Önümüzdeki Salı gününü nasıl edinebilirim?
PHP'de bu kadar basit strtotime('next tuesday');
.
.NET'te benzer bir şeyi nasıl başarabilirim?
Önümüzdeki Salı gününü nasıl edinebilirim?
PHP'de bu kadar basit strtotime('next tuesday');
.
.NET'te benzer bir şeyi nasıl başarabilirim?
Yanıtlar:
Yorumlarda bahsettiğim gibi, "önümüzdeki Salı" ile ifade edebileceğiniz çeşitli şeyler vardır, ancak bu kod size "önümüzdeki Salı günü veya bugün zaten Salı ise" verir:
DateTime today = DateTime.Today;
// The (... + 7) % 7 ensures we end up with a value in the range [0, 6]
int daysUntilTuesday = ((int) DayOfWeek.Tuesday - (int) today.DayOfWeek + 7) % 7;
DateTime nextTuesday = today.AddDays(daysUntilTuesday);
Zaten Salı ise "bir haftalık zaman" vermek istiyorsanız şunu kullanabilirsiniz:
// This finds the next Monday (or today if it's Monday) and then adds a day... so the
// result is in the range [1-7]
int daysUntilTuesday = (((int) DayOfWeek.Monday - (int) today.DayOfWeek + 7) % 7) + 1;
... ya da orijinal formülü kullanabilirsiniz, ancak yarından itibaren:
DateTime tomorrow = DateTime.Today.AddDays(1);
// The (... + 7) % 7 ensures we end up with a value in the range [0, 6]
int daysUntilTuesday = ((int) DayOfWeek.Tuesday - (int) tomorrow.DayOfWeek + 7) % 7;
DateTime nextTuesday = tomorrow.AddDays(daysUntilTuesday);
EDIT: Sadece bu güzel ve çok yönlü yapmak için:
public static DateTime GetNextWeekday(DateTime start, DayOfWeek day)
{
// The (... + 7) % 7 ensures we end up with a value in the range [0, 6]
int daysToAdd = ((int) day - (int) start.DayOfWeek + 7) % 7;
return start.AddDays(daysToAdd);
}
"Bugün veya önümüzdeki 6 gün içinde" değerini elde etmek için:
DateTime nextTuesday = GetNextWeekday(DateTime.Today, DayOfWeek.Tuesday);
"Bugün hariç gelecek Salı" değerini almak için:
DateTime nextTuesday = GetNextWeekday(DateTime.Today.AddDays(1), DayOfWeek.Tuesday);
+7)%7
çözüm oldukça güzel. Kullanmamamın nedeni, bir mikro optimizasyonun biraz olması ve yanlış anlaşılması çok kolay (bazı okunabilirliği feda etmenin yanı sıra).
Bu hile yapmalı:
static DateTime GetNextWeekday(DayOfWeek day)
{
DateTime result = DateTime.Now.AddDays(1);
while( result.DayOfWeek != day )
result = result.AddDays(1);
return result;
}
.AddDays(1)
, ilk satırdan kaldırın , böylece DateTime.Now
kendisini de kontrol edecektir .
Bu soruna daha az ayrıntılı ve daha akıllı / zarif çözümler var, ancak aşağıdaki C # işlevi birçok durum için gerçekten iyi çalışıyor.
/// <summary>
/// Find the closest weekday to the given date
/// </summary>
/// <param name="includeStartDate">if the supplied date is on the specified day of the week, return that date or continue to the next date</param>
/// <param name="searchForward">search forward or backward from the supplied date. if a null parameter is given, the closest weekday (ie in either direction) is returned</param>
public static DateTime ClosestWeekDay(this DateTime date, DayOfWeek weekday, bool includeStartDate = true, bool? searchForward=true)
{
if (!searchForward.HasValue && !includeStartDate)
{
throw new ArgumentException("if searching in both directions, start date must be a valid result");
}
var day = date.DayOfWeek;
int add = ((int)weekday - (int)day);
if (searchForward.HasValue)
{
if (add < 0 && searchForward.Value)
{
add += 7;
}
else if (add > 0 && !searchForward.Value)
{
add -= 7;
}
else if (add == 0 && !includeStartDate)
{
add = searchForward.Value ? 7 : -7;
}
}
else if (add < -3)
{
add += 7;
}
else if (add > 3)
{
add -= 7;
}
return date.AddDays(add);
}
@Jon Skeet iyi cevap.
Önceki Gün için:
private DateTime GetPrevWeekday(DateTime start, DayOfWeek day) {
// The (... - 7) % 7 ensures we end up with a value in the range [0, 6]
int daysToRemove = ((int) day - (int) start.DayOfWeek - 7) % 7;
return start.AddDays(daysToRemove);
}
Teşekkürler!!
DayOfWeek
gibi değerleri değiştirmek olacaktır :int daysToSubtract = -(((int)dateTime.DayOfWeek - (int)day + 7) % 7);
Güncel tarihi dahil etmek veya hariç tutmak için çok basit bir örnek, ilgilendiğiniz tarihi ve günü belirtirsiniz.
public static class DateTimeExtensions
{
/// <summary>
/// Gets the next date.
