问题描述
我试图通过忽略周末来获取两个特定时间实例之间的分钟数.这就是我所做的.
I'm trying to get the number of minutes between two particular time instances by ignoring weekends. This is what I've done.
public static final List<Integer> NON_WORKING_DAYS;
static {
List<Integer> nonWorkingDays = new ArrayList<Integer>();
nonWorkingDays.add(Calendar.SATURDAY);
nonWorkingDays.add(Calendar.SUNDAY);
NON_WORKING_DAYS = Collections.unmodifiableList(nonWorkingDays);
}
public static int getMinsBetween(Date d1, Date d2, boolean onlyBusinessDays)
{
int minsBetween = (int)((d2.getTime() - d1.getTime()) / (1000 * 60));
int minsToSubtract = 0;
if(onlyBusinessDays){
Calendar dateToCheck = Calendar.getInstance();
dateToCheck.setTime(d1);
Calendar dateToCompare = Calendar.getInstance();
dateToCompare.setTime(d2);
//moving the first day of the week to Tues so that a Sat, sun and mon fall in the same week, easy to adjust dates
dateToCheck.setFirstDayOfWeek(Calendar.TUESDAY);
dateToCompare.setFirstDayOfWeek(Calendar.TUESDAY);
//moving the dates out of weekends
if(!isBusinessDay(dateToCheck, NON_WORKING_DAYS)){
dateToCheck.set(Calendar.DAY_OF_WEEK, Calendar.SATURDAY);
dateToCheck.set(Calendar.HOUR, 0);
dateToCheck.set(Calendar.MINUTE, 0);
dateToCheck.set(Calendar.SECOND, 0);
dateToCheck.set(Calendar.MILLISECOND, 0);
}
if(!isBusinessDay(dateToCompare, NON_WORKING_DAYS)){
dateToCompare.set(Calendar.DAY_OF_WEEK, Calendar.MONDAY);
dateToCompare.set(Calendar.HOUR, 0);
dateToCompare.set(Calendar.MINUTE, 0);
dateToCompare.set(Calendar.SECOND, 0);
dateToCompare.set(Calendar.MILLISECOND, 0);
}
for(; dateToCheck.getTimeInMillis() < dateToCompare.getTimeInMillis() ; dateToCheck.add(Calendar.DAY_OF_MONTH, 1)){
if(isBusinessDay(dateToCheck, NON_WORKING_DAYS)){
minsToSubtract = minsToSubtract + 1440;
}
}
minsBetween = minsBetween - minsToSubtract;
}
return minsBetween;
}
private static boolean isBusinessDay(Calendar dateToCheck, List<Integer> daysToExclude){
for(Integer dayToExclude : daysToExclude){
if(dayToExclude != null && dayToExclude == dateToCheck.get(Calendar.DAY_OF_WEEK)) {
return true;
}
else continue;
}
return false;
}
谁能告诉我我的逻辑是否正确，如果不正确，该怎么做?我不太确定当周末月份发生变化时这段代码会如何表现.
Can someone tell me if my logic is correct and if not how to do this? I'm not too sure how this code would behave when the month changes over the weekend.
一些测试用例的预期输出:
Expected output for some test cases:
- 周五下午 6 点，周一早上 6 点 - 应该返回 12 小时
- 周六 12 点，周日 12 点 - 应该返回 0 小时
- 星期六中午 12 点，星期一早上 6 点 - 应该返回 6 小时
推荐答案
我强烈推荐使用 Joda-时间在Java中处理有关日期的任何事情，因为它带有许多有用的功能，可以使代码不那么复杂.
I would highly recommend using Joda-Time for anything concerning date manipulations in Java, because it comes with a lot of helpful functions to make the code less complicated.
此代码使用 JodaTime:
This code uses JodaTime:
public static final List<Integer> NON_WORKING_DAYS;
static {
List<Integer> nonWorkingDays = new ArrayList<Integer>();
nonWorkingDays.add(DateTimeConstants.SATURDAY);
nonWorkingDays.add(DateTimeConstants.SUNDAY);
NON_WORKING_DAYS = Collections.unmodifiableList(nonWorkingDays);
}
public static Minutes getMinsBetween(DateTime d1, DateTime d2,
boolean onlyBusinessDays) {
BaseDateTime startDate = onlyBusinessDays && !isBusinessDay(d1) ?
new DateMidnight(d1) : d1;
BaseDateTime endDate = onlyBusinessDays && !isBusinessDay(d2) ?
new DateMidnight(d2) : d2;
Minutes minutes = Minutes.minutesBetween(startDate, endDate);
if (onlyBusinessDays) {
DateTime d = new DateTime(startDate);
while (d.isBefore(endDate)) {
if (!isBusinessDay(d)) {
Duration dayDuration = new Duration(d, d.plusDays(1));
minutes = minutes.minus(int) dayDuration.getStandardMinutes());
}
d = d.plusDays(1);
}
}
return minutes;
}
private static boolean isBusinessDay(DateTime dateToCheck) {
return !NON_WORKING_DAYS.contains(dateToCheck.dayOfWeek().get());
}
当测试这段代码时，它给出了以下结果:
When this code is tested, it gives the following results:
DateTime d1 = new DateTime(2013, 1, 4, 18, 0); // a Friday, 6 pm
DateTime d2 = new DateTime(2013, 1, 7, 6, 0); // the following Monday, 6 am
Minutes minutes = getMinsBetween(d1, d2, true);
System.out.println(minutes.toStandardHours().getHours()); // outputs "12" (in hours)
d1 = new DateTime(2013, 1, 5, 12, 0); // a Saturday, 12 pm
d2 = new DateTime(2013, 1, 6, 12, 0); // the following Sunday, 12 pm
minutes = getMinsBetween(d1, d2, true);
System.out.println(minutes.toStandardHours().getHours()); // outputs "0" (in hours)
d1 = new DateTime(2013, 1, 5, 12, 0); // a Saturday, 12 pm
d2 = new DateTime(2013, 1, 7, 6, 0); // the following Monday, 6 am
minutes = getMinsBetween(d1, d2, true);
System.out.println(minutes.toStandardHours().getHours()); // outputs "6" (in hours)
我刚刚测试了一个周末换月的案例:从 3 月 29 日星期五(下午 6 点)到 4 月 1 日星期一(早上 6 点):
I just tested a case where the month changes over the weekend: From Friday, March 29th (6pm) to Monday, April 1st (6am):
d1 = new DateTime(2013, 3, 29, 18, 0);
d2 = new DateTime(2013, 4, 1, 6, 0);
minutes = getMinsBetween(d1, d2, true);
System.out.println(minutes.toStandardHours().getHours());
结果是 12 小时，所以它适用于月份变化.
The result is 12 hours, so it works for the month change.
我的第一个解决方案没有正确处理夏令时.我们必须在减去分钟时确定每个实际天的持续时间，因为夏令时更改的天数不会是 24 小时:
My first solution wasn't handling daylight saving times correctly. We have to determine the duration of each actual day when subtracting the minutes because days with a change in daylight saving time will not be 24h:
if (!isBusinessDay(d)) {
Duration dayDuration = new Duration(d, d.plusDays(1));
minutes = minutes.minus(int) dayDuration.getStandardMinutes());
}
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