Question

我的web.xml为

<context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/applicationContext.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>
        <load-on-startup>2</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>dispatcher</servlet-name>
        <url-pattern>/</url-pattern>
    </servlet-mapping>

applicationContext as

<?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:p="http://www.springframework.org/schema/p"
       xmlns:aop="http://www.springframework.org/schema/aop"
       xmlns:tx="http://www.springframework.org/schema/tx"
       xmlns:context="http://www.springframework.org/schema/context"
       xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.1.xsd
       http://www.springframework.org/schema/aop http://www.springframework.org/schema/aop/spring-aop-3.1.xsd
       http://www.springframework.org/schema/tx http://www.springframework.org/schema/tx/spring-tx-3.1.xsd
       http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.1.xsd"
>

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


</beans>

Dispatcher servlet as

    <?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:p="http://www.springframework.org/schema/p"
           xmlns:aop="http://www.springframework.org/schema/aop"
           xmlns:tx="http://www.springframework.org/schema/tx"
           xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.1.xsd
           http://www.springframework.org/schema/aop http://www.springframework.org/schema/aop/spring-aop-3.1.xsd
           http://www.springframework.org/schema/tx http://www.springframework.org/schema/tx/spring-tx-3.1.xsd">

        <bean class="org.springframework.web.servlet.mvc.support.ControllerClassNameHandlerMapping"/>

        <!--
        Most controllers will use the ControllerClassNameHandlerMapping above, but
        for the index controller we are using ParameterizableViewController, so we must
        define an explicit mapping for it.
        -->
        <bean id="urlMapping" class="org.springframework.web.servlet.handler.SimpleUrlHandlerMapping">
            <property name="mappings">
                <props>
                    <prop key="index.htm">indexController</prop>
                </props>
            </property>
        </bean>

        <bean id="viewResolver"
              class="org.springframework.web.servlet.view.InternalResourceViewResolver"
              p:prefix="/WEB-INF/jsp/"
              p:suffix=".jsp" />

        <!--
        The index controller.
        -->
        <bean name="indexController"
              class="org.springframework.web.servlet.mvc.ParameterizableViewController"
              p:viewName="index" />

    </beans>

控制器，如com.mindedges.testapp包

@Controller
@RequestMapping("/sample")
public class SampleController  {

    public String sample(Model model){
        return "sample.jsp";
    }
}

当我点击http://localhost:8084/TestApp/sample。

警告：在名为
的DispatcherServlet中找不到带有URI [/ TestApp / sample]的HTTP请求的映射

当春天开始时，我也看不到Testapp /样本加载的处理程序

我猜我的组件扫描不起作用。为什么？任何其他原因

编辑：调度程序上下文是

<?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:p="http://www.springframework.org/schema/p"
       xmlns:aop="http://www.springframework.org/schema/aop"
       xmlns:tx="http://www.springframework.org/schema/tx"   
       xmlns:context="http://www.springframework.org/schema/context"
       xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.1.xsd
       http://www.springframework.org/schema/aop http://www.springframework.org/schema/aop/spring-aop-3.1.xsd
       http://www.springframework.org/schema/tx http://www.springframework.org/schema/tx/spring-tx-3.1.xsd
       http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.1.xsd
">

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

    <bean class="org.springframework.web.servlet.mvc.support.ControllerClassNameHandlerMapping"/>

    <!--
    Most controllers will use the ControllerClassNameHandlerMapping above, but
    for the index controller we are using ParameterizableViewController, so we must
    define an explicit mapping for it.
    -->
    <bean id="urlMapping" class="org.springframework.web.servlet.handler.SimpleUrlHandlerMapping">
        <property name="mappings">
            <props>
                <prop key="index.htm">indexController</prop>
            </props>
        </property>
    </bean>

    <bean id="viewResolver"
          class="org.springframework.web.servlet.view.InternalResourceViewResolver"
          p:prefix="/WEB-INF/jsp/"
          p:suffix=".jsp" />

    <!--
    The index controller.
    -->
    <bean name="indexController"
          class="org.springframework.web.servlet.mvc.ParameterizableViewController"
          p:viewName="index" />

</beans>

Answer 1

返回没有.jsp文件扩展名的视图名称。视图名称的解析由InternalResourceViewResolver处理。

public String sample(Model model){
    return "sample";
}

您的配置片段：

<bean id="viewResolver"
      class="org.springframework.web.servlet.view.InternalResourceViewResolver"
      p:prefix="/WEB-INF/jsp/"
      p:suffix=".jsp" />

获取视图名称sample并应用适当的前缀（/WEB-INF/jsp/）和后缀（.jsp）以形成用于解析视图的整个视图名称。

另请确保sample.jsp位于/WEB-INF/jsp/sample.jsp。

同样，在使用@RequestMapping注释时，您必须通过包含<mvc:annotation-driven/>在调度程序serlvet的配置中启用这些注释。

您还必须在调度程序配置中添加组件扫描，以便向调度程序servlet注册控制器。目前，您使用以下命令在applicationContext.xml配置文件中执行组件扫描：

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

但是，由于此组件扫描发生在applicationContext文件中，因此注册的bean（假设它们是您的控制器）将不可用于调度程序serlvet。您必须将此配置代码段放在dispatcher-context.xml。

中

许多人想咨询我的previous answers之一，了解调度程序配置和上下文配置之间的差异。

您的dispatcher-context.xml文件中还有一些空格，并且缺少mvc命名空间。我已经纠正了这些问题，并在GitHub Gist中提供了这些问题。

Answer 2

更改请求地图的名称，与其尝试返回的jsp具有不同的名称

警告：在带有名称的DispatcherServlet中，没有找到带有URI []的HTTP请求的映射

2 个答案: