简单的SpringMVC项目创建流程（基于XML文件（了解））

news2024/9/27 17:29:51

1：首先创建一个maven项目，并在pom.xml文件中导入依赖

<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 http://maven.apache.org/maven-v4_0_0.xsd">
  <modelVersion>4.0.0</modelVersion>
  <groupId>com.zql</groupId>
  <artifactId>sm01-hello</artifactId>
  <packaging>war</packaging>
  <version>1.0-SNAPSHOT</version>
  <name>sm01-hello Maven Webapp</name>
  <url>http://maven.apache.org</url>
  <dependencies>
    <!-- https://mvnrepository.com/artifact/javax.servlet/javax.servlet-api -->
    <dependency>
      <groupId>javax.servlet</groupId>
      <artifactId>javax.servlet-api</artifactId>
      <version>4.0.1</version>
      <scope>provided</scope>
    </dependency>
    <!-- https://mvnrepository.com/artifact/org.springframework/spring-webmvc -->
    <dependency>
      <groupId>org.springframework</groupId>
      <artifactId>spring-webmvc</artifactId>
      <version>5.3.32</version>
    </dependency>

    <dependency>
      <groupId>junit</groupId>
      <artifactId>junit</artifactId>
      <version>3.8.1</version>
      <scope>test</scope>
    </dependency>
  </dependencies>
  <build>
    <finalName>sm01-hello</finalName>
  </build>
</project>

最主要的依赖是servlet的依赖，以及spring-webmvc的依赖，注意版本问题，不要导入版本过高，我导入的适用于jdk8

servlet依赖必须4.0.1以下

<!-- https://mvnrepository.com/artifact/javax.servlet/javax.servlet-api -->
<dependency>
  <groupId>javax.servlet</groupId>
  <artifactId>javax.servlet-api</artifactId>
  <version>4.0.1</version>
  <scope>provided</scope>
</dependency>

spring-webmvc 版本不宜过高，5.3就可以

<!-- https://mvnrepository.com/artifact/org.springframework/spring-webmvc -->
<dependency>
  <groupId>org.springframework</groupId>
  <artifactId>spring-webmvc</artifactId>
  <version>5.3.32</version>
</dependency>

这些依赖都是从maven中央仓库下载的，下载地址如下：

Maven Repository: Search/Browse/Explore (mvnrepository.com)

2：要配置webapp目录下的web.xml文件，文件目录如图：

在web.xml文件中写如下代码：设置DispatcherServlet类用于接受请求，设置映射器文件为springmvc.xml

<!DOCTYPE web-app PUBLIC
 "-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN"
 "http://java.sun.com/dtd/web-app_2_3.dtd" >

<web-app>
  <servlet>
    <servlet-name>dispatcherServlet</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <init-param>
      <param-name>contextConfigLocation</param-name>
      <param-value>classpath:springmvc.xml</param-value>
    </init-param>
  </servlet>
    <servlet-mapping>
      <servlet-name>dispatcherServlet</servlet-name>
      <url-pattern>/</url-pattern>
    </servlet-mapping>
</web-app>

3：在resources资源目录下创建springmvc.xml文件，用于UserContriller对象的管理

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
       xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
       xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd">
        <bean id="/user" class="com.zql.controller.UserController"></bean>
</beans>

4：创建UserController类，这个类要实现Controller接口，并重写其中的handRequest方法，返回ModelAndView类型的对象

package com.zql.controller;

import org.springframework.web.servlet.ModelAndView;
import org.springframework.web.servlet.mvc.Controller;

import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

public class UserController implements Controller {
    @Override
    public ModelAndView handleRequest(HttpServletRequest request, HttpServletResponse response) throws Exception {
        ModelAndView mv=new ModelAndView();
        mv.setViewName("/user.jsp");
        mv.addObject("name","zhangsan");
        mv.addObject("age",18);
        return mv;
    }
}

5：配置tomcat

点击右上角的配置