原来是Struts2.0的一个Bug

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Struts2

Struts Socket JSP XML Workflow

开始使用的是Struts2.06版本。在配置Struts.xml的时候，发现了一个问题：

struts.xml代码：

xml 代码

"http://struts.apache.org/dtds/struts-2.0.dtd">
<struts>
<include file="struts-default.xml" />
<package name="default" extends="struts-default">
<interceptors>
<interceptor-stack name="strutsStack">
<interceptor-ref name="exception" />
<interceptor-ref name="alias" />
<interceptor-ref name="servlet-config" />
<interceptor-ref name="prepare" />
<interceptor-ref name="i18n" />
<interceptor-ref name="chain" />
<interceptor-ref name="debugging" />
<interceptor-ref name="profiling" />
<interceptor-ref name="scoped-model-driven" />
<interceptor-ref name="model-driven" />
<interceptor-ref name="fileUpload" />
<interceptor-ref name="checkbox" />
<interceptor-ref name="static-params" />
<interceptor-ref name="params" />
<interceptor-ref name="validation">
<param name="excludeMethods">
input,back,cancel,browse
param>
interceptor-ref>
<interceptor-ref name="workflow">
<param name="excludeMethods">
input,back,cancel,browse
param>
interceptor-ref>
interceptor-stack>
interceptors>
<default-interceptor-ref name="strutsStack" />
package>
<include file="struts_xml/user.xml" />
struts>

这是struts.xml所包含的代码。

xml 代码

"http://struts.apache.org/dtds/struts-2.0.dtd">
<struts>
<package name="user" extends="default">
<action name="userAction"
class="com.leo.controller.UserController" method="userAction">
<result name="success">a.jspresult>
<result name="input">login.jspresult>
action>
<action name="index" class="com.leo.controller.UserController"
method="index">
<result name="success">login.jspresult>
<result name="input">login.jspresult>
action>
<action name="login" class="com.leo.controller.UserController"
method="login">
<result name="success" type="redirect-action">
index.action
result>
<result name="input">login.jspresult>
action>
<action name="logout" class="com.leo.controller.UserController"
method="logout">
<result name="success">login.jspresult>
<result name="input">login.jspresult>
action>
package>
struts>

这种“包”继承关系是合法的，但在struts2.06版本怎么也无法通过，控制台一直报错：Error building results for action ××× in namespace - action 。调试了很久，Webwork文档与书也翻了N遍，还是没有效果。最后实在没有办法了，换成Struts2.09包，竟然通过了。

大家可以试试，其实Struts2的包继承是非常有用的，很多全局拦截器的设置，Action的设置，以及Result-type等，都可以通过包继承来组织你的配置文件。

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Struts2+Spring+Hibernate In Action（一） | Struts2需要注意的一个小地方

2007-10-27 23:00
浏览 5616
评论(6)
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分类:企业架构
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6 楼 hmx_x 2008-10-27

Result-type等，都可

5 楼 dimscar 2007-10-31

真不知道为什么,webwork2.0变成了这样的Struts2.0,真叫晕倒!

