Flutter - 如何找到两个日期之间的年、月和日差异？

Question

我正在寻找一种使用DateTime来解析两个日期以显示差异的方法。 我想把它放在格式上：“X 年，Y 个月，Z 天”。

对于 JS，我们有momentjs库和以下代码::

var a = moment([2015, 11, 29]);
var b = moment([2007, 06, 27]);

var years = a.diff(b, 'year');
b.add(years, 'years');

var months = a.diff(b, 'months');
b.add(months, 'months');

var days = a.diff(b, 'days');

console.log(years + ' years ' + months + ' months ' + days + ' days');
// 8 years 5 months 2 days

dart是否有类似的库可以帮助实现这个用例？

Answer 1

我认为使用 DateTime 不可能轻松地完成您想要的操作。 因此，您可以使用在日期时间处理方面非常强大的https://pub.dev/packages/time_machine package：

import 'package:time_machine/time_machine.dart';

void main() {
  LocalDate a = LocalDate.today();
  LocalDate b = LocalDate.dateTime(DateTime(2022, 1, 2));
  Period diff = b.periodSince(a);
  print("years: ${diff.years}; months: ${diff.months}; days: ${diff.days}");
}

小时/分钟/秒精度：

import 'package:time_machine/time_machine.dart';

void main() {
  LocalDateTime a = LocalDateTime.now();
  LocalDateTime b = LocalDateTime.dateTime(DateTime(2022, 1, 2, 10, 15, 47));
  Period diff = b.periodSince(a);
  print("years: ${diff.years}; months: ${diff.months}; days: ${diff.days}; hours: ${diff.hours}; minutes: ${diff.minutes}; seconds: ${diff.seconds}");
}

Answer 2

您正在寻找的是 DartDateTime class您可以在 moment.js 中接近您想要的

main() {
  var a = DateTime.utc(2015, 11, 29);
  var b = DateTime.utc(2007, 06, 27);

  var years = a.difference(b);
  print(years.inDays ~/365);

}

DateTime 没有 inYears 或 inMonths 选项，但这就是打印中划分年份的原因。 差异 function 以秒为单位返回差异，因此您必须自己处理几天。

Answer 3

您可以在持续时间 class 上编写扩展名来格式化它：

extension DurationExtensions on Duration {
  String toYearsMonthsDaysString() {
    final years = this.inDays ~/ 365
    // You will need a custom logic for the months part, since not every month has 30 days
    final months = (this.inDays ~% 365) ~/ 30
    final days = (this.inDays ~% 365) ~% 30

    return "$years years $months months $days days";
  }
}

用法将是：

final date1 = DateTime()
final date2 = DateTime()
date1.difference(date2).toYearsMonthsDaysString()

Answer 4

您可以像这样使用Jiffy Package

var jiffy1 = Jiffy("2008-10", "yyyy-MM");
var jiffy2 = Jiffy("2007-1", "yyyy-MM");

jiff1.diff(jiffy2, Units.YEAR); // 1
jiff1.diff(jiffy2, Units.YEAR, true);

Answer 5

使用以下代码尝试 intl package：

import 'package:intl/intl.dart';

String startDate = '01/01/2021';
String endDate = '01/01/2022';

final start = DateFormat('dd/MM/yyyy').parse(startDate);
final end = DateFormat('dd/MM/yyyy').parse(endDate);

然后，您可以使用以下代码计算两个日期之间的持续时间：

final duration = end.difference(start);

要获取年、月、日的数量，您可以执行以下操作：

final years = duration.inDays / 365;
final months = duration.inDays % 365 / 30;
final days = duration.inDays % 365 % 30;

最后，您可以使用这些变量以所需格式显示结果：

    final result = '${years.toInt()} years ${months.toInt()} months y ${days.toInt()} days';

Answer 6

final firstDate = DateTime.now();
final secondDate = DateTime(firstDate.year, firstDate.month - 20);

final yearsDifference = firstDate.year - secondDate.year;
final monthsDifference = (firstDate.year - secondDate.year) * 12 +
    firstDate.month - secondDate.month;
final totalDays = firstDate.difference(secondDate).inDays;

简单的方法，不需要包。

Answer 7

我为公历创建了自己的 class，并创建了一种处理此问题的方法，它“逻辑地”计算日期之间的年、月和日之间的差异......我实际上是从头开始创建 class 而不使用任何其他包（包括 DateTime 包），但在这里我使用 DateTime package 来说明这种方法是如何工作的......直到知道它对我来说很好......

