問題描述
我試圖通過忽略周末來獲取兩個特定時間實例之間的分鐘數.這就是我所做的.
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;
}
誰能告訴我我的邏輯是否正確，如果不正確，該怎么做?我不太確定當周末月份發生變化時這段代碼會如何表現.
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.
一些測試用例的預期輸出:
Expected output for some test cases:
- 周五下午 6 點，周一早上 6 點 - 應該返回 12 小時
- 周六 12 點，周日 12 點 - 應該返回 0 小時
- 星期六中午 12 點，星期一早上 6 點 - 應該返回 6 小時
推薦答案
我強烈推薦使用 Joda-時間在Java中處理有關日期的任何事情，因為它帶有許多有用的功能，可以使代碼不那么復雜.
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.
此代碼使用 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());
}
當測試這段代碼時，它給出了以下結果:
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)
我剛剛測試了一個周末換月的案例:從 3 月 29 日星期五(下午 6 點)到 4 月 1 日星期一(早上 6 點):
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());
結果是 12 小時，所以它適用于月份變化.
The result is 12 hours, so it works for the month change.
我的第一個解決方案沒有正確處理夏令時.我們必須在減去分鐘時確定每個實際天的持續時間，因為夏令時更改的天數不會是 24 小時:
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());
}
這篇關于如何使用 Java 日歷忽略周末?的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！