4 楼 jalinhu 2007-10-31

<div class='code_title'>
<div class='code_title'>java 代码
<div class='dp-highlighter'>
<div class='bar'/>
<ol class='dp-j'>
 <li class='alt'>import java.net.*; </li>
 <li class=''>import java.io.*; </li>
 <li class='alt'>import java.util.*; </li>
 <li class=''> </li>
 <li class='alt'>public class UDPServer { </li>
 <li class=''> static final int INPORT = 1711; </li>
 <li class='alt'> private byte[] buf = new byte[1000]; </li>
 <li class=''> private DatagramPacket dp = new DatagramPacket(buf, buf.length); </li>
 <li class='alt'> private DatagramSocket socket; </li>
 <li class=''> public UDPServer() { </li>
 <li class='alt'> try { </li>
 <li class=''> socket = new DatagramSocket(INPORT); </li>
 <li class='alt'> System.out.println("Server started!"); </li>
 <li class=''> while (true) { </li>
 <li class='alt'> socket.receive(dp); </li>
 <li class=''> String rcvd = Dgram.toString(dp) + ",from ddress:" + dp.getAddress() + </li>
 <li class='alt'> ",port:" + dp.getPort(); </li>
 <li class=''> System.out.println(rcvd); </li>
 <li class='alt'> String echoString = "Echoed:" + rcvd; </li>
 <li class=''> DatagramPacket echo = Dgram.toDatagram(echoString, dp.getAddress(), </li>
 <li class='alt'> dp.getPort()); </li>
 <li class=''> socket.send(echo); </li>
 <li class='alt'> } </li>
 <li class=''> } </li>
 <li class='alt'> catch (SocketException e) { </li>
 <li class=''> System.err.println("Can't open socket"); </li>
 <li class='alt'> System.exit(1); </li>
 <li class=''> } </li>
 <li class='alt'> catch (IOException e) { </li>
 <li class=''> System.err.println("Communication error!"); </li>
 <li class='alt'> e.printStackTrace(); </li>
 <li class=''> } </li>
 <li class='alt'> } </li>
 <li class=''> </li>
 <li class='alt'> public static void main(String[] args) { </li>
 <li class=''> new UDPServer(); </li>
 <li class='alt'> } </li>
 <li class=''>} </li>
 <li class='alt'> </li>
 <li class=''>//===================================== </li>
 <li class='alt'>class UDPClient </li>
 <li class=''> extends Thread { </li>
 <li class='alt'> private DatagramSocket s; </li>
 <li class=''> private InetAddress hostAddress; </li>
 <li class='alt'> private byte[] buf = new byte[1000]; </li>
 <li class=''> private DatagramPacket dp = new DatagramPacket(buf, buf.length); </li>
 <li class='alt'> private int id; </li>
 <li class=''> </li>
 <li class='alt'> public UDPClient(int identifier) { </li>
 <li class=''> id = identifier; </li>
 <li class='alt'> try { </li>
 <li class=''> s = new DatagramSocket(); </li>
 <li class='alt'> hostAddress = InetAddress.getByName("localhost"); </li>
 <li class=''> } </li>
 <li class='alt'> catch (UnknownHostException e) { </li>
 <li class=''> System.err.println("Cannot find host"); </li>
 <li class='alt'> System.exit(1); </li>
 <li class=''> } </li>
 <li class='alt'> catch (SocketException e) { </li>
 <li class=''> System.err.println("Can't open Socket"); </li>
 <li class='alt'> e.printStackTrace(); </li>
 <li class=''> System.exit(1); </li>
 <li class='alt'> } </li>
 <li class=''> } </li>
 <li class='alt'> </li>
 <li class=''> public void run() { </li>
 <li class='alt'> try { </li>
 <li class=''> for (int i = 0; i < 1; i++) { //消息数 </li>
 <li class='alt'> String outMessage = "Client #" + </li>
 <li class=''> id + ",message#" + i; </li>
 <li class='alt'> s.send(Dgram.toDatagram(outMessage, hostAddress, UDPServer.INPORT)); </li>
 <li class=''> s.receive(dp); </li>
 <li class='alt'> String rcvd = "Client #" + id + ",rcvd from " + </li>
 <li class=''> dp.getAddress() + "," + </li>
 <li class='alt'> dp.getPort() + ":" + </li>
 <li class=''> Dgram.toString(dp); </li>
 <li class='alt'> System.out.println(rcvd); </li>
 <li class=''> } </li>
 <li class='alt'> } </li>
 <li class=''> catch (IOException e) { </li>
 <li class='alt'> e.printStackTrace(); </li>
 <li class=''> System.exit(1); </li>
 <li class='alt'> } </li>
 <li class=''> } </li>
 <li class='alt'> </li>
 <li class=''> public static void main(String[] args) { </li>
 <li class='alt'> for (int i = 0; i < 1; i++) //客户数量 </li>
 <li class=''> new UDPClient(i).start(); </li>
 <li class='alt'> } </li>
 <li class=''>} </li>
 <li class='alt'> </li>
 <li class=''>class Dgram { </li>
 <li class='alt'> public static DatagramPacket toDatagram( </li>
 <li class=''> String s, InetAddress destIA, int destPort) { </li>
 <li class='alt'> byte[] buf = new byte[s.length() + 1]; </li>
 <li class=''> s.getBytes(0, s.length(), buf, 0); </li>
 <li class='alt'> return new DatagramPacket(buf, buf.length, destIA, destPort); </li>
 <li class=''> } </li>
 <li class='alt'> </li>
 <li class=''> public static String toString(DatagramPacket p) { </li>
 <li class='alt'> return new String(p.getData(), 0, p.getLength()); </li>
 <li class=''> } </li>
 <li class='alt'>} </li>
</ol>
</div>
</div>
</div>

3 楼 Sharp 2007-10-31

了解!

2 楼 totobacoo 2007-10-31

http://www.iteye.com/topic/89919

老问题了

1 楼 topcloud 2007-10-29

怎么才知道啊，哈哈！

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