10

我需要计算两个日期之间的工作日。例如：我们在 7 月 4 日放假（在美国）。所以如果我的日期是 date1 = 07/03/2012 date2 = 07/06/2012

由于 7 月 4 日是假期，因此这些日期应为 1 个工作日。

我有一个下面的方法来计算工作日，它只计算周末而不是假期。还有什么方法可以计算假期....请帮助我。

  public static int getWorkingDaysBetweenTwoDates(Date startDate, Date endDate) {  
    Calendar startCal;  
    Calendar endCal;  
    startCal = Calendar.getInstance();  
    startCal.setTime(startDate);  
    endCal = Calendar.getInstance();  
    endCal.setTime(endDate);  
    int workDays = 0;  

    //Return 0 if start and end are the same  
    if (startCal.getTimeInMillis() == endCal.getTimeInMillis()) {  
        return 0;  
    }  

    if (startCal.getTimeInMillis() > endCal.getTimeInMillis()) {  
        startCal.setTime(endDate);  
        endCal.setTime(startDate);  
    }  

    do {  
        startCal.add(Calendar.DAY_OF_MONTH, 1);  
        if (startCal.get(Calendar.DAY_OF_WEEK) != Calendar.SATURDAY   
       && startCal.get(Calendar.DAY_OF_WEEK) != Calendar.SUNDAY) {  
            ++workDays;  
        }  
    } while (startCal.getTimeInMillis() < endCal.getTimeInMillis());  

    return workDays;  
}
4

5 回答 5

12

由于接受的答案仍然使用过时的Calendar类 - 这是我使用包中提供的更新的 Java 日期和时间 API 的两分钱java.time。

您必须从某个地方获取假期的日期，没有标准的 Java 库。无论如何，这将过于本地化，因为假期在很大程度上取决于您所在的国家和地区（众所周知的假期除外，例如圣诞节或复活节）。
例如，您可以从假期 API 中获取它们。在下面的代码中，我将它们硬编码为 a Setof LocalDates。

爪哇 9

LocalDate startDate = LocalDate.of(2012, 3, 7);
LocalDate endDate = LocalDate.of(2012, 6, 7);

// I've hardcoded the holidays as LocalDates
// and put them in a Set
final Set<LocalDate> holidays = Set.of(
    LocalDate.of(2018, 7, 4)
);
// For the sake of efficiency, I also put the business days into a Set.
// In general, a Set has a better lookup speed than a List.
final Set<DayOfWeek> businessDays = Set.of(
    MONDAY, TUESDAY, WEDNESDAY, THURSDAY, FRIDAY
);

List<LocalDate> allDates =

    // Java 9 provides a method to return a stream with dates from the
    // startdate to the given end date. Note that the end date itself is
    // NOT included.
    startDate.datesUntil(endDate)

        // Retain all business days. Use static imports from
        // java.time.DayOfWeek.*
        .filter(t -> businessDays.contains(t.getDayOfWeek()))

        // Retain only dates not present in our holidays list
        .filter(t -> !holidays.contains(t))

         // Collect them into a List. If you only need to know the number of
         // dates, you can also use .count()
        .collect(Collectors.toList());

爪哇 8

该方法LocalDate.datesUntil在 Java 8 中不可用，因此您必须以不同的方式获取这两个日期之间的所有日期的流。我们必须首先计算使用该ChronoUnit.DAYS.between方法之间的总天数。

long numOfDaysBetween = ChronoUnit.DAYS.between(startDate, endDate);

然后我们必须生成一个整数序列，其长度与开始日期和结束日期之间的天数完全相同，然后从中创建LocalDates，从开始日期开始。

IntStream.iterate(0, i -> i + 1)
    .limit(numOfDaysBetween)
    .mapToObj(startDate::plusDays)

现在我们有了一个Stream<LocalDate>，然后您就可以使用 Java 9 代码的剩余部分了。您还需要替换 的用法Set.of()，因为它在 Java 8 中不可用。可能是new HashSet<>(Arrays.asList(MONDAY...FRIDAY)).

