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Java檢查值是否存在於數據庫中

[英]Java checking whether the value exists in database

你好,我正在研究Java和MySQL。 我的實現如下:

import javax.swing.JFrame;
import javax.swing.JLabel;
import java.sql.*;
import java.util.Scanner;

public class BankingSystem extends JFrame {
    public static void main(String[] args) throws Exception{
        int ur=0;
        int PIN;
        String ID;
        Scanner s=new Scanner(System.in);
        Class.forName("com.mysql.jdbc.Driver");
        String url = "jdbc:mysql://localhost:3306/BankingSystem";
        String user = "root";
        String pass="";
        Connection con = DriverManager.getConnection(url,user,pass);
        Statement st = con.createStatement();
        System.out.println("Enter Your 4 digit PIN");
        PIN=s.nextInt();

        ResultSet rs=st.executeQuery("select * from customerinformation where pin ="+PIN);
//checking for existance of user entered pin
if((rs.getString(1)).equals("")){System.out.println("Invalid PIN");}


        while(rs.next())
        {

            System.out.println(rs.getString(1)+"  "+rs.getString(2)+"  "+rs.getString(3)+"  "+rs.getString(4));

        }

    }

}

但是此代碼不起作用。 給出某種異常錯誤,但是當我刪除包含if語句的行時,它工作正常。如何檢查引腳是否有效。

它給出以下異常:

Exception in thread "main" java.sql.SQLException: Before start of result set
        at com.mysql.jdbc.ResultSet.checkRowPos(ResultSet.java:3624)
        at com.mysql.jdbc.ResultSet.getString(ResultSet.java:1762)
        at BankingSystem.main(BankingSystem.java:22)

嘗試這種方式替換此Statement st = con.createStatement();

String query ="select * from customerinformation where pin =?"
PreparedStatement st =con.prepareStatement("query");
st.setInt(1,PIN);
ResultSet resultSet = st.executeQuery();

您永遠不要在PreparedStatement上使用Statement

if (!resultSet.next() ) {
    System.out.println("resultset does not data");
} else {

    do {
        System.out.println(rs.getString(1)+"  "+
                           rs.getString(2)+"  "+
                          rs.getString(3)+"  "+
                          rs.getString(4));


    } while (resultSet.next());
}
if((rs.getString(1)).equals("")){
   System.out.println("Invalid PIN");
}

沒有rs.next() ,則不可能rs.getString(1)

嘗試這個:

    if(rs.next()){
        System.out.println(rs.getString(1)+"  "+rs.getString(2)+" "+rs.getString(3)+"  "+rs.getString(4));
    } else {
     System.out.println("Invalid PIN");   
    }

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