java - 如何使用 Scanner user_input ResultSet 中的整数变量在 Java 中选择特定的 MySQL 行？

Question

我已经在 MySQL 中设置了表，并相信它们都已正确连接，我只是想显示特定信息。我要求用户输入并使用该整数值作为 result.absolute() 的对象。每当我运行 timeresult 输入时，我都会得到“实际时间”的错误行。是java代码还是某些MySQL设置导致显示不正确的行？ println 中的所有其他信息都是正确的。

public static void main(String[] args) throws Exception {      

    Scanner user_input = new Scanner( System.in );

    String dayinput;
    System.out.print("Choose A Day (1-7): ");
    dayinput = user_input.next();

    //Accessing driver from the JAR file
    Class.forName("com.mysql.jdbc.Driver");

    //Creating a variable for the connection called "con"
    Connection con = DriverManager.getConnection("jdbc:mysql://localhost:3306/axamschedule","root","password");
    //jdbc:mysql://localhost:3306/shows-----> This is the database
    //root is the database username
    //password is the database password

    System.out.println("Connected To Database");

    //Creating query for Sundays Table
    if (dayinput.equals("1")){
            PreparedStatement statement = con.prepareStatement("select * FROM suntimes, days, shows WHERE (days.idday = suntimes.idday and suntimes.idshow = shows.idshow)");

            //Creating a variable to execute query
            ResultSet result = statement.executeQuery();

            System.out.println("---------------------");

            Integer timeinput;
            System.out.print("Choose A Timeslot (1-24): ");
            timeinput = 0;
            timeinput = user_input.nextInt();


            if(result.absolute(timeinput)){

                    System.out.println("Time: " + result.getString("actualtime"));
                    System.out.println("Day: " + result.getString("day"));
                    System.out.println("Title: " + result.getString("title"));
                    System.out.println("Length: " + result.getString("length"));
                    System.out.println("Host: " + result.getString("author"));
            }                      
    }

     user_input.close();
   else{
         System.out.println("No Info");
   }

Answer 1

您获得的结果集未排序。它们返回的顺序由我的 mysql 决定。

尝试

select * FROM suntimes, days, shows WHERE (days.idday = suntimes.idday and suntimes.idshow = shows.idshow) order by actualtime;