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J里web.xml实现错误处理页面的制作404,和500

作者：黑客防线网安JSP教程基地来源：黑客防线网安JSP教程基地浏览次数：0

黑客防线网安网讯：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> ...

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>