servlet2.3新特性之错误处理的问题(鸟人我真的没钱了,惭愧!)

Question

我把下面这段XML拷到web.xml里面就出错!
<error-page>
        <error-code>404</error-code>
        <location>/404.html</location>
    </error-page>    <error-page>
        <exception-type>javax.servlet.ServletException</exception-type>
        <location>/servlet/ErrorDisplay</location>
    </error-page>出错信息是:
... 组件“web-app”在此不允许“error-page”...
那我该怎么处理servlet的错误呀,!我需要把servlet的错误定向一个出错页面呐,!高手快指点!
天气真闷热:<

Answer 1

你的错误在:<location>/servlet/ErrorDisplay</location>!!!
如果你的ErrorDisplay的全名是com.sample.error.ErrorDisplay
那么你在Web.xml中是这样的:
  <servlet>
    <servlet-name>errordisplay</servlet-name>
    <servlet-class>com.sample.error.ErrorDisplay</servlet-class>
  </servlet>......
.....
  <servlet-mapping>
    <servlet-name>errordisplay</servlet-name>
    <url-pattern>/servlet/errordisplay</url-pattern>
  </servlet-mapping>
....
最后是:
    <error-page>
        <exception-type>javax.servlet.ServletException</exception-type>
        <location>/servlet/errordisplay</location>
    </error-page>