更改java版本后“路径login.page不以“/”字符开头的消息异常

"Path login.page does not start with a "/" character" message exception after changing java version

提问by Anca Z

I have a web application which worked with the following: Tomcat 5.0 Struts 1.0 Java 1.5

我有一个使用以下内容的 Web 应用程序：Tomcat 5.0 Struts 1.0 Java 1.5

I have to switch to Java 1.6.When I tried to do that, i received the following stack of exception in my browser:

我必须切换到 Java 1.6。当我尝试这样做时，我在浏览器中收到以下异常堆栈：

javax.servlet.ServletException: Path login.page does not start with a "/" character
    org.apache.struts.chain.ComposableRequestProcessor.process(ComposableRequestProcessor.java:286)
    org.apache.struts.action.ActionServlet.process(ActionServlet.java:1913)
    org.apache.struts.action.ActionServlet.doGet(ActionServlet.java:449)
    javax.servlet.http.HttpServlet.service(HttpServlet.java:743)
    javax.servlet.http.HttpServlet.service(HttpServlet.java:856)
    org.springframework.web.filter.OncePerRequestFilter.doFilter(OncePerRequestFilter.java:72)
    org.apache.jasper.runtime.PageContextImpl.doForward(PageContextImpl.java:712)
    org.apache.jasper.runtime.PageContextImpl.forward(PageContextImpl.java:682)
    org.apache.jsp.index_jsp._jspService(index_jsp.java:45)
    org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:133)
    javax.servlet.http.HttpServlet.service(HttpServlet.java:856)
    org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:311)
    org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:301)
    org.apache.jasper.servlet.JspServlet.service(JspServlet.java:248)
    javax.servlet.http.HttpServlet.service(HttpServlet.java:856)
    org.springframework.web.filter.CharacterEncodingFilter.doFilterInternal(CharacterEncodingFilter.java:78)
    org.springframework.web.filter.OncePerRequestFilter.doFilter(OncePerRequestFilter.java:77)

root cause

根本原因

java.lang.IllegalArgumentException: Path login.page does not start with a "/" character
    org.apache.struts.chain.commands.servlet.PerformForward.handleAsForward(PerformForward.java:107)
    org.apache.struts.chain.commands.servlet.PerformForward.perform(PerformForward.java:96)
    org.apache.struts.chain.commands.AbstractPerformForward.execute(AbstractPerformForward.java:54)
    org.apache.struts.chain.commands.ActionCommandBase.execute(ActionCommandBase.java:51)
    org.apache.commons.chain.impl.ChainBase.execute(ChainBase.java:190)
    org.apache.commons.chain.generic.LookupCommand.execute(LookupCommand.java:304)
    org.apache.commons.chain.impl.ChainBase.execute(ChainBase.java:190)
    org.apache.struts.chain.ComposableRequestProcessor.process(ComposableRequestProcessor.java:283)
    org.apache.struts.action.ActionServlet.process(ActionServlet.java:1913)
    org.apache.struts.action.ActionServlet.doGet(ActionServlet.java:449)
    javax.servlet.http.HttpServlet.service(HttpServlet.java:743)
    javax.servlet.http.HttpServlet.service(HttpServlet.java:856)
    org.springframework.web.filter.OncePerRequestFilter.doFilter(OncePerRequestFilter.java:72)
    org.apache.jasper.runtime.PageContextImpl.doForward(PageContextImpl.java:712)
    org.apache.jasper.runtime.PageContextImpl.forward(PageContextImpl.java:682)
    org.apache.jsp.index_jsp._jspService(index_jsp.java:45)
    org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:133)
    javax.servlet.http.HttpServlet.service(HttpServlet.java:856)
    org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:311)
    org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:301)
    org.apache.jasper.servlet.JspServlet.service(JspServlet.java:248)
    javax.servlet.http.HttpServlet.service(HttpServlet.java:856)
    org.springframework.web.filter.CharacterEncodingFilter.doFilterInternal(CharacterEncodingFilter.java:78)
    org.springframework.web.filter.OncePerRequestFilter.doFilter(OncePerRequestFilter.java:77)

回答by polygenelubricants

It looks like a request is made to "login.page", which as the exception indicates, does not start with a "/". Try requesting for "/login.page"instead.

看起来像是对 发出的请求"login.page"，正如异常所指示的那样，它不以 开头"/"。尝试请求"/login.page"代替。

Most servlet libraries require that pathnames begins with a "/". For example: javax.servlet.ServletContext#getRequestDispatcher:

大多数 servlet 库要求路径名以"/". 例如：javax.servlet.ServletContext#getRequestDispatcher：

The pathname must begin with a "/"and is interpreted as relative to the current context root. Use getContextto obtain a RequestDispatcherfor resources in foreign contexts.

路径名必须以 a 开头，"/"并被解释为相对于当前上下文根。使用getContext获得RequestDispatcher的国外环境资源。

回答by prathapd

add the tile request processor class and remove the conten type attribute from controller tag

添加 tile 请求处理器类并从控制器标签中删除内容类型属性

<controller processorClass="org.apache.struts.tiles.TilesRequestProcessor" />

<message-resources parameter="EMSMessageResources" null="false" ></message-resources>

<plug-in className="org.apache.struts.tiles.TilesPlugin" >
    <set-property property="definitions-config" value="/WEB-INF/ems-tiles-defs.xml" />
    <set-property property="moduleAware" value="true" />
</plug-in>

回答by Mordrekai

Check this page.

检查此页面。

http://wiki.apache.org/struts/StrutsUpgradeNotes12to13

http://wiki.apache.org/struts/StrutsUpgradeNotes12to13

Obviously you're using stuts tiles. If you're using Struts 1.3, you should add this to your web.xml file:

显然你正在使用 stuts 瓷砖。如果您使用的是 Struts 1.3，您应该将其添加到您的 web.xml 文件中：

<init-param>
    <param-name>chainConfig</param-name>
    <param-value>org/apache/struts/tiles/chain-config.xml</param-value>
</init-param>

回答by Ajay

Solved

解决了

Whenever new Jsp file is added new entry goes in respective

每当添加新的 Jsp 文件时，相应的新条目就会进入

"struts-config_* .xml" and "tiles-defs_* .xml" files.

“ struts-config_* .xml”和“ tiles-defs_* .xml”文件。