于 2018-11-26T09:47:45.763 回答
9

正如您所提到的，假设您有一个包含所有假期的列表。

ArrayList<Integer> holidays = ...

只需在您的if条件中添加一个条件do-while：

do {
          startCal.add(Calendar.DAY_OF_MONTH, 1);
          if (startCal.get(Calendar.DAY_OF_WEEK) != Calendar.SATURDAY
          && startCal.get(Calendar.DAY_OF_WEEK) != Calendar.SUNDAY
          && !holidays.contains((Integer) startCal.get(Calendar.DAY_OF_YEAR))) {
              ++workDays;
          }
} while (startCal.getTimeInMillis() < endCal.getTimeInMillis());

为简单起见，我假设holiday包含格式与Calendar.DAY_OF_YEAR.

于 2012-06-01T16:53:35.090 回答
8

纳格日期

您可以使用Nager.Date项目的JSON API。它支持美国、加拿大和欧洲。您可以将每年可用的数据保存在自己的数据库中。

例子

//https://github.com/FasterXML/jackson-databind/
ObjectMapper mapper = new ObjectMapper();
MyValue value = mapper.readValue(new URL("http://date.nager.at/api/v1/get/US/2017"), PublicHoliday[].class);

PublicHoliday.class

public class PublicHoliday
{
    public String date;
    public String localName;
    public String name;
    public String countryCode;
    public Boolean fixed;
    public Boolean countyOfficialHoliday;
    public Boolean countyAdministrationHoliday;
    public Boolean global;
    public String[] counties;
    public int launchYear;
}

检索到的示例 JSON 数据。

[
  {
    "date": "2017-01-01",
    "localName": "New Year's Day",
    "name": "New Year's Day",
    "countryCode": "US",
    "fixed": true,
    "countyOfficialHoliday": true,
    "countyAdministrationHoliday": true,
    "global": true,
    "counties": null,
    "launchYear": null
  },
  {
    "date": "2017-01-16",
    "localName": "Martin Luther King, Jr. Day",
    "name": "Martin Luther King, Jr. Day",
    "countryCode": "US",
    "fixed": true,
    "countyOfficialHoliday": true,
    "countyAdministrationHoliday": true,
    "global": true,
    "counties": null,
    "launchYear": null
  },
  {
    "date": "2017-01-20",
    "localName": "Inauguration Day",
    "name": "Inauguration Day",
    "countryCode": "US",
    "fixed": true,
    "countyOfficialHoliday": true,
    "countyAdministrationHoliday": true,
    "global": false,
    "counties": [
      "US-DC",
      "US-LA",
      "US-MD",
      "US-VA"
    ],
    "launchYear": null
  },
  {
    "date": "2017-02-20",
    "localName": "Washington's Birthday",
    "name": "Presidents' Day",
    "countryCode": "US",
    "fixed": true,
    "countyOfficialHoliday": true,
    "countyAdministrationHoliday": true,
    "global": true,
    "counties": null,
    "launchYear": null
  },
  {
    "date": "2017-05-29",
    "localName": "Memorial Day",
    "name": "Memorial Day",
    "countryCode": "US",
    "fixed": true,
    "countyOfficialHoliday": true,
    "countyAdministrationHoliday": true,
    "global": true,
    "counties": null,
    "launchYear": null
  },
  {
    "date": "2017-07-04",
    "localName": "Independence Day",
    "name": "Independence Day",
    "countryCode": "US",
    "fixed": true,
    "countyOfficialHoliday": true,
    "countyAdministrationHoliday": true,
    "global": true,
    "counties": null,
    "launchYear": null
  },
  {
    "date": "2017-09-04",
    "localName": "Labor Day",
    "name": "Labor Day",
    "countryCode": "US",
    "fixed": true,
    "countyOfficialHoliday": true,
    "countyAdministrationHoliday": true,
    "global": true,
    "counties": null,
    "launchYear": null
  },
  {
    "date": "2017-09-09",
    "localName": "Columbus Day",
    "name": "Columbus Day",
    "countryCode": "US",
    "fixed": true,
    "countyOfficialHoliday": true,
    "countyAdministrationHoliday": true,
    "global": false,
    "counties": [
      "US-AL",
      "US-AZ",
      "US-CO",
      "US-CT",
      "US-DC",
      "US-GA",
      "US-ID",
      "US-IL",
      "US-IN",
      "US-IA",
      "US-KS",
      "US-KY",
      "US-LA",
      "US-ME",
      "US-MD",
      "US-MA",
      "US-MS",
      "US-MO",
      "US-MT",
      "US-NE",
      "US-NH",
      "US-NJ",
      "US-NM",
      "US-NY",
      "US-NC",
      "US-OH",
      "US-OK",
      "US-PA",
      "US-RI",
      "US-SC",
      "US-TN",
      "US-UT",
      "US-VA",
      "US-WV"
    ],
    "launchYear": null
  },
  {
    "date": "2017-11-10",
    "localName": "Veterans Day",
    "name": "Veterans Day",
    "countryCode": "US",
    "fixed": false,
    "countyOfficialHoliday": true,
    "countyAdministrationHoliday": true,
    "global": true,
    "counties": null,
    "launchYear": null
  },
  {
    "date": "2017-12-23",
    "localName": "Thanksgiving Day",
    "name": "Thanksgiving Day",
    "countryCode": "US",
    "fixed": true,
    "countyOfficialHoliday": true,
    "countyAdministrationHoliday": true,
    "global": true,
    "counties": null,
    "launchYear": 1863
  },
  {
    "date": "2017-12-25",
    "localName": "Christmas Day",
    "name": "Christmas Day",
    "countryCode": "US",
    "fixed": true,
    "countyOfficialHoliday": true,
    "countyAdministrationHoliday": true,
    "global": true,
    "counties": null,
    "launchYear": null
  }
]
于 2017-02-09T21:27:10.937 回答
3

