使用java8的排序列表

我有一个类似于

ArrayList

---

您可以使用Java8流来实现如下所示的相同功能

List<String> sortedList = list.stream().sorted().collect(Collectors.toList());

List-sortedList=List.stream（）.sorted（）.collect（Collectors.toList（））；

用于创建所需的模式，以便可以将日期字符串解析为

用于比较解析的日期字符串上的

DateFrequency

对象

演示：

import java.time.LocalDate;
import java.time.format.DateTimeFormatter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.List;

class DateFrequency {
    private String date;
    private String frequency;

    public DateFrequency(String date, String frequency) {
        this.date = date;
        this.frequency = frequency;
    }

    public String getDate() {
        return date;
    }

    @Override
    public String toString() {
        return "DateFrequency [date=" + date + ", frequency=" + frequency + "]";
    }
}

public class Main {
    public static void main(String[] args) {
        DateTimeFormatter formatter = DateTimeFormatter.ofPattern("dd/MM/yyyy");

        List<DateFrequency> list = new ArrayList<DateFrequency>();
        list.add(new DateFrequency("05/10/2020", "60-DAYS"));
        list.add(new DateFrequency("05/10/2020", "30-DAYS"));
        list.add(new DateFrequency("05/11/2020", "30-DAYS"));
        list.add(new DateFrequency("05/12/2020", "60-DAYS"));
        list.add(new DateFrequency("05/11/2020", "90-DAYS"));

        // Sort on dates in ascending order
        Collections.sort(list, Comparator.comparing((df) -> LocalDate.parse(df.getDate(), formatter)));

        // Display the sorted result
        for (DateFrequency df : list) {
            System.out.println(df);
        }
    }
}

DateFrequency [date=05/10/2020, frequency=60-DAYS]
DateFrequency [date=05/10/2020, frequency=30-DAYS]
DateFrequency [date=05/11/2020, frequency=30-DAYS]
DateFrequency [date=05/11/2020, frequency=90-DAYS]
DateFrequency [date=05/12/2020, frequency=60-DAYS]

假设：

日期字符串的格式为

dd/MM/yyyy

。如果没有，请在对象中相应地更改它，

formatter

仅在日期按升序排序时才需要排序

---

尝试使用比较器。解决方案取决于

DateFrequency

是什么。这是已经在5个地方解释过的非常基本的东西。在询问之前进行搜索，以比任何人在此处键入新答案更快地获得好答案。不要将日期和频率作为字符串保存在对象中。日期使用

LocalDate

，频率使用

int

或某种枚举类型；你也可以考虑<代码>期限<代码>。当您需要显示数据时，您可以始终以用户友好的方式格式化数据。问题不是字符串列表。我没有收到像2020年10月5日、“30天”或2020年10月5日、“60天”或2020年11月5日、“30天”或2020年11月11日、“90天”这样的订单@SubrahmanyamGuntupalli——我假设记录只需要按日期排序，而不需要按频率排序。这是正确的假设吗？我需要按频率和日期列出顺序

import java.time.LocalDate;
import java.time.format.DateTimeFormatter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.List;

class DateFrequency {
    private String date;
    private String frequency;

    public DateFrequency(String date, String frequency) {
        this.date = date;
        this.frequency = frequency;
    }

    public String getDate() {
        return date;
    }

    @Override
    public String toString() {
        return "DateFrequency [date=" + date + ", frequency=" + frequency + "]";
    }
}

public class Main {
    public static void main(String[] args) {
        DateTimeFormatter formatter = DateTimeFormatter.ofPattern("dd/MM/yyyy");

        List<DateFrequency> list = new ArrayList<DateFrequency>();
        list.add(new DateFrequency("05/10/2020", "60-DAYS"));
        list.add(new DateFrequency("05/10/2020", "30-DAYS"));
        list.add(new DateFrequency("05/11/2020", "30-DAYS"));
        list.add(new DateFrequency("05/12/2020", "60-DAYS"));
        list.add(new DateFrequency("05/11/2020", "90-DAYS"));

        // Sort on dates in ascending order
        Collections.sort(list, Comparator.comparing((df) -> LocalDate.parse(df.getDate(), formatter)));

        // Display the sorted result
        for (DateFrequency df : list) {
            System.out.println(df);
        }
    }
}

DateFrequency [date=05/10/2020, frequency=60-DAYS]
DateFrequency [date=05/10/2020, frequency=30-DAYS]
DateFrequency [date=05/11/2020, frequency=30-DAYS]
DateFrequency [date=05/11/2020, frequency=90-DAYS]
DateFrequency [date=05/12/2020, frequency=60-DAYS]