/// </summary>
/// <param name="date">The date to inspected.</param>
/// <param name="dayOfWeek">The day of week you want to get.</param>
/// <param name="exclDate">if set to <c>true</c> the current date will be excluded and include next occurrence.</param>
/// <returns></returns>
public static DateTime GetNextDate(this DateTime date, DayOfWeek dayOfWeek, bool exclDate = true)
{
//note: first we need to check if the date wants to move back by date - Today, + diff might move it forward or backwards to Today
//eg: date - Today = 0 - 1 = -1, so have to move it forward
var diff = dayOfWeek - date.DayOfWeek;
var ddiff = date.Date.Subtract(DateTime.Today).Days + diff;
//note: ddiff < 0 : date calculates to past, so move forward, even if the date is really old, it will just move 7 days from date passed in
//note: ddiff >= (exclDate ? 6 : 7) && diff < 0 : date is into the future, so calculated future weekday, based on date
if (ddiff < 0 || ddiff >= (exclDate ? 6 : 7) && diff < 0)
diff += 7;
//note: now we can get safe values between 0 - 6, especially if past dates is being used
diff = diff % 7;
//note: if diff is 0 and we are excluding the date passed, we will add 7 days, eg: 1 week
diff += diff == 0 & exclDate ? 7 : 0;
return date.AddDays(diff);
}
}
bazı test senaryoları
[TestMethod]
public void TestNextDate()
{
var date = new DateTime(2013, 7, 15);
var start = date;
//testing same month - forwardOnly
Assert.AreEqual(start = start.AddDays(1), date.GetNextDate(DayOfWeek.Tuesday)); //16
Assert.AreEqual(start = start.AddDays(1), date.GetNextDate(DayOfWeek.Wednesday)); //17
Assert.AreEqual(start = start.AddDays(1), date.GetNextDate(DayOfWeek.Thursday)); //18
Assert.AreEqual(start = start.AddDays(1), date.GetNextDate(DayOfWeek.Friday)); //19
Assert.AreEqual(start = start.AddDays(1), date.GetNextDate(DayOfWeek.Saturday)); //20
Assert.AreEqual(start = start.AddDays(1), date.GetNextDate(DayOfWeek.Sunday)); //21
Assert.AreEqual(start.AddDays(1), date.GetNextDate(DayOfWeek.Monday)); //22
//testing same month - include date
Assert.AreEqual(start = date, date.GetNextDate(DayOfWeek.Monday, false)); //15
Assert.AreEqual(start = start.AddDays(1), date.GetNextDate(DayOfWeek.Tuesday, false)); //16
Assert.AreEqual(start.AddDays(1), date.GetNextDate(DayOfWeek.Wednesday, false)); //17
//testing month change - forwardOnly
date = new DateTime(2013, 7, 29);
start = date;
Assert.AreEqual(start = start.AddDays(1), date.GetNextDate(DayOfWeek.Tuesday)); //30
Assert.AreEqual(start = start.AddDays(1), date.GetNextDate(DayOfWeek.Wednesday)); //31
Assert.AreEqual(start = start.AddDays(1), date.GetNextDate(DayOfWeek.Thursday)); //2013/09/01-month increased
Assert.AreEqual(start.AddDays(1), date.GetNextDate(DayOfWeek.Friday)); //02
//testing year change
date = new DateTime(2013, 12, 30);
start = date;
Assert.AreEqual(start = start.AddDays(1), date.GetNextDate(DayOfWeek.Tuesday)); //31
Assert.AreEqual(start = start.AddDays(1), date.GetNextDate(DayOfWeek.Wednesday)); //2014/01/01 - year increased
Assert.AreEqual(start = start.AddDays(1), date.GetNextDate(DayOfWeek.Thursday)); //02
}
Ayrıca bir uzantı olabilir , hepsi bağlıdır
public static class DateTimeExtensions
{
public static IEnumerable<DateTime> Next(this DateTime date, DayOfWeek day)
{
// This loop feels expensive and useless, but the point is IEnumerable
while(true)
{
if (date.DayOfWeek == day)
{
yield return date;
}
date = date.AddDays(1);
}
}
}
kullanım
var today = DateTime.Today;
foreach(var monday in today.Next(DayOfWeek.Monday))
{
Console.WriteLine(monday);
Console.ReadKey();
}
Şimdi oneliner lezzetinde - bir parametre olarak parametre olarak geçirmeniz gerektiğinde.
DateTime.Now.AddDays(((int)yourDate.DayOfWeek - (int)DateTime.Now.DayOfWeek + 7) % 7).Day
Bu özel durumda:
DateTime.Now.AddDays(((int)DayOfWeek.Tuesday - (int)DateTime.Now.DayOfWeek + 7) % 7).Day
Amaç C Versiyonu:
+(NSInteger) daysUntilNextWeekday: (NSDate*)startDate withTargetWeekday: (NSInteger) targetWeekday
{
NSInteger startWeekday = [[NSCalendar currentCalendar] component:NSCalendarUnitWeekday fromDate:startDate];
return (targetWeekday - startWeekday + 7) % 7;
}