简体   繁体   English

java.sql.SQLException:mysql中结果集错误结束后

[英]java.sql.SQLException: After end of result set error in mysql

I tried to read the data from database and store the ResultSet in ArrayList . 我试图从数据库中读取数据并将ResultSet存储在ArrayList However, I receive the error java.sql.SQLException: After end of result set when I print the arraylist. 但是,我收到错误java.sql.SQLException: After end of result set打印数组列表时java.sql.SQLException: After end of result set I have checked that I don't have call my result.next() twice according to this link . 我检查了是否没有根据此链接两次调用result.next() Anyone have better way to store the result set? 有人有更好的方法来存储结果集吗? Then I can print it out at the location I want. 然后,我可以在所需的位置将其打印出来。

Statement stmt=con.createStatement();  
            ResultSet rsVehicle=stmt.executeQuery("select * from vehicle");  
            while(rsVehicle.next()) {  
                vehicleList.add(rsVehicle);
            }
            System.out.println(vehicleList.size());

            for(int i=0;i<vehicleList.size();i++) {
                System.out.println(vehicleList.get(i).getInt(1)+"  "+vehicleList.get(i).getString(2)+"  "+vehicleList.get(i).getString(3)+"  "+
                        vehicleList.get(i).getString(4)+"  "+vehicleList.get(i).getString(5)+"  "+vehicleList.get(i).getString(6)+"  "+
                        vehicleList.get(i).getString(7)+"  "+vehicleList.get(i).getString(8)+"  "+vehicleList.get(i).getString(9)+"  "+
                        vehicleList.get(i).getString(10)+"  "+vehicleList.get(i).getString(11)+"  "+vehicleList.get(i).getString(12)+"  "+
                        vehicleList.get(i).getString(13)+"  "+vehicleList.get(i).getString(14)+"  "+vehicleList.get(i).getString(15));
            }

            PreparedStatement ps;
            ps =con.prepareStatement("SELECT * from job WHERE date = ?");
            ps.setString(1, date);
            ResultSet rsJob=ps.executeQuery();  
            while(rsJob.next()) {  
                jobList.add(rsJob);
            }
            System.out.println(jobList.size());

            for(int i=0;i<jobList.size();i++) {
                System.out.println(jobList.get(i).getInt(1)+"  "+jobList.get(i).getString(2)+"  "+jobList.get(i).getString(3)+"  "+
                        jobList.get(i).getString(4)+"  "+jobList.get(i).getString(5)+"  "+jobList.get(i).getString(6)+"  "+
                        jobList.get(i).getString(7)+"  "+jobList.get(i).getString(8)+"  "+jobList.get(i).getString(9)+"  "+
                        jobList.get(i).getString(10)+"  "+jobList.get(i).getString(11)+"  "+jobList.get(i).getString(12)+"  "+
                        jobList.get(i).getString(13)+"  "+jobList.get(i).getString(14)+"  "+jobList.get(i).getString(15));
            }

In your code you add the result set to a list and then you iterate over that list to get the data, but the correct flow is: 在代码中,将结果集添加到列表中,然后遍历该列表以获取数据,但是正确的流程是:

  • Extract data from the result set and put them in a list. 从结果集中提取数据并将其放在列表中。
  • After you can iterate the list to print the results. 之后,您可以遍历列表以打印结果。

The code should be something similar to: 该代码应类似于:

// Create a new list of vehicles
List<Vehicle> vehicleList = new ArrayList<>();
Statement stmt=con.createStatement();  
ResultSet rsVehicle=stmt.executeQuery("select * from vehicle");  
while(rsVehicle.next()) {  
    // Create a new vehicle for each record returned from the query
    Vehicle vehicle = new Vehicle();

    // Set to the vehicle each property taken from that record
    vehicle.setBrand(rsVehicle.getString(1));
    ...

    // Add that vehicle to the list
    vehicleList.add(vehicle);
}

...

// Iterate the list and print each field you need
for (Vehicle vehicle: vehicleList) {
   // PRint each vehicle
   System.out.println("Vechicle brand: " + vehicle.getBrand());
}

Note: If you redefine the method toString of the class Vehicle you can call directly System.out.println(vehicle); 注意:如果您将Vehicle类的方法重新定义为toString ,则可以直接调用System.out.println(vehicle);

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM