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JSP里web.xml实现错误处理页面的制作404,和500

 2009-02-08 16:35:44 来源:WEB开发网   
核心提示:web.xml文件增加如下代码查看复制到剪切板打印<error-page> <error-code>404</error-code> <location>/building.jsp</location> </error-page> <

web.xml文件增加如下代码查看复制到剪切板打印
<error-page> 
     <error-code>404</error-code> 
     <location>/building.jsp</location> 
</error-page> 
<error-page> 
     <error-code>500</error-code> 
     <location>/error.jsp</location> 
</error-page> 

   <error-page>
         <error-code>404</error-code>
         <location>/building.jsp</location>
     </error-page>
     <error-page>
         <error-code>500</error-code>
         <location>/error.jsp</location>
     </error-page>building.jsp查看复制到剪切板打印
<%@ page language="java" contentType="text/html; charset=GBK" isErrorPage="true" pageEncoding="GBK"%> 
<% 
  response.setStatus(HttpServletResponse.SC_OK); 
%> 
对不起,您请求的页面没有找到! 

<%@ page language="java" contentType="text/html; charset=GBK" isErrorPage="true" pageEncoding="GBK"%>
<%
  response.setStatus(HttpServletResponse.SC_OK);
%>
对不起,您请求的页面没有找到!error.jsp查看复制到剪切板打印
<%@ page language="java" contentType="text/html; charset=GBK" isErrorPage="true" pageEncoding="GBK"%> 
<%@ page import="java.io.*,java.util.*"%> 
<%response.setStatus(HttpServletResponse.SC_OK); 
 
    %> 
<body> 
程序发生了错误,有可能该页面正在调试或者是设计上的缺陷. 
 
你可以选择 
 <a href=<%=request.getContextPath()+"/forum/new.jsp" %>>反馈</a> 
提醒我... 或者 
<a href="javascript:history.go(-1)">返回上一页</a> 
<hr width=80%> 
<h2><font color=#DB1260>JSP Error Page</font></h2> 
 
<p>An exception was thrown: <b> <%=exception.getClass()%>:<%=exception.getMessage()%></b></p> 
<% 
System.out.PRintln("Header...."); 
Enumeration<String> e = request.getHeaderNames(); 
String key; 
while(e.hasMoreElements()){ 
  key = e.nextElement(); 
  System.out.println(key+"="+request.getHeader(key)); 
} 
System.out.println("Attribute...."); 
e = request.getAttributeNames(); 
while(e.hasMoreElements()){ 
  key = e.nextElement(); 
  System.out.println(key+"="+request.getAttribute(key)); 
} 
 
System.out.println("arameter...."); 
e = request.getParameterNames(); 
while(e.hasMoreElements()){ 
  key = e.nextElement(); 
  System.out.println(key+"="+request.getParameter(key)); 
} 
%> 
111<%=request.getAttribute("javax.servlet.forward.request_uri") %> 
 
<%=request.getAttribute("javax.servlet.forward.servlet_path") %> 
 
<p>With the following stack trace:</p> 
<pre> 
<%exception.printStackTrace(); 
    ByteArrayOutputStream ostr = new ByteArrayOutputStream(); 
    exception.printStackTrace(new PrintStream(ostr)); 
    out.print(ostr); 
   %> 
</pre> 
<hr width=80%> 
</body> 

Tags:JSP web xml

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