我没有任何代码示例或类似的东西，但我做了一些搜索，发现了这个 Stack Overflow 线程，其中有一些指向 Web 服务的链接，可以为你返回假期日期，这可能会帮助你到达你想要的地方需要：国庆节网络服务

该线程中的最佳答案链接到此 Web 服务： http: //www.holidaywebservice.com/

我不确定为这类事情使用网络服务是否是矫枉过正，但肯定有更好的方法。抱歉，我不是最有经验的程序员，所以我无法为您提供尽可能多的帮助。

于 2012-06-01T16:37:37.510 回答
1

我发现最快的方法是使用一些数学（借用SO post）： 

public static int totalBusinessDaysBetween(LocalDate start, LocalDate end) {
    Objects.requireNonNull(start, "Start date must not be null");
    Objects.requireNonNull(end, "End date must not be null");
    long daysBetweenWithoutWeekends = calculateNumberOfDaysBetweenMinusWeekends(start, end);
    final Set<LocalDate> holidayForYearRange = getUSFederalHolidayForYearRange(start.getYear(), end.getYear());
    for (LocalDate localDate : holidayForYearRange) {
        if (localDate.isAfter(start) && localDate.isBefore(end)) {
            daysBetweenWithoutWeekends--;
        }
    }
    return (int) daysBetweenWithoutWeekends;
}

和：

private static long calculateNumberOfDaysBetweenMinusWeekends(LocalDate start, LocalDate end) {

    final DayOfWeek startW = start.getDayOfWeek();
    final DayOfWeek endW = end.getDayOfWeek();

    final long days = ChronoUnit.DAYS.between(start, end);
    final long daysWithoutWeekends = days - 2 * ((days + startW.getValue()) / 7);

    //adjust for starting and ending on a Sunday:
    return daysWithoutWeekends + (startW == DayOfWeek.SUNDAY ? 1 : 0) + (endW == DayOfWeek.SUNDAY ? 1 : 0);
}

我在这里把更完整的东西放在一起。

于 2018-11-29T14:12:12.093 回答