamically by combining strings and key names.

3. Final Answer:

   Rename keys dynamically in object types -> Option B

4. Quick Check:

   Mapped type + template literals = dynamic key renaming [OK]

Hint: Mapped types + template literals = dynamic key renaming [OK]

Common Mistakes:

• Confusing runtime behavior with type transformations
• Thinking it changes object values instead of keys
• Assuming it creates variables or runs code

2.

Which of the following is the correct syntax to rename keys using a mapped type with template literals?

type RenameKeys<T> = { [K in keyof T as ???]: T[K] }

easy

A. `new_${string & K}`

B. `new_${K}`

C. `new_${number & K}`

D. `new_${any & K}`

Solution

1. Step 1: Understand key remapping syntax

   Use as with template literals and intersect key type with string to ensure keys are strings.

2. Step 2: Check each option

   `new_${string & K}` uses string & K which is correct; others use incorrect types like number or any.

3. Final Answer:

   `new_${string & K}` -> Option A

4. Quick Check:

   Use string & K for safe key renaming [OK]

Hint: Use string & K to rename keys safely in template literals [OK]

Common Mistakes:

• Using K directly without string intersection
• Using number or any instead of string & K
• Missing backticks for template literals

3.

What is the resulting type of the following code?

type User = { name: string; age: number };
type PrefixedUser = { [K in keyof User as `get${Capitalize}`]: () => User[K] };

medium

A. { name: () => string; age: () => number }

B. { getName: string; getAge: number }

C. { getname: () => string; getage: () => number }

D. { getName: () => string; getAge: () => number }

Solution

1. Step 1: Analyze key transformation

   Keys 'name' and 'age' are capitalized and prefixed with 'get', becoming 'getName' and 'getAge'.

2. Step 2: Analyze value transformation

   Values become functions returning the original property types: () => string and () => number.

3. Final Answer:

   { getName: () => string; getAge: () => number } -> Option D

4. Quick Check:

   Mapped keys capitalized + function values = { getName: () => string; getAge: () => number } [OK]

Hint: Capitalize keys and prefix with 'get' for function types [OK]

Common Mistakes:

• Not capitalizing keys
• Returning values instead of functions
• Using original keys without prefix

4.

Identify the error in this mapped type using template literals:

type Flags = { isActive: boolean; isAdmin: boolean };
type PrefixedFlags = { [K in keyof Flags as `flag_${K}`]: Flags[K] };

medium

A. Key remapping must use Capitalize utility

B. Cannot use template literals in mapped types

C. Missing string intersection with K in template literal

D. Flags[K] is invalid syntax

Solution

1. Step 1: Check key remapping syntax

   Template literal keys must intersect K with string: string & K to ensure keys are strings.

2. Step 2: Identify missing intersection

   The code uses `flag_${K}` without string & K, causing a type error.

3. Final Answer:

   Missing string intersection with K in template literal -> Option C

4. Quick Check:

   Use string & K in template literals to fix error [OK]

Hint: Always intersect K with string in template literals [OK]

Common Mistakes:

• Omitting string & K intersection
• Thinking Capitalize is mandatory
• Believing template literals can't be used here

5.

Given this type, how would you create a mapped type that prefixes keys with on and capitalizes them, but only for keys starting with handle?

type Handlers = { handleClick: () => void; handleHover: () => void; reset: () => void };

hard

A. type Prefixed = { [K in keyof Handlers as K extends `handle${string}` ? `on${Capitalize}` : never]: Handlers[K] };

B. type Prefixed = { [K in keyof Handlers as `on${Capitalize}`]: Handlers[K] };

C. type Prefixed = { [K in keyof Handlers as K extends string ? `on${Capitalize<K>}` : never]: Handlers[K] };

D. type Prefixed = { [K in keyof Handlers as K extends `on${string}` ? `handle${Capitalize}` : never]: Handlers[K] };

Solution

1. Step 1: Filter keys starting with 'handle'

   Use conditional type K extends `handle${string}` to select only keys starting with 'handle'.

2. Step 2: Rename keys with 'on' prefix and capitalize

   Use template literal `on${Capitalize}` to rename keys accordingly.

3. Step 3: Assign original value types

   Keep the original function types from Handlers[K].

4. Final Answer:

   type Prefixed = { [K in keyof Handlers as K extends `handle${string}` ? `on${Capitalize}` : never]: Handlers[K] }; -> Option A

5. Quick Check:

   Conditional key remapping with prefix and capitalize = type Prefixed = { [K in keyof Handlers as K extends `handle${string}` ? `on${Capitalize}` : never]: Handlers[K] }; [OK]

Hint: Use conditional types to filter keys before renaming [OK]

Common Mistakes:

• Not filtering keys by prefix
• Applying rename to all keys
• Incorrect conditional syntax