【Java】使用 IDEA 快速生成 SpringBoot 模块-编程知识

【Java】使用 IDEA 快速生成 SpringBoot 模块

在 pom.xml 更改为 spring initializr 配置之后的 pom.xml

<?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>3.2.0</version><relativePath/> <!-- lookup parent from repository --></parent><groupId>com.example</groupId><artifactId>module1</artifactId><version>0.0.1-SNAPSHOT</version><name>module1</name><description>Demo project for Spring Boot</description><properties><java.version>17</java.version></properties><dependencies><dependency><groupId>org.springframework.boot</groupId><artifactId>spring-boot-starter-web</artifactId></dependency><dependency><groupId>org.springframework.boot</groupId><artifactId>spring-boot-starter-test</artifactId><scope>test</scope></dependency></dependencies><build><plugins><plugin><groupId>org.springframework.boot</groupId><artifactId>spring-boot-maven-plugin</artifactId></plugin></plugins></build></project>

创建入口类

package com.example;import org.springframework.boot.SpringApplication;
import org.springframework.boot.autoconfigure.SpringBootApplication;@SpringBootApplication
public class Module1App {public static void main(String[] args) {SpringApplication.run(Module1App.class, args);}
}