当我加载我的应用程序时,我只得到404资源未找到错误..根本没有登录tomcat ...
在这里你可以看到我的项目配置:
这是我的web.xml文件:
<?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>Spring Web MVC Application</display-name>
<servlet>
<servlet-name>mvc-dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>mvc-dispatcher</servlet-name>
<url-pattern>/pages/*</url-pattern>
<!--url-pattern>*.jsp</url-pattern-->
</servlet-mapping>
<urlrewrite default-match-type="wildcard">
<rule>
<from>/</from>
<to>/pages/</to>
</rule>
<rule>
<from>/**</from>
<to>/pages/$1</to>
</rule>
<outbound-rule>
<from>/pages/**</from>
<to>/$1</to>
</outbound-rule>
</urlrewrite>
<filter>
<filter-name>urlRewriteFilter</filter-name>
<filter-class>org.tuckey.web.filters.urlrewrite.UrlRewriteFilter</filter-class>
</filter>
<filter-mapping>
<filter-name>urlRewriteFilter</filter-name>
<url-pattern>/*</url-pattern>
</filter-mapping>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/mvc-dispatcher-servlet.xml,
/WEB-INF/spring-security.xml,
/WEB-INF/spring-database.xml
</param-value>
</context-param>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<!-- Spring Security -->
<filter>
<filter-name>springSecurityFilterChain</filter-name>
<filter-class>
org.springframework.web.filter.DelegatingFilterProxy
</filter-class>
</filter>
<filter-mapping>
<filter-name>springSecurityFilterChain</filter-name>
<url-pattern>/*</url-pattern>
</filter-mapping>
<welcome-file-list>
<welcome-file>/index.jsp</welcome-file>
</welcome-file-list>
</web-app>
tiles.xml文件
<?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/pages/layout.jsp">
<put-attribute name="title" value="" />
<put-attribute name="header" value="/WEB-INF/pages/header.jsp" />
<put-attribute name="menu" value="/WEB-INF/pages/menu.jsp" />
<put-attribute name="body" value="" />
<put-attribute name="footer" value="/WEB-INF/pages/footer.jsp" />
</definition>
<definition name="contact" extends="base.definition">
<put-attribute name="title" value="Contact Manager" />
<put-attribute name="body" value="/WEB-INF/pages/prueba2.jsp" />
</definition>
</tiles-definitions>
mvc-dispatcher-servlet.xml
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:context="http://www.springframework.org/schema/context"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:mvc="http://www.springframework.org/schema/mvc"
xsi:schemaLocation="
http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc-3.2.xsd
http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-3.2.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context-3.2.xsd">
<context:component-scan base-package="com.mkyong.common.controller" />
<bean
class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<property name="prefix">
<value>/WEB-INF/pages/</value>
</property>
<property name="suffix">
<value>.jsp</value>
</property>
</bean>
<bean id="messageSource"
class="org.springframework.context.support.ResourceBundleMessageSource">
<property name="basenames">
<list>
<value>mymessages</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
</value>
</property>
</bean>
<bean id="tilesConfigurer"
class="org.springframework.web.servlet.view.tiles3.TilesConfigurer">
<property name="definitions">
<list>
<value>/WEB-INF/tiles.xml</value>
</list>
</property>
</bean>
<mvc:annotation-driven />
<mvc:resources mapping="/resources/**" location="/resources/" />
<context:annotation-config />
</beans>
弹簧security.xml文件
<beans:beans xmlns="http://www.springframework.org/schema/security"
xmlns:beans="http://www.springframework.org/schema/beans" 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-3.2.xsd
http://www.springframework.org/schema/security
http://www.springframework.org/schema/security/spring-security-3.0.3.xsd">
<!--import resource="../database/spring-database.xml"/-->
<http auto-config="true" access-denied-page="/accessDenied">
<intercept-url pattern="/login*" access="IS_AUTHENTICATED_ANONYMOUSLY"/>
<intercept-url pattern="/*" access="ROLE_USER" />
<form-login login-page="/login" default-target-url="/prueba2"
authentication-failure-url="/loginfailed.jsp"/>
<logout logout-success-url="/logout" />
</http>
<authentication-manager>
<authentication-provider>
<jdbc-user-service data-source-ref="dataSource"
users-by-username-query="
select username,password, enabled
from users where USERNAME=?"
authorities-by-username-query="
select u.username, ur.authority from users u, user_roles ur
where u.user_id = ur.user_id and u.username =? "
/>
</authentication-provider>
</authentication-manager>
</beans:beans>
这是我的控制者:
@Controller
@RequestMapping("/prueba2")
public class controller2 extends AbstractController {
Stock stock=new Stock();
List stockList=new ArrayList<Stock>();
ApplicationContext appContext = new ClassPathXmlApplicationContext("spring/config/BeanLocations.xml");
StockBo stockBo = (StockBo)appContext.getBean("stockBo");
/*@RequestMapping(method = RequestMethod.GET)
public Stock returnCustomer(ModelMap model) {
stock=stockBo.findByStockCode("7668");
model.addAttribute("miStock", stock);
return stock;
}*/
protected ModelAndView handleRequestInternal (HttpServletRequest req, HttpServletResponse res) throws Exception{
User user = (User) SecurityContextHolder.getContext().getAuthentication().getPrincipal();
String name = user.getUsername();
Map<String, Object> model = new HashMap<String, Object>();
stockList=stockBo.findAll();
stock=stockBo.findByStockCode("7668");
model.put("listaStock",stockList);
model.put("miStock", stock);
model.put("nombreUsuario", name);
System.out.println("lista objetos--->" + stockList.toString());
return new ModelAndView( "prueba2", "model", model );
}
public StockBo getStockBo() {
return stockBo;
}
public void setStockBo(StockBo stockBo) {
this.stockBo = stockBo;
}
}
答案 0 :(得分:1)
我不直接知道你问题的答案,但这是我接近它的方法:
首先,如果我正确理解您的问题,您希望转到网址http://localhost:8080/SpringExample
会将您重定向到登录页面,但这不会发生(您看到的是404)。我假设您已经检查过您的应用程序实际部署到SpringExample上下文。
您可能会遇到几个可能影响URL解释和重定向的事情。首先,你有Spring,它将以/ pages /开头的任何请求映射到Spring调度程序。接下来,你有一些URL重写,因为它是在过滤器中完成的,应该在spring dispatcher servlet之前进行。你还有一个加载所有弹簧上下文配置的监听器。第三,你有一个在过滤器中实现的Spring Security。最后,您已经获得了tile配置,如果找不到资源,理论上也可能导致404错误,尽管看起来你的确定没问题。
这非常复杂,如果事情没有以正确的顺序发生,你就会遇到问题。我要做的是去除每个组件,然后开始逐个添加它们。首先,除欢迎文件配置外,取出所有内容,看看是否可以通过转到您的网址到达/index.jsp。然后重新添加Spring,看看你是否仍然可以使用它。然后添加URL重定向,然后添加安全性,然后添加切片。这有助于您缩小问题范围。
答案 1 :(得分:0)
<bean id="viewResolver" class="org.springframework.web.servlet.view.tiles3.TilesViewResolver"/>
<bean id="tilesConfigurer"
class="org.springframework.web.servlet.view.tiles3.TilesConfigurer">
<property name="definitions">
<list>
<value>/WEB-INF/tiles.xml</value>
</list>
</property>
</bean>
尝试编写代码=&gt;
答案 2 :(得分:0)
无需编写此代码瓷砖集成就足够了 / WEB-INF /页/
</property>
<property name="suffix">
<value>.jsp</value>
</property>
</bean>