如何实现struts2拦截器失败 自动跳转

Question

例如在浏览器里输入一个我并没有配置的action 动作，正常情况控制台会显示如下信息，请问如何实现跳转到我的主页上去
08:17:56,156 ERROR [Dispatcher] Could not find action or result
There is no Action mapped for namespace / and action name login. - [unknown location]
at com.opensymphony.xwork2.DefaultActionProxy.prepare(DefaultActionProxy.java:186)
at org.apache.struts2.impl.StrutsActionProxyFactory.createActionProxy(StrutsActionProxyFactory.java:41)
at org.apache.struts2.dispatcher.Dispatcher.serviceAction(Dispatcher.java:494)
at org.apache.struts2.dispatcher.FilterDispatcher.doFilter(FilterDispatcher.java:419)
at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:235)
at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206)
at org.apache.struts2.dispatcher.ActionContextCleanUp.doFilter(ActionContextCleanUp.java:99)
at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:235)
at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206)
at org.jboss.web.tomcat.filters.ReplyHeaderFilter.doFilter(ReplyHeaderFilter.java:96)
at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:235)
at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206)
at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:230)
at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:175)
at org.jboss.web.tomcat.security.SecurityAssociationValve.invoke(SecurityAssociationValve.java:179)
at org.jboss.web.tomcat.security.JaccContextValve.invoke(JaccContextValve.java:84)
at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:127)
at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:102)
at org.jboss.web.tomcat.service.jca.CachedConnectionValve.invoke(CachedConnectionValve.java:157)
at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:109)
at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:262)
at org.apache.coyote.http11.Http11Processor.process(Http11Processor.java:844)
at org.apache.coyote.http11.Http11Protocol$Http11ConnectionHandler.process(Http11Protocol.java:583)
at org.apache.tomcat.util.net.JIoEndpoint$Worker.run(JIoEndpoint.java:446)
at java.lang.Thread.run(Thread.java:619)

Answer 1

<global-results>
     <result name="root">/index.jsp</result>
</global-results>
<global-exception-mappings>
     <exception-mapping result="root" exception="java.lang.Exception" />
</global-exception-mappings>

Answer 2

配置下input result，
form表单中action的值为login,这和你的struts.xml文件中的值一样

Answer 3

可以用struts2声明式异常处理 如2楼代码

Answer 4

    <action name="**">  
        <result>/index.jsp</result>  
    </action>  

Answer 5

同意，找不到action,给它个默认的

Answer 6

试试看看看，貌似所有的action 都给拦截掉了，
而我只想拦截不存在的action