java - 我如何检查 joda-time 是给定日期的第一周，第二周，......还是一个月的最后一周？

Question

例如:

• 2016 年 2 月 5 日 - 第一周，
• 2016 年 2 月 12 日- 第二周，
• 2016 年 2 月 28 日- 上周

Answer 1

嗯，您不太清楚如何定义一个月中的星期几。对于下面的讨论，我假设您谈论的是从星期一开始的几周(例如 ISO-8601 标准和大多数欧洲国家/地区)。

在统计周数时如何处理月的开始和结束，有两种可能的定义方式。

由于星期一的一周开始不一定与一个月的第一天相同，一周可以从上个月开始，也可以属于下个月。

JDK 类 SimpleDateFormat 及其字段模式符号 w(以及新的 JSR-310 字段 WeekFields.weekOfMonth())使用以下策略:

If the first day of month is falling on Monday to Thursday then the associated week has at least 4 days in current month and will be counted as week 1 otherwise as week 0 (zero). Consistently the last day of month will always use an incrementing number even if it belongs to first week of next month.

与该定义相反，CLDR 日期-时间-模式规范和 ISO-8601 几乎没有提及月中周上下文中的细节。然而，这些标准并没有对他们描述另一种策略的一年中的哪一周保持沉默。和 CLDR explicitly says about week-of-month (第 8.4 节):

8.4 Week of Year

Values calculated for the Week of Year field range from 1 to 53 for the Gregorian calendar (they may have different ranges for other calendars). Week 1 for a year is the first week that contains at least the specified minimum number of days from that year. Weeks between week 1 of one year and week 1 of the following year are numbered sequentially from 2 to 52 or 53 (if needed). For example, January 1, 1998 was a Thursday. If the first day of the week is MONDAY and the minimum days in a week is 4 (these are the values reflecting ISO 8601 and many national standards), then week 1 of 1998 starts on December 29, 1997, and ends on January 4, 1998. However, if the first day of the week is SUNDAY, then week 1 of 1998 starts on January 4, 1998, and ends on January 10, 1998. The first three days of 1998 are then part of week 53 of 1997.

Values are similarly calculated for the Week of Month.

在 2016 年 2 月 29 日应用的两种策略之间的差异将是:

• 根据 JDK 第 5 周
• 根据 CLDR/ISO 第 1 周

---

现在我提出一个Joda-Time 的解决方案。

public static void main(String[] args) throws Throwable {

    System.out.println(getWeekOfMonth(false)); // CLDR/ISO-spec
    // 1 for today=2016-02-05
    // 2 for today=2016-02-12
    // 4 for today=2016-02-28
    // 1 for today=2016-02-29

    System.out.println(getWeekOfMonth(true)); // JDK-behaviour
    // 1 for today=2016-02-05
    // 2 for today=2016-02-12
    // 4 for today=2016-02-28
    // 5 for today=2016-02-29

}

private static int getWeekOfMonth(boolean bounded) {
    int weekOfMonth;
    LocalDate today = LocalDate.now();

    LocalDate first = today.dayOfMonth().withMinimumValue();
    int dowFirst = first.getDayOfWeek();

    if (dowFirst <= DateTimeConstants.THURSDAY) {
        // we are in week 1 and go to Monday as start of week
        first = first.minusDays(dowFirst - DateTimeConstants.MONDAY);

        // first try: we determine the week of current month
        weekOfMonth = Days.daysBetween(first, today).getDays() / 7 + 1;

        if (!bounded) {
            // edge case: are we in first week of next month?
            LocalDate next = first.plusMonths(1);
            int dowNext = next.getDayOfWeek();

            if (dowNext <= DateTimeConstants.THURSDAY) {
                next = next.minusDays(dowNext - DateTimeConstants.MONDAY);
                if (!next.isAfter(today)) {
                    weekOfMonth = 1;
                }
            }
        }
    } else if (bounded) {
        weekOfMonth = 0;
    } else {
        // we are in last week of previous month so let's check the start of previous month
        LocalDate previous = first.minusMonths(1);
        int dowPrevious = previous.getDayOfWeek();

        if (dowPrevious <= DateTimeConstants.THURSDAY) {
            previous = previous.minusDays(dowPrevious - DateTimeConstants.MONDAY);
        } else {
            previous = previous.plusDays(DateTimeConstants.MONDAY - dowPrevious + 7);
        }

        weekOfMonth = Days.daysBetween(previous, today).getDays() / 7 + 1;
    }

    return weekOfMonth;
}

我希望这对您来说不会太复杂。

---

顺便说一下，如果您对适用于早于 Java-8 的平台的简单替代方案看起来像什么感兴趣:

Time4J (我的图书馆)

private static int time4j(boolean bounded) { // supports both definitions
  PlainDate today = SystemClock.inLocalView().today(); // using system timezone
    return today.get(
          (bounded ? Weekmodel.ISO.boundedWeekOfMonth() : Weekmodel.ISO.weekOfMonth()));
}

Threeten-BP (backport of Java-8):

private static int threeten() { // only JDK-definition (code similar to Java-8)
    org.threeten.bp.LocalDate today = org.threeten.bp.LocalDate.now();
    return today.get(WeekFields.ISO.weekOfMonth());
}

Old JDK:

private static int oldJDK() { // only JDK-definition
    GregorianCalendar gcal = new GregorianCalendar();
    gcal.setMinimalDaysInFirstWeek(4);
    gcal.setFirstDayOfWeek(Calendar.MONDAY);
    return gcal.get(Calendar.WEEK_OF_MONTH);
}

如您所见，使用这些替代方案可以很容易地将基础周模型更改为非 ISO 案例(如美国周)。如果您希望在 Joda-Time 中实现这一点，那么我将重写所提供的 Joda 解决方案的任务留给您。

由于以下关于主题的评论而更新:

所以整个事情都是关于每月的星期几。 Joda-Time 也不支持开箱即用的这个元素/字段。对不起。但是你也许可以研究 necessary algorithm对于其他库中使用的此类字段。

Time4J 中的一个演示示例，用于对评论中提到的 rfc2445 规则进行建模:

    PlainDate dtStart = PlainDate.of(2016, Month.FEBRUARY, 4);
    int count = 5;
    Weekday byday = Weekday.FRIDAY; // first
    int interval = 1;
    CalendarUnit frequency = CalendarUnit.MONTHS;

    List<PlainDate> sequence = new ArrayList<>(count);
    PlainDate wim = dtStart;
    for (int i = 0; i < count; i++) {
        wim = wim.with(PlainDate.WEEKDAY_IN_MONTH.setToFirst(byday));
        sequence.add(wim);
        wim = wim.with(PlainDate.DAY_OF_MONTH, 1).plus(interval, frequency);
    }
    if (!sequence.isEmpty() && !sequence.get(0).equals(dtStart)) {
        sequence.remove(0); // Not quite sure - do you need another condition?
    }
    System.out.println(sequence); // [2016-03-04, 2016-04-01, 2016-05-06, 2016-06-03]

在 Java-8 中，也有通过 specialized adjuster 的支持。因此您可以使用 java.time.LocalDate 轻松地将给定的演示示例转移到 Java-8。