What is the recommended project structure for spri

问题:

I'm a beginner with spring boot. I'm involved in the beginning of a project where we would build rest services using spring boot. Could you please advise the recommended directory structure to follow when building a project that will just expose rest services?

回答1:

You do not need to do anything special to start. Start with a normal java project, either maven or gradle or IDE project layout with starter dependency.

You need just one main class, as per guide here and rest...

There is no constrained package structure. Actual structure will be driven by your requirement/whim and the directory structure is laid by build-tool / IDE

You can follow same structure that you might be following for a Spring MVC application.

You can follow either way

• A project is divided into layers:
  

  for example: DDD style

  • Service layer : service package contains service classes
  • DAO/REPO layer : dao package containing dao classes
  • Entity layers

  
  or
  

  any layer structure suitable to your problem for which you are writing problem.
  
  

• A project divided into modules or functionalities or features and A module is divided into layers like above

I prefer the second, because it follows Business context. Think in terms of concepts.

What you do is dependent upon how you see the project. It is your code organization skills.

回答2:

config - class which will read from property files

cache - caching mechanism class files

constants - constant defined class

controller - controller class

exception - exception class

model - pojos classes will be present

security - security classes

service - Impl classes

util - utility classes

validation - validators classes

bootloader - main class

回答3:

I have an example which I have been using for couple years. Please take a look as a reference.

https://github.com/bigzidane/springboot-rest-h2-swagger

回答4:

There is a somehow recommended directory structure mentioned at https://docs.spring.io/spring-boot/docs/current/reference/html/using-boot-structuring-your-code.html

You can create a api folder and put your controllers there.

If you have some configuration beans, put them in a separate package too.

回答5:

Though this question has an accepted answer, still I would like to share my project structure for RESTful services.

src/main/java
    +- com
        +- example
            +- Application.java
            +- ApplicationConstants.java
                +- configuration
                |   +- ApplicationConfiguration.java
                +- controller
                |   +- ApplicationController.java
                +- dao
                |   +- impl
                |   |   +- ApplicationDaoImpl.java
                |   +- ApplicationDao.java
                +- dto
                |   +- ApplicationDto.java
                +- service
                |   +- impl
                |   |   +- ApplicationServiceImpl.java
                |   +- ApplicationService.java
                +- util
                |   +- ApplicationUtils.java
                +- validation
                |   +- impl
                |   |   +- ApplicationValidationImpl.java
                |   +- ApplicationValidation.java

DAO = Data Access Object.
DTO = Data Transfer Object.

回答6:

Please use Spring Tool Suite (Eclipse-based development environment that is customized for developing Spring applications).
Create a Spring Starter Project, it will create the directory structure for you with the spring boot maven dependencies.

回答7:

Use Link-1 to generate a project. this a basic project for learning. you can understand the folder structure. Use Link-2 for creating a basic Spring boot project. 1: http://start.spring.io/ 2: https://projects.spring.io/spring-boot/

Create a gradle/maven project Automatically src/main/java and src/main/test will be created. create controller/service/Repository package and start writing the code.

-src/main/java(source folder) ---com.package.service(package) ---ServiceClass(Class) ---com.package.controller(package) ---ControllerClass(Class)