根据 String DateTime 对 List<T> 进行排序

Question

我想根据字符串形式的UTC DateTime对列表进行降序排序。

我的班级

    data class CartEntity( val itemId: String,  var itemName: String, var createdDate: String)

在这个 createdDate 中是 "2020-07-28T14:28:52.877Z"

我尝试过的

    const val UTC_FORMAT = "yyyy-MM-dd'T'HH:mm:ss.SSS'Z'"

    Collections.sort(list, object : Comparator<CartEntity> {
        var f: DateFormat =
            SimpleDateFormat(
                AppConstants.UTC_FORMAT, Locale.ENGLISH
            )
    
        override fun compare(o1: CartEntity?, o2: CartEntity?): Int {
            return try {
                val firstitem = f.parse(o1?.createdDate!!)
                val seconditem = f.parse(o2?.createdDate!!)
                firstitem.compareTo(seconditem)
            } catch (e: ParseException) {
                throw IllegalArgumentException(e)
            }
        }
    
    })

但仍然按降序排序并没有按预期工作

Answer 1

这可以使用以下方法来完成。

import java.time.ZonedDateTime;
import java.util.Arrays;
import java.util.Comparator;
import java.util.List;
import java.util.stream.Collectors;

public class Tester {

    static class Car {

        private String date;

        String getDate() {
            return this.date;
        }

        void setDate(String date) {
            this.date = date;
        }

        @Override
        public String toString() {
            return date;
        }

        // Imagine if you keep date not as strings
        ZonedDateTime toDateTime(){
            return ZonedDateTime.parse(this.date);
        }
    }

    public static void main(String[] args) {

        Car first = new Car();
        first.setDate("2020-07-28T14:28:52.877Z");

        Car second = new Car();
        second.setDate("2010-07-28T14:28:52.877Z");

        Car third = new Car();
        third.setDate("2021-07-28T14:28:52.877Z");

        List<Car> cars = Arrays.asList(first, second, third);

        List<Car> sorted = cars.stream().sorted(Comparator.nullsLast(
                (current, next) -> {
                    ZonedDateTime currentDate = ZonedDateTime.parse(current.getDate());
                    ZonedDateTime nextDate = ZonedDateTime.parse(next.getDate());

                    return nextDate.compareTo(currentDate);
                })).collect(Collectors.toList());
    }
}

如果您将日期保留为常规 Date 对象，则使用它们会更容易，并且上面的代码可以简化一点。

List<Car> sorted = cars.stream()
                               .sorted(Comparator.comparing(Car::toDateTime, Comparator.nullsLast(Comparator.reverseOrder())))
                               .collect(Collectors.toList());

Answer 2

你的代码工作得很好@Charwaka。

data class CartEntity(val itemId: String, var itemName: String, var createdDate: String)
const val UTC_FORMAT = "yyyy-MM-dd'T'HH:mm:ss.SSS'Z'"
fun main() {
    val list = mutableListOf<CartEntity>().apply {
        add(CartEntity(itemId = "", itemName = "", createdDate = "2020-07-28T14:28:52.877Z"))
        add(CartEntity(itemId = "", itemName = "", createdDate = "2020-09-28T14:28:52.877Z"))
        add(CartEntity(itemId = "", itemName = "", createdDate = "2020-08-28T14:28:52.877Z"))
        add(CartEntity(itemId = "", itemName = "", createdDate = "2020-04-28T14:28:52.877Z"))
    }
    Collections.sort(list, object : Comparator<CartEntity> {
        var f: DateFormat = SimpleDateFormat(UTC_FORMAT, Locale.ENGLISH)

        override fun compare(o1: CartEntity, o2: CartEntity): Int {
            return try {
                val firstitem = f.parse(o1.createdDate)
                val seconditem = f.parse(o2.createdDate)
                seconditem.compareTo(firstitem)
            } catch (e: ParseException) {
                throw IllegalArgumentException(e)
            }
        }
    })
    list
}

输出：

CartEntity(itemId=, itemName=, createdDate=2020-09-28T14:28:52.877Z), 
CartEntity(itemId=, itemName=, createdDate=2020-08-28T14:28:52.877Z), 
CartEntity(itemId=, itemName=, createdDate=2020-07-28T14:28:52.877Z), 
CartEntity(itemId=, itemName=, createdDate=2020-04-28T14:28:52.877Z)

@Animesh Sahu 的代码也可以使用！

val 格式：DateFormat = SimpleDateFormat(AppConstants.UTC_FORMAT, Locale.ENGLISH) list.sortByDescending { format.parse(it.createdDate) }

Answer 3

在 kotlin 风格中，您可以使用标准库函数（以下习语）：

从您要排序的列表中创建一个新的排序列表：

fun main() {
    val list = listOf<CartEntity>(
        CartEntity(itemId = "", itemName = "", createdDate = "2020-07-28T14:28:52.877Z"),
        CartEntity(itemId = "", itemName = "", createdDate = "2020-09-28T14:28:52.877Z"),
        CartEntity(itemId = "", itemName = "", createdDate = "2020-08-28T14:28:52.877Z"),
        CartEntity(itemId = "", itemName = "", createdDate = "2020-04-28T14:28:52.877Z"),
    )

    val format: DateFormat = SimpleDateFormat(AppConstants.UTC_FORMAT, Locale.ENGLISH)

    val sortedList = list.sortedByDescending { format.parse(it.createdDate) }

    println(sortedList)  // `sortedList` is sorted out list, the `list` is holding the original order
}

对原始列表进行排序（列表应该是可变的）：

fun main() {
    val list = mutableListOf<CartEntity>(
        CartEntity(itemId = "", itemName = "", createdDate = "2020-07-28T14:28:52.877Z"),
        CartEntity(itemId = "", itemName = "", createdDate = "2020-09-28T14:28:52.877Z"),
        CartEntity(itemId = "", itemName = "", createdDate = "2020-08-28T14:28:52.877Z"),
        CartEntity(itemId = "", itemName = "", createdDate = "2020-04-28T14:28:52.877Z"),
    )

    val format: DateFormat = SimpleDateFormat(AppConstants.UTC_FORMAT, Locale.ENGLISH)
    list.sortByDescending { format.parse(it.createdDate) }

    println(list)  // `list` is sorted out list
}