我正在跟随this教程使用Apache Tiles 3我的项目servlet-context.xml是:
<?xml version="1.0" encoding="UTF-8"?>
<beans:beans xmlns="http://www.springframework.org/schema/mvc"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:beans="http://www.springframework.org/schema/beans"
xmlns:context="http://www.springframework.org/schema/context"
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/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 -->
<annotation-driven />
<!-- Handles HTTP GET requests for /resources/** by efficiently serving
up static resources in the ${webappRoot}/resources directory -->
<resources mapping="/resources/**" location="/resources/" />
<!-- Resolves views selected for rendering by @Controllers to .jsp resources
in the /WEB-INF/views directory -->
<beans:bean
class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<beans:property name="prefix" value="/WEB-INF/views/" />
<beans:property name="suffix" value=".jsp" />
</beans:bean>
<context:component-scan base-package="ali.arshad.soomro" />
<bean id="tilesConfigurer" class="org.springframework.web.servlet.view.tiles3.TilesConfigurer">
<property name="definitions">
<list>
<value>/WEB-INF/defs/general.xml</value>
</list>
</property>
</bean>
<bean id="viewResolver" class="org.springframework.web.servlet.view.UrlBasedViewResolver">
<property name="viewClass" value="org.springframework.web.servlet.view.tiles3.TilesView"
/> </bean>
</beans:beans>
这里我面临错误
此行找到多个注释: - 配置问题:找不到元素[bean] Offending资源的BeanDefinitionParser:file [G:/ Spring / java-blog / src / main / webapp / WEB-INF /弹簧/ appServlet / servlet的context.xml中] - 找不到元素[bean]的BeanDefinitionParser - cvc-complex-type.2.4.c:匹配的通配符是strict,但是没有为元素'bean'找到声明。
在第class="org.springframework.web.servlet.view.tiles3.TilesConfigurer"
行和错误
cvc-complex-type.2.4.c:匹配的通配符是strict,但是没有找到元素'bean'的声明。
在第class="org.springframework.web.servlet.view.UrlBasedViewResolver"
行。
pom.xml依赖项是
<dependency>
<groupId>org.apache.tiles</groupId>
<artifactId>tiles-api</artifactId>
<version>3.0.3</version>
</dependency>
<dependency>
<groupId>org.apache.tiles</groupId>
<artifactId>tiles-core</artifactId>
<version>3.0.3</version>
</dependency>
<dependency>
<groupId>org.apache.tiles</groupId>
<artifactId>tiles-jsp</artifactId>
<version>3.0.3</version>
</dependency>
<dependency>
<groupId>org.apache.tiles</groupId>
<artifactId>tiles-servlet</artifactId>
<version>3.0.3</version>
</dependency>
<dependency>
<groupId>org.apache.tiles</groupId>
<artifactId>tiles-template</artifactId>
<version>3.0.3</version>
</dependency>
更新
在下面的答案中提出建议后,我在servlet-contex.xml中作为
<beans:bean class="org.springframework.web.servlet.view.tiles3.TilesViewResolver">
<beans:property name="viewClass" value="org.springframework.web.servlet.view.tiles3.TilesView"></beans:property>
<beans:property name="order" value="0"></beans:property>
</beans:bean>
<beans:bean class="org.springframework.web.servlet.view.tiles3.TilesConfigurer" id="tilesConfigurer">
<beans:property name="definitions" value="/WEB-INF/spring/tiles.xml"> </beans:property>
</beans:bean>
现在我收到此错误
第属性:class bean类的完全限定名称,除非它仅用作child的父定义 bean定义。
数据类型:字符串
class="org.springframework.web.servlet.view.tiles3.TilesViewResolver"
行
在第class="org.springframework.web.servlet.view.tiles3.TilesConfigurer"
行
任何人都可以建议我解决这些错误吗?
答案 0 :(得分:1)
问题在于命名空间。您应该拥有以bean开头的所有标签