無法檢查已有用戶名

-1

import java.io.IOException; 
import java.io.PrintWriter; 
import java.sql.Connection; 
import java.sql.PreparedStatement; 
import java.sql.ResultSet; 
import java.sql.SQLException; 
import java.sql.Statement; 

import javax.servlet.RequestDispatcher; 
import javax.servlet.ServletException; 
import javax.servlet.annotation.WebServlet; 
import javax.servlet.http.HttpServlet; 
import javax.servlet.http.HttpServletRequest; 
import javax.servlet.http.HttpServletResponse; 

@WebServlet(name = "SimpleServlet", urlPatterns = { "/SimpleServlet" }) 
public class SimpleServlet extends HttpServlet { 

    private static final long serialVersionUID = 1L; 

    public void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { 

     response.setContentType("text/html"); 
     PrintWriter out = response.getWriter(); 

     String username = request.getParameter("username").toString(); 
     String password = request.getParameter("password").toString(); 
     String repassword = request.getParameter("repassword").toString(); 
     String ph = request.getParameter("phone").toString(); 
     double phone = Double.parseDouble(ph); 

     String address = request.getParameter("address").toString(); 

     int status = register(username, password, repassword, phone, address); 

     if (status > 0) { 

      RequestDispatcher rd = request.getRequestDispatcher("Register_success.jsp"); 
      rd.include(request, response); 
     } else { 
      RequestDispatcher rd = request.getRequestDispatcher("Register_fail.jsp"); 
      rd.include(request, response); 
     } 

     out.close(); 
    } 

    static int status = 0; 

    @SuppressWarnings("null") 
    public static int register(String username, String password, String repassword, double phone, String address) { 

     PreparedStatement ps; 
     Connection con; 
     con = GetCon.getCon(); 
     try { 
      Statement st = con.createStatement(); 
      ResultSet rs = st.executeQuery("SELECT username FROM newuser where username='" + username + "'"); 

      while(rs.next()) 
      if (rs.getString("username").equalsIgnoreCase(username)) { 
       System.out.println(username); 
       HttpServletResponse response = null; 
       response.sendRedirect("UserAlreadyExist.jsp"); 
      } else { 

       ps = con.prepareStatement("Insert into newuser values(?,?,?,?,?)"); 
       ps.setString(1, username); 
       ps.setString(2, password); 
       ps.setString(3, repassword); 
       ps.setDouble(4, phone); 
       ps.setString(5, address); 

       status = ps.executeUpdate(); 
      } 
     } catch (SQLException e) { 

      e.printStackTrace(); 
     } catch (IOException e) { 

      e.printStackTrace(); 
     } 
     return status; 

    } 
}

您好，我在執行上述代碼時遇到問題。當我們輸入現有的用戶名時，它會拋出一個NullPointerException異常。請有人幫我找到錯誤。該代碼可以使用新用戶名，但與現有用戶無關。謝謝！無法檢查已有用戶名

來源

2016-01-04 pushkar zagade

閱讀堆棧跟蹤，並使用調試器。 YOu應該能夠找到引發異常的確切線路，然後解決問題 – jhamon

您能否指出您面臨問題的線路？ –

建議：'HttpServletResponse的響應= NULL; response.sendRedirect是（「UserAlreadyExist.jsp」）;'響應爲空 – jhamon

的response是你register方法

HttpServletResponse response = null; 
response.sendRedirect("UserAlreadyExist.jsp");

使用來自doPost(HttpServletRequest request, HttpServletResponse response)的response對象，以便將用戶重定向空。

您將需要一個參數添加到register方法。

你的IDE顯示警告幫助你。不要使用@SuppressWarnings("null")

來源

2016-01-04 09:42:50 jhamon

無法檢查已有用戶名

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