第七节——spring中web层快速布置

Posted 2021-10-22 想学习安全的小白

快速配置

1. 导入坐标

<dependency>
  <groupId>javax.servlet</groupId>
  <artifactId>javax.servlet-api</artifactId>
  <version>4.0.1</version>
</dependency>
<dependency>
  <groupId>javax.servlet.jsp</groupId>
  <artifactId>javax.servlet.jsp-api</artifactId>
  <version>2.3.3</version>
</dependency>
<dependency>
  <groupId>org.springframework</groupId>
  <artifactId>spring-context</artifactId>
  <version>5.3.10</version>
</dependency>

2. 在main-java-demo1-dao目录下创建UserDaoImpl.class模仿数据调用层

public class UserDaoImpl {
    public void show(){
        System.out.println("qwe");
    }
}

3. 在main-java-demo1-service目录下创建UserServiceImpl.class建立业务层

public class UserServiceImpl {
    private UserDaoImpl userDao;

    public void setUserDao(UserDaoImpl userDao) {
        this.userDao = userDao;
    }

    public void show(){
        userDao.show();
    }
}

4. 在main-resources目录下创建applicationContext.xml文件，配置bean

<bean id="userDao" class="demo1.dao.UserDaoImpl"/>
<bean id="userService" class="demo1.service.UserServiceImpl">
    <property name="userDao" ref="userDao"/>
</bean>

5. 在main-java-demo1-web目录下创建UserServlet.class文件，充当控制层

public class UserServlet extends HttpServlet {
    @Override
    protected void doGet(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
        ApplicationContext app = new ClassPathXmlApplicationContext("applicationContext.xml");
        UserServiceImpl userService = (UserServiceImpl) app.getBean("userService");
        userService.show();
    }
}

6. 编辑main-webapp-WEB-INF目录下的web.xml文件配置servlet以及url

<servlet>
  <servlet-name>UserServlet</servlet-name>
  <servlet-class>demo1.web.UserServlet</servlet-class>
</servlet>
<servlet-mapping>
  <servlet-name>UserServlet</servlet-name>
  <url-pattern>/userServlet</url-pattern>
</servlet-mapping>

7. 开启tomcat，访问地址：http://localhost:8080/demo02_war/userServlet，看到控制台有输出