如何在 Spring MVC 和 Tomcat 中配置映射

Question

我正在尝试配置 SpringMVC 项目，但编译时出现错误

WARN : org.springframework.web.servlet.PageNotFound - No mapping found for HTTP request with URI [/test/home.jsp] in DispatcherServlet with name 'appServlet'

我这里有一个 jsp 页面： /test/src/main/webapp/WEB-INF/views/home.jsp

这是我的代码： /test/src/main/webapp/WEB-INF/web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>
            /WEB-INF/spring/root-context.xml
        </param-value>
    </context-param>
    <listener>
        <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
    </listener>
    <servlet>
        <servlet-name>appServlet</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <init-param>
            <param-name>contextConfigLocation</param-name>
            <param-value>/WEB-INF/spring/appServlet/servlet-context.xml</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>

    <servlet-mapping>
        <servlet-name>appServlet</servlet-name>
        <url-pattern>/</url-pattern>
    </servlet-mapping>
    <filter>
        <filter-name>charsetFilter</filter-name>
        <filter-class>org.springframework.web.filter.CharacterEncodingFilter</filter-class>
        <init-param>
            <param-name>encoding</param-name>
            <param-value>UTF-8</param-value>
        </init-param>
        <init-param>
            <param-name>forceEncoding</param-name>
            <param-value>true</param-value>
        </init-param>
    </filter>
    <filter-mapping>
        <filter-name>charsetFilter</filter-name>
        <url-pattern>/*</url-pattern>
    </filter-mapping>
</web-app>

/test/src/main/webapp/WEB-INF/spring/appServlet/servlet-context.xml

<?xml version="1.0" encoding="UTF-8"?>
<beans:beans xmlns="http://www.springframework.org/schema/mvc"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:beans="http://www.springframework.org/schema/beans"
    xsi:schemaLocation="
        http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc-3.0.xsd
        http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.0.xsd">
    <annotation-driven />
    <resources mapping="/resources/**" location="/resources/" />
    <beans:bean
        name="test"
        class="org.springframework.web.servlet.view.InternalResourceViewResolver">
        <beans:property name="prefix" value="/WEB-INF/views/" />
        <beans:property name="suffix" value=".jsp" />
    </beans:bean>
    <beans:import resource="controllers.xml" />
</beans:beans>

Answer 1

不要使用 jsp 扩展名，例如直接将其映射为 html

<display-name>appServlet</display-name>
    <servlet>
    <servlet-name>appServlet</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <load-on-startup>1</load-on-startup>
  </servlet>
  <servlet-mapping>
    <servlet-name>appServlet</servlet-name>
    <url-pattern>*.html</url-pattern>
  </servlet-mapping>

，然后在不带扩展名的情况下在控制器映射文件中：

@RequestMapping(value="/home.html", method=RequestMethod.GET)
    public ModelAndView indexView(){
        ModelAndView mv = new ModelAndView("test/home");
        return mv;
    }

在此示例中，一旦用户点击 /home.html，调度程序 servlet 将解析它并检索 /test/home.jsp

希望它有帮助。