此处的其他答案请参阅Joda API。我想使用.java.time
假设今天的日期是2015年11月26日星期四，当我向它添加2个工作日时，我希望结果为2015年11月30日星期一。
我正在研究我自己的实现，但如果已经存在一些东西，那就太好了！
编辑：
除了循环之外，还有没有办法做到这一点？
我试图派生一个这样的函数：
Y = f(X1,X2) where
Y is actual number of days to add,
X1 is number of business days to add,
X2 is day of the week (1-Monday to 7-Sunday)
然后给出和（从日期的星期几派生），我们可以找到然后使用的方法。X1X2YplusDays()LocalDate
到目前为止，我还没有能够推导出它，它并不一致。谁能确认循环直到添加所需的工作日数是唯一的方法？
以下方法将天数逐个相加，跳过周末，以获得workdays
public LocalDate add(LocalDate date, int workdays) {
if (workdays < 1) {
return date;
}
LocalDate result = date;
int addedDays = 0;
while (addedDays < workdays) {
result = result.plusDays(1);
if (!(result.getDayOfWeek() == DayOfWeek.SATURDAY ||
result.getDayOfWeek() == DayOfWeek.SUNDAY)) {
++addedDays;
}
}
return result;
}
经过一番摆弄，我想出了一种算法来计算要加减的工作日数。
/**
* @param dayOfWeek
* The day of week of the start day. The values are numbered
* following the ISO-8601 standard, from 1 (Monday) to 7
* (Sunday).
* @param businessDays
* The number of business days to count from the day of week. A
* negative number will count days in the past.
*
* @return The absolute (positive) number of days including weekends.
*/
public long getAllDays(int dayOfWeek, long businessDays) {
long result = 0;
if (businessDays != 0) {
boolean isStartOnWorkday = dayOfWeek < 6;
long absBusinessDays = Math.abs(businessDays);
if (isStartOnWorkday) {
// if negative businessDays: count backwards by shifting weekday
int shiftedWorkday = businessDays > 0 ? dayOfWeek : 6 - dayOfWeek;
result = absBusinessDays + (absBusinessDays + shiftedWorkday - 1) / 5 * 2;
} else { // start on weekend
// if negative businessDays: count backwards by shifting weekday
int shiftedWeekend = businessDays > 0 ? dayOfWeek : 13 - dayOfWeek;
result = absBusinessDays + (absBusinessDays - 1) / 5 * 2 + (7 - shiftedWeekend);
}
}
return result;
}
使用示例：
LocalDate startDate = LocalDate.of(2015, 11, 26);
int businessDays = 2;
LocalDate endDate = startDate.plusDays(getAllDays(startDate.getDayOfWeek().getValue(), businessDays));
System.out.println(startDate + (businessDays > 0 ? " plus " : " minus ") + Math.abs(businessDays)
+ " business days: " + endDate);
businessDays = -6;
endDate = startDate.minusDays(getAllDays(startDate.getDayOfWeek().getValue(), businessDays));
System.out.println(startDate + (businessDays > 0 ? " plus " : " minus ") + Math.abs(businessDays)
+ " business days: " + endDate);
输出示例：
2015-11-26 加 2 个工作日： 2015-11-30
2015-11-26减去6个工作日： 2015-11-18
下面是一个同时支持正天数和负天数的版本，并将操作公开为 .这允许你写：TemporalAdjuster
LocalDate datePlus2WorkingDays = date.with(addWorkingDays(2));
法典：
/**
* Returns the working day adjuster, which adjusts the date to the n-th following
* working day (i.e. excluding Saturdays and Sundays).
* <p>
* If the argument is 0, the same date is returned if it is a working day otherwise the
* next working day is returned.
*
* @param workingDays the number of working days to add to the date, may be negative
*
* @return the working day adjuster, not null
*/
public static TemporalAdjuster addWorkingDays(long workingDays) {
return TemporalAdjusters.ofDateAdjuster(d -> addWorkingDays(d, workingDays));
}
private static LocalDate addWorkingDays(LocalDate startingDate, long workingDays) {
if (workingDays == 0) return nextOrSameWorkingDay(startingDate);
LocalDate result = startingDate;
int step = Long.signum(workingDays); //are we going forward or backward?
for (long i = 0; i < Math.abs(workingDays); i++) {
result = nextWorkingDay(result, step);
}
return result;
}
private static LocalDate nextOrSameWorkingDay(LocalDate date) {
return isWeekEnd(date) ? nextWorkingDay(date, 1) : date;
}
private static LocalDate nextWorkingDay(LocalDate date, int step) {
do {
date = date.plusDays(step);
} while (isWeekEnd(date));
return date;
}
private static boolean isWeekEnd(LocalDate date) {
DayOfWeek dow = date.getDayOfWeek();
return dow == SATURDAY || dow == SUNDAY;
}
