Type-safe API response handling in Typescript - Mini Project: Build & Apply

Type-safe API response handling

📖 Scenario: You are building a small TypeScript program that fetches user data from an API. You want to make sure the data you get back matches the expected structure so your program is safe and doesn't crash.

🎯 Goal: Create a type-safe way to handle an API response for user data using TypeScript interfaces and type guards.

📋 What You'll Learn

Create an interface for the user data with exact fields

Create a variable to hold a sample API response object

Write a type guard function to check if an object matches the user interface

Use the type guard to safely print the user's name

💡 Why This Matters

🌍 Real World

APIs often return data that might not be exactly what your program expects. Using type-safe checks helps avoid bugs and crashes.

💼 Career

Type-safe API handling is a key skill for frontend and backend developers working with TypeScript to build reliable applications.

1

Define the User interface

Create a TypeScript interface called User with these exact fields: id as a number, name as a string, and email as a string.

Typescript

// Your code here

Hint

Use the interface keyword and define the three fields with their types.

2

Create a sample API response object

Create a constant variable called apiResponse and assign it this exact object: { id: 101, name: "Alice", email: "[e-mail]" }.

Typescript

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

// Your code here

Hint

Use const to create the variable and assign the exact object.

3

Write a type guard function

Write a function called isUser that takes a parameter obj of type any and returns a boolean. The function should check if obj has id as a number, name as a string, and email as a string. Use this signature: function isUser(obj: any): obj is User.

Typescript

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

const apiResponse = { id: 101, name: "Alice", email: "[e-mail]" };

// Your code here

Hint

Check each property with typeof and make sure obj is not null.

4

Use the type guard to print the user's name

Write an if statement that uses isUser(apiResponse) to check the type. Inside the if, write console.log to print exactly User name: followed by the user's name.

Typescript

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

const apiResponse = { id: 101, name: "Alice", email: "[e-mail]" };

function isUser(obj: any): obj is User {
  return typeof obj === "object"
    && obj !== null
    && typeof obj.id === "number"
    && typeof obj.name === "string"
    && typeof obj.email === "string";
}

// Your code here

Hint

Use a template string inside console.log to print the user's name.

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