Question

Spring 3 with Tiles 2.2.2 - 获取“Servlet spring目前不可用”。我试图用Spring 3和Tiles 2.2制作一个hello world样本。但是我收到了上述错误

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" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" id="WebApp_ID" version="2.5">
  <display-name>Tiles</display-name>
  <welcome-file-list>
    <welcome-file>index.html</welcome-file>
    <welcome-file>index.htm</welcome-file>
    <welcome-file>index.jsp</welcome-file>
    <welcome-file>default.html</welcome-file>
    <welcome-file>default.htm</welcome-file>
    <welcome-file>default.jsp</welcome-file>
  </welcome-file-list>

  <servlet>
        <servlet-name>spring</servlet-name>
        <servlet-class>
            org.springframework.web.servlet.DispatcherServlet
        </servlet-class>
        <load-on-startup>1</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>spring</servlet-name>
        <url-pattern>*.html</url-pattern>
    </servlet-mapping>

</web-app>

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="baselayout"  template="/WEB-INF/jsp/layout.jsp">
        <put-attribute name="title" value="Spring3 Tile Example" />
        <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="welcome" extends="baselayout">
        <put-attribute name="title" value="Hello Word" />
        <put-attribute name="body" value="/WEB-INF/jsp/hello.jsp" />
    </definition>

</tiles-definitions>

弹簧servlet.xml中

<?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:context="http://www.springframework.org/schema/context"
    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">

    <context:component-scan base-package="org.uftwf.controller" />


    <bean id="viewResolver"
        class="org.springframework.web.servlet.view.UrlBasedViewResolver">
        <property name="viewClass">
            <value>
                org.springframework.web.servlet.view.tiles2.TilesView
            </value>
        </property>
    </bean>
    <bean id="tilesConfigurer"
        class="org.springframework.web.servlet.view.tiles2.TilesConfigurer">
        <property name="definitions">
            <list>
                <value>/WEB-INF/tiles.xml</value>
            </list>
        </property>
    </bean>
</beans>

Answer 1

你的web.xml必须加载spring-servlet.xml
比如..

<servlet>
    <servlet-name>spring</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <init-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>classpath:spring-servlet.xml</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>

Spring 3 with Tiles 2.2.2 - 获取“Servlet spring目前无法使用”

1 个答案: