问题
I am facing this error while I am clicking on a link to display a page. I'm using Spring MVC Tiles.
Error:
javax.servlet.ServletException: Could not resolve view with name 'contact' in servlet with name 'dispatcher'
Below is the code.
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>
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
<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>
回答1:
You don't need the following in your viewResolver
definition:
<property name="prefix" value="/WEB-INF/jsp/" />
<property name="suffix" value=".jsp" />
Remove it
来源:https://stackoverflow.com/questions/29921589/java-could-not-resolve-view-with-name-in-spring-mvc