Java：无法在Spring MVC中使用名称解析视图

Question

我在点击链接以显示页面时遇到此错误。我正在使用Spring MVC Tiles。

错误：

javax.servlet.ServletException：在名为 '调度'

下面是代码的servlet无法解析视图名称为 '接触'。

tiles.xml

<?xml version="1.0" encoding="UTF-8" ?> 
<!DOCTYPE tiles-definitions PUBLIC 
     "-//Apache Software Foundation//DTD Tiles Configuration 2.0//EN" 
     "http://tiles.apache.org/dtds/tiles-config_2_0.dtd"> 
<tiles-definitions> 
    <definition name="base.definition" 
     template="/WEB-INF/jsp/layout.jsp"> 
     <put-attribute name="title" value="" /> 
     <put-attribute name="header" value="/WEB-INF/jsp/header.jsp" /> 
     <put-attribute name="menu" value="/WEB-INF/jsp/menu.jsp" /> 
     <put-attribute name="body" value="" /> 
     <put-attribute name="footer" value="/WEB-INF/jsp/footer.jsp" /> 
    </definition> 

    <definition name="contact" extends="base.definition"> 
     <put-attribute name="title" value="Contact Manager" /> 
     <put-attribute name="body" value="/WEB-INF/jsp/contact.jsp" /> 
    </definition> 

    <definition name="hello" extends="base.definition"> 
     <put-attribute name="title" value="Hello Spring MVC" /> 
     <put-attribute name="body" value="/WEB-INF/jsp/hello.jsp" /> 
    </definition> 

</tiles-definitions> 

的web.xml

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

     <servlet> 
     <servlet-name>dispatcher</servlet-name> 
     <servlet-class> 
     org.springframework.web.servlet.DispatcherServlet 
     </servlet-class> 
     <load-on-startup>1</load-on-startup> 
    </servlet> 

    <servlet-mapping> 
     <servlet-name>dispatcher</servlet-name> 
     <url-pattern>/</url-pattern> 
    </servlet-mapping> 

    <context-param> 
     <param-name>contextConfigLocation</param-name> 
     <param-value>/WEB-INF/dispatcher-servlet.xml</param-value> 
    </context-param> 

</web-app> 

servlet.xml中

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

    <bean id="viewResolver" class="org.springframework.web.servlet.view.UrlBasedViewResolver"> 
     <property name="prefix" value="/WEB-INF/jsp/" /> 
     <property name="suffix" value=".jsp" /> 
     <property name="viewClass" value = "org.springframework.web.servlet.view.tiles2.TilesView"/> 
    </bean> 

    <bean class="org.springframework.web.servlet.view.tiles2.TilesConfigurer"> 
     <property name="definitions"> 
      <list> 
       <value>/WEB-INF/tiles.xml</value> 
      </list> 
     </property> 
    </bean> 

</beans> 

Answer 1

你并不需要在你的viewResolver定义如下：

<property name="prefix" value="/WEB-INF/jsp/" /> 
    <property name="suffix" value=".jsp" /> 

删除它