当前位置: 代码迷 >> Java Web开发 >> if(rs.next()),每次提示都是java.lang.NullPointerException. System.out.println(rs1);输出为null.为什么呀.多谢.新年快乐
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if(rs.next()),每次提示都是java.lang.NullPointerException. System.out.println(rs1);输出为null.为什么呀.多谢.新年快乐

热度:802   发布时间:2016-04-17 12:29:32.0
if(rs.next()),每次提示都是java.lang.NullPointerException. System.out.println(rs1);输出为null.为什么呀.谢谢.新年快乐!
userName=new String(request.getParameter("userName").getBytes("8859_1"));
userPassword=new String(request.getParameter("userPassword").getBytes("8859_1"));
userEmail=request.getParameter("userEmail");
sql2="insert into user(userName,userPassword,userEmail) values('"+ userName +"','"+ userPassword +"','"+ userEmail +"')";
rs1 = registerBean.executeUpdate(sql2);
System.out.println(rs1);
if(!rs1.next()){

rs1.close();
try{
registerBean.closeStmt();
registerBean.closeConn();
msg="the user is existed";
session.putValue("register_message",msg);

------解决方案--------------------
Java code
registerBean.executeUpdate(sql2);
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