如何對LocalDateTime進行排序並獲取最大值和最小值？

任何人都可以共享邏輯來排序LocalDateTime並從域對象列表中獲取最大和最小記錄。如何對LocalDateTime進行排序並獲取最大值和最小值？

域對象 -

private int employeeNumber; 
private LocalDateTime updatedDate;

第一個記錄

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

最大輸出：2,2017-10-06 23：25：37 + 0000

敏輸出：1,2016-07-09 00：00：00 + 0000

來源

2017-10-09 Vigneshwaran

假設您的域對象名稱爲「員工」並且具有List<Employee>作爲域對象列表。然後，您可以按如下方式通過比較器來使用Java 8排序功能。隨後，您將根據需要提取Min和Max。

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

查找以下代碼示例以供參考。

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); 
    } 
}

來源

2017-10-09 03:03:48 Damith

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

// Converting the hashset to streams and producing min and max out of it 
System.out.println("Min: " + dates.stream().min(LocalDateTime::compareTo)); 
System.out.println("Max: " + dates.stream().max(LocalDateTime::compareTo));

輸出：

最小：2016-07-09T00：00
最大：2017-10-06T23：25：37

你可以很容易適應這個核心邏輯現在你的情況。

來源

2017-10-09 02:33:00 everton

如何對LocalDateTime進行排序並獲取最大值和最小值？

回答

相關問題