Question

如果我有线

<context:component-scan base-package="uni.inso" />
<mvc:annotation-driven />

在我的dispatcher-servlet.xml中，我将从我的tomcat中收到此错误消息：

WARN "http-bio-8080"-exec-1 servlet.PageNotFound:1108 - No mapping found for HTTP request with URI [/inso/home] in DispatcherServlet with name 'dispatcher'

但是，如果我在我的web-application-config中有这两行，它可以正常工作。此外，InternalViewResolver不会在dispatcher-servlet-xml中工作，但在web-application-config中也可以。为什么会这样？

我得到了以下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">

    <servlet>
        <servlet-name>dispatcher</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>        
        <init-param>
            <param-name>contextConfigLocation</param-name>
            <param-value>/WEB-INF/web-application-config.xml</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>dispatcher</servlet-name>
        <url-pattern>/</url-pattern>
    </servlet-mapping>
</web-app>

这是我的dispatcher-servlet.xml

<?xml version="1.0" encoding="UTF-8"?><beans xmlns="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xmlns:context="http://www.springframework.org/schema/context"
    xmlns:mvc="http://www.springframework.org/schema/mvc"
    xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
    http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.0.xsd
    http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc-3.0.xsd"
    default-lazy-init="true">
    <!-- DispatcherServlet Context: defines this servlet's request-processing infrastructure -->

    <!-- Scans within the base package of the application for @Components to configure as beans -->
    <!-- @Controller, @Service, @Configuration, etc. -->
    <context:component-scan base-package="uni.inso" />

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

    <bean id="viewResolver" class="org.springframework.web.servlet.view.InternalResourceViewResolver">
        <property name="prefix">
            <value>/WEB-INF/views/</value>
        </property>
        <property name="suffix">
            <value>.jsp</value>
        </property>
  </bean>
</beans>

这是我的web-application-config.xml

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xmlns:mvc="http://www.springframework.org/schema/mvc"
    xmlns:context="http://www.springframework.org/schema/context"
    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
        http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.0.xsd">

  <bean id="myDataSource" class="org.apache.commons.dbcp.BasicDataSource" destroy-method="close">
    <property name="driverClassName" value="org.postgresql.Driver"/>
    <property name="url" value="jdbc:postgrsql://localhost:8080/open"/>
    <property name="username" value="po"/>
    <property name="password" value="ha"/>
  </bean>

  <bean id="mySessionFactory" class="org.springframework.orm.hibernate4.LocalSessionFactoryBean">
    <property name="dataSource" ref="myDataSource"/>
    <property name="hibernateProperties">
        <props>
            <prop key="hibernate.dialect">org.hibernate.dialect.PostgreSQLDialect</prop>
        </props>
    </property>
  </bean> 
</beans>

Answer 1

也许你真的想在你的web.xml中做这样的事情：

<context-param>
    <param-name>contextConfigLocation</param-name>
    <param-value>/WEB-INF/web-application-config.xml</param-value>
</context-param>

<listener>
    <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>

<servlet>
    <servlet-name>dispatcher</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <init-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/dispatcher-servlet.xml</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>

Answer 2

我认为你没有在你的包中定义任何控制器servlet，因此它无法为任何URL找到任何类型的映射。

@Controller
public class HelloController{
    @RequestMapping(value = "/hello", method = RequestMethod.GET)
     public String printHello(ModelMap model) {
        model.addAttribute("message", "Hello Spring MVC Framework!");
      return "hello";
     }
}

您也可以参考此Tutorial。

Answer 3

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
    xmlns:context="http://www.springframework.org/schema/context"
    xmlns:mvc="http://www.springframework.org/schema/mvc"
     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-3.0.xsd
        http://www.springframework.org/schema/context 
        http://www.springframework.org/schema/context/spring-context-3.0.xsd
        http://www.springframework.org/schema/mvc
        http://www.springframework.org/schema/mvc/spring-mvc-3.0.xsd ">

       <mvc:annotation-driven></mvc:annotation-driven>

    <context:component-scan base-package="com.neel" />

    <bean
        class="org.springframework.web.servlet.view.InternalResourceViewResolver">
        <property name="prefix" value="/WEB-INF/pages/" />
        <property name="suffix" value=".jsp" />
    </bean>


<mvc:resources location="/WEB-INF/resources/" mapping="/resources/**"/>
</beans>

spring dispatcher-servlet.xml不能正常工作

3 个答案: