Spring Initializr for project creation in Spring Boot - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to specify the project type in Spring Initializr.

Spring Boot

spring init --build=[1] myproject

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

Agradle

Bmaven

Cjar

Dwar

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'jar' or 'war' as the build type instead of build system type.

Confusing Gradle and Maven project types.

The --build option specifies the build system type. maven is a common choice for Spring Boot projects.

2fill in blank

medium

Complete the code to add the web dependency using Spring Initializr.

Spring Boot

spring init --dependencies=[1] mywebapp

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

Aweb

Bdata-jpa

Csecurity

Dbatch

Attempts:

3 left

💡 Hint

Common Mistakes

Choosing unrelated dependencies like 'batch' or 'security' when web support is needed.

Forgetting to specify dependencies at all.

The web dependency adds Spring Web support, enabling you to build web applications.

3fill in blank

hard

Fix the error in the command to generate a Gradle project with Java version 17.

Spring Boot

spring init --build=[1] --java-version=17 myapp

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

Amaven

Bgradle-kotlin

Cgradle

Dmaven-kotlin

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'maven' instead of 'gradle' for Gradle projects.

Confusing Kotlin DSL options with standard Gradle.

To create a Gradle project, use --build=gradle. The command specifies Java version 17 and focuses on the correct build tool.

4fill in blank

hard

Fill both blanks to create a project with group ID and artifact ID.

Spring Boot

spring init --groupId=[1] --artifactId=[2] demoapp

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

Acom.example

Bdemoapp

Corg.project

Dmyapp

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping group ID and artifact ID values.

Using invalid formats for group ID.

The --groupId is usually a domain-like string such as com.example. The --artifactId is the project name, here demoapp.

5fill in blank

hard

Fill all three blanks to generate a project with package name, Java version, and packaging type.

Spring Boot

spring init --package-name=[1] --java-version=[2] --packaging=[3] myservice

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

Acom.mycompany.service

B17

Cjar

Dwar

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'war' packaging when 'jar' is expected.

Incorrect package name format.

The package name is usually a domain-like string such as com.mycompany.service. Java version 17 is a modern choice. Packaging jar is common for Spring Boot apps.