判断是否闰年的方法：

    static bool leapYear(DateTime date) {
    if(date.year%4 == 0) {
      if(date.year%100 == 0){
        return date.year%400 == 0;
      }
      return true;
    }
    return false;
  }

这是计算年、月和日两个日期之间差异的方法。 它将结果放入整数列表中：

     static List<int> differenceInYearsMonthsDays(DateTime dt1, DateTime dt2) {
    List<int> simpleYear = [31,28,31,30,31,30,31,31,30,31,30,31];
    if(dt1.isAfter(dt2)) {
      DateTime temp = dt1;
      dt1 = dt2;
      dt2 = temp;
    }
    int totalMonthsDifference = ((dt2.year*12) + (dt2.month - 1)) - ((dt1.year*12) + (dt1.month - 1));
    int years = (totalMonthsDifference/12).floor();
    int months = totalMonthsDifference%12;
    late int days;
    if(dt2.day >= dt1.day) {days = dt2.day - dt1.day;}
    else {
      int monthDays = dt2.month == 3
          ? (leapYear(dt2)? 29: 28)
          : (dt2.month - 2 == -1? simpleYear[11]: simpleYear[dt2.month - 2]);
      int day = dt1.day;
      if(day > monthDays) day = monthDays;
      days = monthDays - (day - dt2.day);
      months--;
    }
    if(months < 0) {
      months = 11;
      years--;
    }
    return [years, months, days];
  }

计算两个日期在月和天之间的差异的方法：

static List<int> differenceInMonths(DateTime dt1, DateTime dt2){
    List<int> inYears = differenceInYearsMonthsDays(dt1, dt2);
    int difMonths = (inYears[0]*12) + inYears[1];
    return [difMonths, inYears[2]];
  }

以天为单位计算两个日期之间差异的方法：

static int differenceInDays(DateTime dt1, DateTime dt2) {
    if(dt1.isAfter(dt2)) {
      DateTime temp = dt1;
      dt1 = dt2;
      dt2 = temp;
    }
    return dt2.difference(dt1).inDays;
  }

用法示例：

void main() {
  DateTime date1 = DateTime(2005, 10, 3);
  DateTime date2 = DateTime(2022, 1, 12);
  List<int> diffYMD = GregorianDate.differenceInYearsMonthsDays(date1, date2);
  List<int> diffMD = GregorianDate.differenceInMonths(date1, date2);
  int diffD = GregorianDate.differenceInDays(date1, date2);

  print("The difference in years, months and days: ${diffYMD[0]} years, ${diffYMD[1]} months, and ${diffYMD[2]} days.");
  print("The difference in months and days: ${diffMD[0]} months, and ${diffMD[1]} days.");
  print("The difference in days: $diffD days.");
}

output：

The difference in years, months and days: 16 years, 3 months, and 9 days.
The difference in months and days: 195 months, and 9 days.
The difference in days: 5945 days.

Answer 8

DateTime 年差是特定的 function，如下所示：

  static int getDateDiffInYear(DateTime dateFrom, DateTime dateTo) {
    int sign = 1;
    if (dateFrom.isAfter(dateTo)) {
      DateTime temp = dateFrom;
      dateFrom = dateTo;
      dateTo = temp;
      sign = -1;
    }
    int years = dateTo.year - dateFrom.year;
    int months = dateTo.month - dateFrom.month;
    if (months < 0) {
      years--;
    } else {
      int days = dateTo.day - dateFrom.day;
      if (days < 0) {
        years--;
      }
    }
    return years * sign;
  }

Answer 9

DateTime a = DateTime(2015, 11, 29);
DateTime b = DateTime(2007, 06, 27);
int days = a.difference(b).inDays;

Answer 10

答案是肯定的，您可以使用 Dart 中的 DateTime class 轻松实现。 参见：https://api.dart.dev/stable/2.8.3/dart-core/DateTime-class.html

例子

void main() {
  var moonLanding = DateTime(1969,07,20)
  var marsLanding = DateTime(2024,06,10);
  var diff = moonLanding.difference(marsLanding);

  print(diff.inDays.abs());
  print(diff.inMinutes.abs());
  print(diff.inHours.abs());
}

输出：20049 28870560 481176

Answer 11

difHour = someDateTime.difference(DateTime.now()).inHours;
    difMin = (someDateTime.difference(DateTime.now()).inMinutes)-(difHour*60);

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