Question

任何人都可以共享逻辑来对LocalDateTime进行排序，并从域对象列表中获取最大和最小记录。

域对象 -

private int employeeNumber;
private LocalDateTime updatedDate;

第一记录

1.1,2016-07-09 00:00:00+0000
2.2,2017-10-06 23:25:37+0000

最大输出：2,2017-10-06 23:25:37 + 0000

最低产出：1,2016-07-09 00：00：00 + 0000

Answer 1

您的简化数据集

sql = "UPDATE `Camper` SET `amountPaid` = ? where id like ?";

ps = conn.prepareStatement(sql);
ps.setInt(1, newAmountPaid);
ps.setString(2, playerId + "%");

ps.executeUpdate();

将hashset转换为stream并从中生成min和max

DateTimeFormatter fmt = DateTimeFormatter.ofPattern("yyyy-MM-dd HH:mm:ssxx");
dates.add(LocalDateTime.parse("2016-07-09 00:00:00+0000", fmt));
dates.add(LocalDateTime.parse("2017-10-06 23:25:37+0000", fmt));

输出：

Min：2016-07-09T00：00
Max：2017-10-06T23：25：37

您现在可以轻松地将此核心逻辑调整到您的方案中。

Answer 2

假设您的域对象名称为“Employee”，并将List<Employee>作为域对象列表。然后，您可以通过传递Comparator来使用Java 8排序功能，如下所示。随后，您可以根据需要提取Min和Max。

employeeList.sort((Employee e1, Employee e2) -> e1.getUpdatedDate().compareTo(e2.getUpdatedDate()));

找到以下代码示例供您参考。

import java.time.LocalDateTime;
import java.time.format.DateTimeFormatter;
import java.util.Arrays;
import java.util.List;

public class Main {
    static DateTimeFormatter format = DateTimeFormatter.ofPattern("yyyy-MM-dd HH:mm:ssxx");

    static Employee emp1 = new Employee(1, LocalDateTime.parse("2017-10-06 23:25:37+0000", format));
    static Employee emp2 = new Employee(2, LocalDateTime.parse("2016-07-09 00:00:00+0000", format));
    static Employee emp3 = new Employee(3, LocalDateTime.parse("2017-10-06 23:25:38+0000", format));
    static Employee emp4 = new Employee(4, LocalDateTime.parse("2016-07-09 00:00:00+0000", format));

    static List<Employee> employeeList = Arrays.asList(new Employee[] { emp1, emp2, emp3, emp4 });

    public static void main(String[] args) {
        employeeList.sort((Employee e1, Employee e2) -> e1.getUpdatedDate().compareTo(e2.getUpdatedDate()));

        System.out.println("Max:" + employeeList.get(employeeList.size() - 1));
        System.out.println("Min:" + employeeList.get(0));

        System.out.println("==== Complete list ===");
        employeeList.forEach(e -> System.out.println(e));

    }
}

class Employee {
    private int employeeNumber;
    private LocalDateTime updatedDate;

    public Employee() {
    }

    public Employee(int employeeNumber, LocalDateTime updatedDate) {
        this.employeeNumber = employeeNumber;
        this.updatedDate = updatedDate;
    }

    public int getEmployeeNumber() {
        return employeeNumber;
    }

    public void setEmployeeNumber(int employeeNumber) {
        this.employeeNumber = employeeNumber;
    }

    public LocalDateTime getUpdatedDate() {
        return updatedDate;
    }

    public void setUpdatedDate(LocalDateTime updatedDate) {
        this.updatedDate = updatedDate;
    }

    public String toString() {
        return String.format("Employee Number: %s, Updated Date: %s", this.employeeNumber, this.updatedDate);
    }
}

如何对LocalDateTime进行排序并获得最大值和最小值？

2 个答案: