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分享如何修正这个错误.菜鸟求教
用户列表 id 用户名 年龄 操作 操作
c c 3 删除 修改
b b 2 删除 修改
a a 1 删除 修改
----------------------------------------------------------------------------
<%@ 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 userinfo where id="+id;
int n = stmt.executeUpdate(sql);
out.println(n);
conn.close();
%>
-------------------------------------------------------------------------
org.apache.jasper.JasperException: Exception in JSP: /doDelete.jsp:6
3: <!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
4: <%@include file="DBconn.jsp" %>
5: <%
6: int id = Integer.parseInt(request.getParameter("id"));
7:
8:
9:
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: "c"
java.lang.NumberFormatException.forInputString(Unknown Source)
java.lang.Integer.parseInt(Unknown Source)
java.lang.Integer.parseInt(Unknown Source)
org.apache.jsp.doDelete_jsp._jspService(doDelete_jsp.java:63)
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.
请问各位大哥line6这句语句要怎么改才能让程序顺利运行啊...
c c 3 删除 修改
b b 2 删除 修改
a a 1 删除 修改
----------------------------------------------------------------------------
<%@ 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 userinfo where id="+id;
int n = stmt.executeUpdate(sql);
out.println(n);
conn.close();
%>
-------------------------------------------------------------------------
org.apache.jasper.JasperException: Exception in JSP: /doDelete.jsp:6
3: <!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
4: <%@include file="DBconn.jsp" %>
5: <%
6: int id = Integer.parseInt(request.getParameter("id"));
7:
8:
9:
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: "c"
java.lang.NumberFormatException.forInputString(Unknown Source)
java.lang.Integer.parseInt(Unknown Source)
java.lang.Integer.parseInt(Unknown Source)
org.apache.jsp.doDelete_jsp._jspService(doDelete_jsp.java:63)
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.
请问各位大哥line6这句语句要怎么改才能让程序顺利运行啊...
...全文
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shilixiu 2007-10-29
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你的id值c是不是int型，执行Integer.parseInt(request.getParameter("id")); 当然会出错了
String id = request.getParameter("id");
Statement stmt = conn.createStatement();
String sql = "delete from userinfo where id='"+id+"'";
int n = stmt.executeUpdate(sql);
out.println(n);
conn.close();
String id = request.getParameter("id");
Statement stmt = conn.createStatement();
String sql = "delete from userinfo where id='"+id+"'";
int n = stmt.executeUpdate(sql);
out.println(n);
conn.close();
llj0004 2007-10-29
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那个"删除"链接, 有个参数是"id"
它应该是数字
它应该是数字
Aosoth 2007-10-28
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为什么会取不到值啊。各位大哥拉小弟一把吧
Aosoth 2007-10-28
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老大们可以给点代码吗.谢谢了.
Aosoth 2007-10-28
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老大们可以给点代码吗.急啊...
Aosoth 2007-10-28
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如果直接用
String id = request.getParameter("id");
是可以取到值但到后面
int n = stmt.executeUpdate(sql);
这里又出错了.
后来我用
try{
id=Integer.parseInt(request.getParameter("id"));
}catch(Exception e){
id=0;}
倒是可以运行了.不过点一个删除.却把表里所有的东西都给删了.太郁闷了...
String id = request.getParameter("id");
是可以取到值但到后面
int n = stmt.executeUpdate(sql);
这里又出错了.
后来我用
try{
id=Integer.parseInt(request.getParameter("id"));
}catch(Exception e){
id=0;}
倒是可以运行了.不过点一个删除.却把表里所有的东西都给删了.太郁闷了...
yztommyhc 2007-10-28
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int id = Integer.parseInt(request.getParameter("id")); //这边有问题.
你先打印以下request.getParameter("id")这个值,会发现是"c" ,注意是字符串,不是单个的字符.
请问对它怎么进行parseInt操作?结果又是什么?
你先打印以下request.getParameter("id")这个值,会发现是"c" ,注意是字符串,不是单个的字符.
请问对它怎么进行parseInt操作?结果又是什么?
