Why JPA matters for database access in Spring Boot - Test Your Understanding

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a JPA repository interface.

Spring Boot

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

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

AList

BService

CJpaRepository

DComponent

Attempts:

3 left

💡 Hint

Common Mistakes

Using List instead of JpaRepository

Using Service or Component which are unrelated

The JpaRepository interface provides JPA related methods for database access.

2fill in blank

medium

Complete the code to annotate an entity class for JPA.

Spring Boot

@[1]
public class Product {
  @Id
  private Long id;
}

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

AComponent

BRepository

CService

DEntity

Attempts:

3 left

💡 Hint

Common Mistakes

Using @Component or @Service instead of @Entity

Forgetting @Entity causes no table mapping

The @Entity annotation marks the class as a JPA entity mapped to a database table.

3fill in blank

hard

Fix the error in the repository method to find by name.

Spring Boot

List<User> findBy[1](String name);

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

AName

Bname

Cusername

DNameIgnoreCase

Attempts:

3 left

💡 Hint

Common Mistakes

Using lowercase 'name' instead of 'Name'

Using unrelated field names

In Spring Data JPA, query method names use 'findBy' followed by the capitalized property name, so findByName is correct if the field is 'name'.

4fill in blank

hard

Complete the code to create a JPA query method that finds products cheaper than a price.

Spring Boot

List<Product> findByPrice[1](Double price);

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

ALessThan

BGreaterThan

C, Pageable pageable

Attempts:

3 left

💡 Hint

Common Mistakes

Using GreaterThan instead of LessThan

Adding unnecessary parameters

findByPriceLessThan finds products with price less than the given value. No extra parameter is needed here.

5fill in blank

hard

Fill all three blanks to write a JPA query method that finds users by email containing a keyword and orders by id descending.

Spring Boot

List<User> findByEmail[1][2]OrderById[3](String keyword);

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

AContaining

BAsc

CDesc

DIgnoreCase

Attempts:

3 left

💡 Hint

Common Mistakes

Using Asc instead of Desc for ordering

Forgetting IgnoreCase for case-insensitive search

The method findByEmailContainingIgnoreCaseOrderByIdDesc finds users whose email contains the keyword, orders results by id descending, ignoring case.

Practice

1.

Why is JPA important when working with databases in Spring Boot?

easy

A. It requires you to write all SQL queries manually.

B. It lets you work with database data as Java objects instead of SQL.

C. It only works with NoSQL databases.

D. It replaces the need for a database entirely.

Solution

1. Step 1: Understand JPA's role in database access

   JPA maps database tables to Java objects, so you can use Java code to handle data instead of SQL.

2. Step 2: Compare options to JPA's purpose

   JPA does not require manual SQL for all queries (many are auto-generated), works primarily with relational databases, and does not eliminate the need for a database.

3. Final Answer:

   It lets you work with database data as Java objects instead of SQL. -> Option B

4. Quick Check:

   JPA = Java objects for database [OK]

Hint: JPA means Java objects, not raw SQL [OK]

Common Mistakes:

• Thinking JPA eliminates the database
• Believing JPA only works with NoSQL
• Assuming you must write all SQL manually

2.

Which of the following is the correct way to declare a JPA entity class in Spring Boot?

?

easy

A. @Entity public class User { private Long id; private String name; }

B. public class User { @Entity private Long id; private String name; }

C. @Table public class User { private Long id; private String name; }

D. @Entity public interface User { Long getId(); String getName(); }

Solution

1. Step 1: Identify correct JPA entity annotation usage

   The @Entity annotation must be placed on the class to mark it as a JPA entity.

2. Step 2: Check class structure

   @Entity public class User { private Long id; private String name; } correctly uses @Entity on a class with fields. public class User { @Entity private Long id; private String name; } misplaces @Entity on a field, @Table is for table naming rather than marking an entity, and an interface is not valid for JPA entities.

3. Final Answer:

   @Entity public class User { private Long id; private String name; } -> Option A

4. Quick Check:

   @Entity on class = correct entity [OK]

Hint: @Entity goes on class, not fields or interfaces [OK]

Common Mistakes:

• Putting @Entity on fields instead of class
• Using interface instead of class for entity
• Confusing @Entity with @Table annotation

3.

Given this Spring Data JPA repository interface:

public interface UserRepository extends JpaRepository<User, Long> {}

What happens when you call userRepository.findAll()?

medium

A. It returns null because no query is written.

B. It throws a compile-time error because findAll() is not defined.

C. It deletes all User records from the database.

D. It returns a list of all User objects from the database.

Solution

1. Step 1: Understand JpaRepository methods

   JpaRepository provides built-in methods like findAll() that return all records as Java objects.

2. Step 2: Analyze the method call

   Calling findAll() returns a list of all User entities from the database, no error or deletion occurs.

3. Final Answer:

   It returns a list of all User objects from the database. -> Option D

4. Quick Check:

   findAll() = list of all entities [OK]

Hint: JpaRepository has findAll() ready to use [OK]

Common Mistakes:

• Thinking findAll() needs manual query
• Confusing findAll() with delete methods
• Expecting null instead of empty list

4.

What is wrong with this JPA entity code snippet?

@Entity
public class Product {
  @Id
  private Long id;
  private String name;

  public Product(String name) {
    this.name = name;
  }
}

medium

A. The class should be abstract to be a JPA entity.

B. The @Id annotation should be on the class, not the field.

C. Missing no-argument constructor required by JPA.

D. The field 'name' must be annotated with @Column.

Solution

1. Step 1: Recall JPA entity constructor rules

   JPA requires a public or protected no-argument constructor to create entity instances.

2. Step 2: Check the provided constructors

   The class only has a constructor with a parameter, so the no-argument constructor is missing.

3. Final Answer:

   Missing no-argument constructor required by JPA. -> Option C

4. Quick Check:

   No-arg constructor needed for JPA [OK]

Hint: Always add a no-arg constructor for JPA entities [OK]

Common Mistakes:

• Thinking @Id goes on class
• Believing @Column is mandatory for all fields
• Assuming abstract class is needed

5.

You want to fetch all users whose name starts with 'A' using Spring Data JPA. Which repository method signature should you add?

hard

A. List<User> findByNameStartingWith(String prefix);

B. List<User> findAllByNameContains(String prefix);

C. List<User> findUsersByNameLike(String prefix);

D. List<User> getUsersWhereNameStartsWith(String prefix);

Solution

1. Step 1: Understand Spring Data JPA query method naming

   Spring Data JPA supports method names like findByNameStartingWith to generate queries automatically.

2. Step 2: Evaluate method signatures

   List<User> findByNameStartingWith(String prefix); uses the correct 'findByNameStartingWith' pattern. List<User> findAllByNameContains(String prefix); uses 'Contains' which matches anywhere, 'Like' is not a valid method keyword, and getUsersWhereNameStartsWith uses an invalid naming pattern.

3. Final Answer:

   List<User> findByNameStartingWith(String prefix); -> Option A

4. Quick Check:

   Method name pattern = findByNameStartingWith [OK]

Hint: Use 'findByNameStartingWith' for prefix queries [OK]

Common Mistakes:

• Using invalid method names not supported by Spring Data
• Confusing 'Contains' with 'StartingWith'
• Trying to write custom queries instead of method names