一个用JSP+mysql建立用户登陆,查询.删除等功能的程序.各位高手来帮忙看下吧

<%@ page language="java" contentType="text/html; charset=UTF-8"
    pageEncoding="UTF-8"%>
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<%@include file="DBconn.jsp" %>
<%
int id = Integer.parseInt(request.getParameter("id"));//取得连接状态对象
Statement stmt = conn.createStatement();
String sql = "delete from useradmin where userid="+id;
int n = stmt.executeUpdate(sql);
out.println(n);
conn.close();
%>
-----------------------------
HTTP Status 500 - --------------------------------------------------------------------------------type Exception reportmessage description The server encountered an internal error () that prevented it from fulfilling this request.exception org.apache.jasper.JasperException: Exception in JSP: /userEdit.jsp:74: <%@include file="DBconn.jsp" %>
5: <%//取得连接状态对象
6: Statement stmt = conn.createStatement();
7: int id = Integer.parseInt(request.getParameter("id"));
8: //取得查询的结果集
9: ResultSet rs = stmt.executeQuery("SELECT * from useradmin where userid="+id);
10: rs.next();
Stacktrace:
org.apache.jasper.servlet.JspServletWrapper.handleJspException(JspServletWrapper.java:506)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:395)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:314)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:264)
javax.servlet.http.HttpServlet.service(HttpServlet.java:802)
root cause java.lang.NumberFormatException: For input string: "aaa"
java.lang.NumberFormatException.forInputString(Unknown Source)
java.lang.Integer.parseInt(Unknown Source)
java.lang.Integer.parseInt(Unknown Source)
org.apache.jsp.userEdit_jsp._jspService(userEdit_jsp.java:65)
org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:97)
javax.servlet.http.HttpServlet.service(HttpServlet.java:802)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:334)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:314)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:264)
javax.servlet.http.HttpServlet.service(HttpServlet.java:802)
note The full stack trace of the root cause is available in the Apache Tomcat/5.5.20 logs.报错.显示第七行也是就7: int id = Integer.parseInt(request.getParameter("id"));这句语句出错.可我弄了半天也搞不定.求高手指点.还有修改资料的代码
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"
"http://www.w3.org/TR/html4/loose.dtd">
<%@page  language="java" pageEncoding="gb2312"%>
<%@include file="DBconn.jsp" %>
<%//取得连接状态对象
Statement stmt = conn.createStatement();
int id = Integer.parseInt(request.getParameter("id"));
//取得查询的结果集
ResultSet rs = stmt.executeQuery("SELECT * from useradmin where userid="+id);
rs.next();
%>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=gb2312">
<title>无标题文档</title>
<style type="text/css">

</style>
</head><body>
<form name="form1" method="post" action="doEdit.jsp">
  <ul align="center">
    <div align="center">用户修改</div>
  </ul>
  <p align="center">帐号：
    <input name="username" type="text" id="username" value="<%=rs.getObject("userid") %>">
</p>
  <p align="center">密码：
    <input name="password" type="password" id="password" value="<%=rs.getObject("password") %>">
</p>
  <p align="center">年龄：
    <input name="age" type="text" id="age" value="<%=rs.getObject("age") %>">
  </p>
  <p align="center">专业：
    <input name="speciablity" type="text" id="speciablity" value="<%=rs.getObject("speciablity") %>">
</p>
  <p align="center">
  <input type="hidden" name="id" value="<%=id%>">
    <input type="submit" name="Submit" value="提交">
           <input type="submit" name="Submit2" value="重置">
</p>
  </form>
</body>
</html>
<%conn.close(); %>
运行结果也是报同一个错
HTTP Status 500 - --------------------------------------------------------------------------------type Exception reportmessage description The server encountered an internal error () that prevented it from fulfilling this request.exception org.apache.jasper.JasperException: Exception in JSP: /userEdit.jsp:74: <%@include file="DBconn.jsp" %>
5: <%//取得连接状态对象
6: Statement stmt = conn.createStatement();
7: int id = Integer.parseInt(request.getParameter("id"));
8: //取得查询的结果集
9: ResultSet rs = stmt.executeQuery("SELECT * from useradmin where userid="+id);
10: rs.next();
Stacktrace:
org.apache.jasper.servlet.JspServletWrapper.handleJspException(JspServletWrapper.java:506)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:395)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:314)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:264)
javax.servlet.http.HttpServlet.service(HttpServlet.java:802)
root cause java.lang.NumberFormatException: null
java.lang.Integer.parseInt(Unknown Source)
java.lang.Integer.parseInt(Unknown Source)
org.apache.jsp.userEdit_jsp._jspService(userEdit_jsp.java:65)
org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:97)
javax.servlet.http.HttpServlet.service(HttpServlet.java:802)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:334)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:314)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:264)
javax.servlet.http.HttpServlet.service(HttpServlet.java:802)
note The full stack trace of the root cause is available in the Apache Tomcat/5.5.20 logs.
--------------------------------------------------------------------------------
求高手帮忙啊.

解决方案 »

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你的id是“aaa”, 没办法转成 int .
int id = Integer.parseInt(request.getParameter("id"));
在你的程序里面会执行
int id = Integer.parseInt(“aaa”);  //因为取到的是“aaa”,这个是不能转成int的，所以报错Stacktrace:
org.apache.jasper.servlet.JspServletWrapper.handleJspException(JspServletWrapper.java:506)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:395)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:314)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:264)
javax.servlet.http.HttpServlet.service(HttpServlet.java:802)
root cause java.lang.NumberFormatException: For input string: "aaa"
试着自己分析异常信息。祝进步！