入门案例开发步骤
①:创建新模块,选择Spring初始化,并配置模块相关基础信息
②:选择当前模块需要使用的技术集
③:开发控制器类
@RestController
@RequestMapping("/books")
public class BookController {
@GetMapping("/{id}")
public String getById(@PathVariable Integer id) {
System.out.println("id ==> " + id);
return "hello , spring boot! ";
}
}
④:运行自动生成的Application类
最简SpringBoot程序所包含的基础文件
<?xml version="1.0" encoding="UTF-8"?>
<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 https://maven.apache.org/xsd/maven-4.0.0.xsd">
<modelVersion>4.0.0</modelVersion>
<parent>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-parent</artifactId>
<version>2.5.0</version>
</parent>
<groupId>com.shell</groupId>
<artifactId>springboot-01-quickstart</artifactId>
<version>0.0.1-SNAPSHOT</version>
<dependencies>
<dependency>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-web</artifactId>
</dependency>
</dependencies>
</project>
@SpringBootApplication
public class Application {
public static void main(String[] args) {
SpringApplication.run(Application.class, args);
}
}
注意事项:
基于idea开发SpringBoot程序需要确保联网且能够加载到程序框架结构
文章评论