struts2.0中关于namespace到底如何指定

Question

一个JSP页面action到一个Action实现类中，然后还知道一个namespace指定唯一一个Action
但是为什么我在随意制定了一个namespace为/hello，结果还是能成功forward到结果页面
但是在控制台下却抛出了一个异常，现在我想要解决这个异常该怎么办？实在是想不明白。
异常链如下：
com.opensymphony.xwork2.config.ConfigurationException: There is no Action mapped for namespace [/] and action name [login] associated with context path [/servlet3.0].
at com.opensymphony.xwork2.DefaultActionProxy.prepare(DefaultActionProxy.java:185) ~[xwork-core-2.3.24.jar:?]
at org.apache.struts2.impl.StrutsActionProxy.prepare(StrutsActionProxy.java:63) ~[struts2-core-2.3.24.jar:2.3.24]
at org.apache.struts2.impl.StrutsActionProxyFactory.createActionProxy(StrutsActionProxyFactory.java:37) ~[struts2-core-2.3.24.jar:2.3.24]
at com.opensymphony.xwork2.DefaultActionProxyFactory.createActionProxy(DefaultActionProxyFactory.java:58) ~[xwork-core-2.3.24.jar:?]
at org.apache.struts2.dispatcher.Dispatcher.serviceAction(Dispatcher.java:554) [struts2-core-2.3.24.jar:2.3.24]
at org.apache.struts2.dispatcher.ng.ExecuteOperations.executeAction(ExecuteOperations.java:81) [struts2-core-2.3.24.jar:?]
at org.apache.struts2.dispatcher.ng.filter.StrutsPrepareAndExecuteFilter.doFilter(StrutsPrepareAndExecuteFilter.java:99) [struts2-core-2.3.24.jar:2.3.24]
at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:239) [catalina.jar:8.0.24]
at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206) [catalina.jar:8.0.24]
at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:219) [catalina.jar:8.0.24]
at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:106) [catalina.jar:8.0.24]
at org.apache.catalina.authenticator.AuthenticatorBase.invoke(AuthenticatorBase.java:502) [catalina.jar:8.0.24]
at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:142) [catalina.jar:8.0.24]
at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:79) [catalina.jar:8.0.24]
at org.apache.catalina.valves.AbstractAccessLogValve.invoke(AbstractAccessLogValve.java:617) [catalina.jar:8.0.24]
at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:88) [catalina.jar:8.0.24]
at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:518) [catalina.jar:8.0.24]
at org.apache.coyote.http11.AbstractHttp11Processor.process(AbstractHttp11Processor.java:1091) [tomcat-coyote.jar:8.0.24]
at org.apache.coyote.AbstractProtocol$AbstractConnectionHandler.process(AbstractProtocol.java:668) [tomcat-coyote.jar:8.0.24]
at org.apache.tomcat.util.net.NioEndpoint$SocketProcessor.doRun(NioEndpoint.java:1527) [tomcat-coyote.jar:8.0.24]
at org.apache.tomcat.util.net.NioEndpoint$SocketProcessor.run(NioEndpoint.java:1484) [tomcat-coyote.jar:8.0.24]
at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1142) [?:1.8.0_45]
at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:617) [?:1.8.0_45]
at org.apache.tomcat.util.threads.TaskThread$WrappingRunnable.run(TaskThread.java:61) [tomcat-util.jar:8.0.24]
at java.lang.Thread.run(Thread.java:745) [?:1.8.0_45]
是不是在指定并且我在

标签中是这样写的：
namespace/xxxAction，不然会出现404错误。

Answer 1

写个 /后面自定义名称