Mapped type with template literals in Typescript

Introduction

Mapped types with template literals let you create new object types by changing keys using string patterns. This helps make your code flexible and reusable.

When you want to create a new object type by adding prefixes or suffixes to existing keys.

When you need to transform all keys of an object type to a different naming style.

When you want to generate types for related properties automatically, like 'get' or 'set' methods.

When you want to enforce consistent naming patterns across object keys.

When you want to avoid repeating similar type definitions with slight key changes.

Syntax

Typescript

type NewType<T> = {
  [K in keyof T as `prefix_${string & K}_suffix`]: T[K]
};

The as keyword lets you rename keys using template literal strings.

Use string & K to ensure the key is treated as a string for template literals.

Examples

This creates a new type where keys are prefixed with 'get_'. For example, 'name' becomes 'get_name'.

Typescript

type User = {
  name: string;
  age: number;
};

type PrefixedUser = {
  [K in keyof User as `get_${string & K}`]: User[K];
};

This adds '_flag' suffix to each key, like 'isActive_flag'.

Typescript

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

type SuffixFlags = {
  [K in keyof Flags as `${string & K}_flag`]: Flags[K];
};

This converts all keys to uppercase using the built-in Uppercase utility type.

Typescript

type Options = {
  debug: boolean;
  verbose: boolean;
};

type UppercaseOptions = {
  [K in keyof Options as Uppercase<string & K>]: Options[K];
};

Sample Program

This program creates a new type where each key from Person is renamed with a 'readonly_' prefix and made readonly. Then it creates an object of that type and prints it.

Typescript

type Person = {
  firstName: string;
  lastName: string;
  age: number;
};

type ReadonlyPerson = {
  readonly [K in keyof Person as `readonly_${string & K}`]: Person[K];
};

const person: ReadonlyPerson = {
  readonly_firstName: "Alice",
  readonly_lastName: "Smith",
  readonly_age: 30
};

console.log(person);

Important Notes

Mapped types with template literals only work with keys that can be converted to strings.

You can combine this with modifiers like readonly or ? (optional) for more control.

Template literal types are powerful for creating dynamic and descriptive keys in types.

Summary

Mapped types with template literals let you rename keys dynamically in object types.

This helps create flexible and reusable type patterns without repeating code.

Use as with template literals and string & K to transform keys safely.

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