Type-safe API response handling in Typescript - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define a type for the API response.

Typescript

type ApiResponse = {
  data: string[];
  status: [1];
};

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

Aboolean

Bstring

Cnumber

Dany

Attempts:

3 left

💡 Hint

Common Mistakes

Using string instead of number for status code.

Using any which loses type safety.

The status field is a number representing HTTP status codes.

2fill in blank

medium

Complete the function signature to specify the return type as a Promise of ApiResponse.

Typescript

function fetchData(): [1] {
  return fetch('/api/data').then(res => res.json());
}

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

Aany

BApiResponse

Cvoid

DPromise<ApiResponse>

Attempts:

3 left

💡 Hint

Common Mistakes

Returning ApiResponse directly without Promise.

Using void which means no return value.

The function returns a Promise that resolves to ApiResponse.

3fill in blank

hard

Fix the error in the type assertion to ensure the response is treated as ApiResponse.

Typescript

const response = await fetch('/api/data');
const data = (await response.json()) as [1];

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

AApiResponse

Bany

Cunknown

Dstring

Attempts:

3 left

💡 Hint

Common Mistakes

Using any which disables type checking.

Using string which does not match the object structure.

Use ApiResponse to safely assert the JSON matches the expected type.

4fill in blank

hard

Fill both blanks to create a type guard function that checks if an object is ApiResponse.

Typescript

function isApiResponse(obj: any): obj is [1] {
  return obj && typeof obj.status === [2];
}

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

AApiResponse

Bstring

Cnumber

Dboolean

Attempts:

3 left

💡 Hint

Common Mistakes

Checking status as string instead of number.

Not using the correct type guard syntax.

The type guard returns true if obj has a status property of type number.

5fill in blank

hard

Fill all three blanks to filter valid ApiResponse objects from an array.

Typescript

const responses: any[] = [...];
const validResponses = responses.filter((item): item is [1] => {
  return item && typeof item.status === [2] && Array.isArray(item.[3]);
});

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

AApiResponse

Bnumber

Cdata

Dstring

Attempts:

3 left

💡 Hint

Common Mistakes

Checking wrong property names.

Using string instead of number for status type check.

This filter keeps only objects that match ApiResponse by checking status is a number and data is an array.

Practice

1. What is the main benefit of using type-safe API response handling in TypeScript?

easy

A. It makes the API faster to respond.

B. It removes the need for testing API responses.

C. It automatically fixes bugs in the API.

D. It helps catch errors early by defining expected data shapes.

Solution

1. Step 1: Understand type safety purpose

   Type-safe handling means defining what data should look like before using it.

2. Step 2: Identify the benefit

   This helps catch mistakes early, like missing or wrong data types, before running the program.

3. Final Answer:

   It helps catch errors early by defining expected data shapes. -> Option D

4. Quick Check:

   Type safety = Early error detection [OK]

Hint: Type safety means catching errors before running code [OK]

Common Mistakes:

• Thinking type safety improves API speed
• Believing it fixes bugs automatically
• Assuming it replaces testing

2. Which TypeScript syntax correctly defines a generic function to fetch API data with a type-safe response?

easy

A. function fetchData(url: string): Promise<any> { /* ... */ }

B. function fetchData<T>(url: string): Promise<T> { /* ... */ }

C. function fetchData<T>(url: string): T { /* ... */ }

D. function fetchData(url: string): T { /* ... */ }

Solution

1. Step 1: Understand generic function syntax

   Generic functions use <T> to represent a type parameter, allowing flexible typing.

2. Step 2: Check return type for async fetch

   Fetching data returns a Promise of type T, so the return type must be Promise<T>.

3. Final Answer:

   function fetchData<T>(url: string): Promise<T> { /* ... */ } -> Option B

4. Quick Check:

   Generic + Promise = function fetchData<T>(url: string): Promise<T> { /* ... */ } [OK]

Hint: Generic fetch returns Promise of type T [OK]

Common Mistakes:

• Returning T directly instead of Promise<T>
• Missing generic <T> in function definition
• Using Promise<any> loses type safety

3. Given this TypeScript code, what will be the output?

interface User { id: number; name: string; }

async function getUser(): Promise<User> {
  return { id: 1, name: "Alice" };
}

getUser().then(user => console.log(user.name));

medium

A. { id: 1, name: "Alice" }

B. 1

C. "Alice"

D. undefined

Solution

1. Step 1: Analyze the getUser function

   It returns a Promise resolving to an object with id 1 and name "Alice".

2. Step 2: Check the then callback

   The callback logs user.name, which is "Alice".

3. Final Answer:

   "Alice" -> Option C

4. Quick Check:

   user.name = "Alice" [OK]

Hint: Promise resolves to object; log property name [OK]

Common Mistakes:

• Logging user instead of user.name
• Confusing id with name
• Expecting entire object printed

4. Identify the error in this TypeScript code for type-safe API response handling:

interface ApiResponse { success: boolean; data: string; }

async function fetchApi(): Promise<ApiResponse> {
  const response = await fetch('https://api.example.com');
  const json = await response.json();
  return json;
}

medium

A. Missing type assertion or validation for json before returning.

B. The fetch call should not use await.

C. The interface ApiResponse is missing a constructor.

D. The function should return void instead of Promise<ApiResponse>.

Solution

1. Step 1: Check how json is handled

   json is returned directly without confirming it matches ApiResponse shape.

2. Step 2: Understand type safety requirement

   TypeScript needs either a type assertion or runtime validation to ensure json fits ApiResponse.

3. Final Answer:

   Missing type assertion or validation for json before returning. -> Option A

4. Quick Check:

   Validate json shape before return [OK]

Hint: Always validate or assert API response type [OK]

Common Mistakes:

• Assuming fetch json is automatically typed
• Ignoring need for runtime checks
• Misunderstanding async/await usage

5. You want to create a type-safe function that fetches a list of users from an API and returns only users with an active status. Which approach correctly applies type-safe API response handling?

hard

A. Define interface User with 'active' boolean, fetch data as Promise, then filter users by active before returning.

B. Fetch data as Promise<any>, filter users by active without type checking, then cast to User[].

C. Define interface User without 'active', fetch data as Promise, then filter users by active.

D. Fetch data as Promise, return data without filtering to keep type safety.

Solution

1. Step 1: Define User interface including 'active' property

   To filter by active status, the User type must include an 'active' boolean field.

2. Step 2: Fetch data as Promise<User[]> and filter

   Fetching with correct type ensures type safety; filtering users by active keeps the result accurate and typed.

3. Final Answer:

   Define interface User with 'active' boolean, fetch data as Promise<User[]>, then filter users by active before returning. -> Option A

4. Quick Check:

   Correct interface + typed fetch + filter = Define interface User with 'active' boolean, fetch data as Promise, then filter users by active before returning. [OK]

Hint: Include all needed fields in interface before filtering [OK]

Common Mistakes:

• Casting any to User[] without checks
• Filtering on missing properties in interface
• Returning unfiltered data losing requirements