org.springframework.web.servlet.PageNotFound noHandlerFound-未找到具有URI的HTTP请求的映射

Question

我知道有上千个这样的问题，但是我是盲目的，找不到错误。对于.jsp页面的此设置，一切正常（指向http：// localhost：8082 /
spring
/后，

令人惊讶的是，http://localhost:8082/spring/WEB- INF/spring/static/index.jsp页面未加载后）。我是新手，所以可能我误会了我在做什么/

当我切换到<property name="suffix" value=".html" />（我有2个文件index.jsp和index.html）时，无法从http://localhost:8082/spring/该http://localhost:8082/spring/WEB- INF/spring/static/index.html链接或该链接加载页面。

mar 14, 2015 8:38:07 PM org.springframework.web.servlet.PageNotFound noHandlerFound
WARNING: No mapping found for HTTP request with URI [/spring/WEB-INF/spring/static/index.html] in DispatcherServlet with name 'appServlet'

servlet-context.xml：

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns:mvc="http://www.springframework.org/schema/mvc"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xmlns="http://www.springframework.org/schema/beans"
    xmlns:context="http://www.springframework.org/schema/context"
    xmlns:tx="http://www.springframework.org/schema/tx"
    xsi:schemaLocation="http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc.xsd
        http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd
        http://www.springframework.org/schema/tx http://www.springframework.org/schema/tx/spring-tx.xsd
        http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context.xsd">

    <!-- DispatcherServlet Context: defines this servlet's request-processing infrastructure -->

    <!-- Enables the Spring MVC @Controller programming model -->
    <mvc:annotation-driven />

    <!-- Handles HTTP GET requests for /resources/** by efficiently serving up static resources in the ${webappRoot}/resources directory -->
    <mvc:resources mapping="/resources/**" location="/resources/" />

    <!-- Resolves views selected for rendering by @Controllers to .jsp resources in the /WEB-INF/views directory -->
    <bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
        <property name="prefix" value="/WEB-INF/spring/static/" />
        <property name="suffix" value=".html" />
    </bean>

    <context:component-scan base-package="net.codejava.spring" />

    <bean id="dataSource" class="org.apache.commons.dbcp2.BasicDataSource" destroy-method="close">
        <property name="driverClassName" value="com.mysql.jdbc.Driver"/>
        <property name="url" value="jdbc:mysql://localhost:3306/usersdb"/>
        <property name="username" value="root"/>
        <property name="password" value="1234"/>
    </bean>

    <bean id="sessionFactory"
        class="org.springframework.orm.hibernate4.LocalSessionFactoryBean">
        <property name="dataSource" ref="dataSource" />
        <property name="configLocation" value="classpath:hibernate.cfg.xml" />
    </bean>

    <tx:annotation-driven />
    <bean id="transactionManager"
        class="org.springframework.orm.hibernate4.HibernateTransactionManager">
        <property name="sessionFactory" ref="sessionFactory" />
    </bean>

    <bean id="userDao" class="net.codejava.spring.dao.UserDAOImpl">
        <constructor-arg>
            <ref bean="sessionFactory" />
        </constructor-arg>
    </bean>     
</beans>

家用控制器：

@Controller
public class HomeController {

    @Autowired
    private UserDAO userDao;

    @RequestMapping(value="/")
    public ModelAndView home() {
        List<User> listUsers = userDao.list();
        //ModelAndView model = new ModelAndView("home");
        ModelAndView model = new ModelAndView("index");
        model.addObject("userList", listUsers);
        return model;
    }

}

web.xml：

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="3.0" xmlns="http://java.sun.com/xml/ns/javaee"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"
    xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd">

    <!-- The definition of the Root Spring Container shared by all Servlets 
        and Filters -->
    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/spring/root-context.xml</param-value>
    </context-param>


    <!-- Creates the Spring Container shared by all Servlets and Filters -->
    <listener>
        <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
    </listener>

    <!-- Processes application requests -->
    <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>

</web-app>

Answer 1

首先-
您的web.xml不仅仅是您放入文件中的内容。例如，在Tomcat中，您的web.xml文件与Tomcat默认的web.xml合并（请参阅conf/web.xmlTomcat的目录）。您可以在其中找到以形式的URL
*.jsp映射到JSP Servlet。

Java Web Apps按照servlet的方式工作（您将它们映射到web.xml中的URL）。例如，您映射DispatchServlet到的URL
/。这个分派servlet“神奇地”与Spring一起工作并加载您的控制器-并“在内部”处理它们的映射。那就是为什么你使用过@RequestMapping(value="/")。

在控制器内部，您基本上会命令将请求分派到具有名称的视图index-然后解析该名称InternalResourceViewResolver以形成/WEB- INF/spring/static/+ index+ 的路径.jsp-就像您在上下文的描述符html" target="_blank">文件中配置的一样。

基本上，这发出一个内部调度请求（不完全是，但您现在可以这样考虑）-然后在第一个请求时使用完全相同的规则对其进行处理-例如，被*.jspservlet
捕获然后进行处理。但是，无论如何/WEB-INF/spring/static/index.html都无法处理类似的请求-找不到匹配的规则。

要回答您的问题-您可以将任何HTML放在JSP文件中-使用任何非必需的JSP标签。但是，如果您想真正地使用纯HTML文件而不进行任何分析-
然后考虑使用静态资源-您已经在其中进行了设置<mvc:resources mapping="/resources/**" location="/resources/" />-这基本上会将/ resources /目录的内容暴露给世界，而无需进行任何处理。