Database and app orchestration in Spring Boot - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define a Spring Boot application entry point.

Spring Boot

import org.springframework.boot.SpringApplication;
import org.springframework.boot.autoconfigure.SpringBootApplication;

@SpringBootApplication
public class Application {
    public static void main(String[] args) {
        SpringApplication.[1](Application.class, args);
    }
}

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

Astart

Blaunch

Crun

Dexecute

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'start' or 'launch' instead of 'run' causes errors.

Forgetting to pass the application class and args.

The SpringApplication.run() method starts the Spring Boot application.

2fill in blank

medium

Complete the code to inject a JPA repository into a service class.

Spring Boot

import org.springframework.stereotype.Service;
import org.springframework.beans.factory.annotation.[1];

@Service
public class UserService {

    private final UserRepository userRepository;

    @Autowired
    public UserService(UserRepository userRepository) {
        this.userRepository = userRepository;
    }

    // service methods
}

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AAutowired

BInject

CResource

DQualifier

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'Inject' without proper imports.

Confusing 'Qualifier' with injection annotation.

The @Autowired annotation tells Spring to inject the repository automatically.

3fill in blank

hard

Fix the error in the repository interface declaration.

Spring Boot

import org.springframework.data.jpa.repository.JpaRepository;

public interface UserRepository extends JpaRepository<[1], Long> {
}

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AUserDTO

BUserEntity

CUserController

DUserService

Attempts:

3 left

💡 Hint

Common Mistakes

Using service or controller classes instead of entity.

Using DTO classes which are not entities.

The repository must extend JpaRepository with the entity class, here UserEntity.

4fill in blank

hard

Fill both blanks to create a REST controller method that fetches a user by ID.

Spring Boot

import org.springframework.web.bind.annotation.*;

@RestController
@RequestMapping("/users")
public class UserController {

    private final UserService userService;

    public UserController(UserService userService) {
        this.userService = userService;
    }

    @GetMapping("/[1]")
    public UserEntity getUserById(@PathVariable [2] id) {
        return userService.findById(id);
    }
}

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Aid

BuserId

CLong

DString

Attempts:

3 left

💡 Hint

Common Mistakes

Mismatching path variable name and parameter name.

Using wrong data type like String for ID.

The path variable name should match the placeholder in the URL, here 'id'. The parameter type is Long for the user ID.

5fill in blank

hard

Fill all three blanks to define a Spring Data JPA query method that finds users by their email.

Spring Boot

import org.springframework.data.jpa.repository.JpaRepository;
import java.util.Optional;

public interface UserRepository extends JpaRepository<UserEntity, Long> {
    Optional<UserEntity> findBy[1]([2] [3]);
}

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AEmail

BString

Cemail

DId

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong method name like findById instead of findByEmail.

Using wrong parameter type or name.

The method name must be findByEmail. The parameter type is String and the parameter name is email.