I have for example Monday, Thursday and Sunday as draw dates. What would be the way to find the closest dayOfWeek to Datetime.Now with Linq Lambda? I accomplished this with normal function, but I would like to know how to do it with Linq Lambda?
Example of my approach:
public static DateTime GetSoonestDrawDate(this DateTime from, IEnumerable<LotteryDrawDate> drawDates)
{
int hour = 0;
int minute = 0;
bool todayDrawOnly = true;
int difference = 7;
int tempDifference = 0;
int todayDay = (int)from.DayOfWeek;
int drawCutOffDay = 0;
if (todayDay == 0)
{
todayDay = 7;
}
var tempCutOffTime = new TimeSpan(23, 59, 59);
foreach (var drawDate in drawDates)
{
// DayId is DayOfWeek
drawCutOffDay = drawDate.DayId;
if (drawCutOffDay < todayDay)
{
drawCutOffDay += 7;
}
tempDifference = drawCutOffDay - todayDay;
var cutOffTime = new TimeSpan();
cutOffTime = TimeSpan.Parse(drawDate.CutOffTime);
if ((difference > tempDifference) || difference == 0)
{
// draw date is not today
if (tempDifference != 0)
{
difference = tempDifference;
hour = cutOffTime.Hours;
minute = cutOffTime.Minutes;
todayDrawOnly = false;
}
// same day, cutOffTime still available
if ((tempDifference == 0 && cutOffTime > from.TimeOfDay))
{
// we use tempCutOffTime to compare newest cutOffTime in case we have more cutOffTimes on the same day
// in that case we have to select the soonest cutOffTime of the day
if (cutOffTime < tempCutOffTime)
{
difference = tempDifference;
hour = cutOffTime.Hours;
minute = cutOffTime.Minutes;
todayDrawOnly = true;
tempCutOffTime = cutOffTime;
}
}
// same day selected only, but cutOffTime passed, so we add another week only in case there is only one draw date and draw date is on the same day
else if (tempDifference == 0 && cutOffTime < from.TimeOfDay && todayDrawOnly == true)
{
if (cutOffTime < tempCutOffTime)
{
difference = 7;
hour = cutOffTime.Hours;
minute = cutOffTime.Minutes;
todayDrawOnly = true;
tempCutOffTime = cutOffTime;
}
}
}
}
from = from.AddDays(difference);
DateTime soonest = new DateTime(from.Year, from.Month, from.Day, hour, minute, 0);
return soonest;
}
Because you didn't show how LotteryDrawDate
looks like, I prepared a little sample using DayOfWeek
only. You have to extend that to look at time part by your own.
public static DateTime GetSoonestDrawDate(this DateTime from, IEnumerable<DayOfWeek> drawDates)
{
var realDrawDates = drawDates.SelectMany(x => new[] { (int)x, (int)x + 7 }).OrderBy(x => x);
var difference = realDrawDates.SkipWhile(x => x < (int)from.DayOfWeek).First() - (int)from.DayOfWeek;
return from.AddDays(difference);
}
Little test code:
var drawDates = new[] { DayOfWeek.Monday, DayOfWeek.Wednesday, DayOfWeek.Saturday };
for (int i = 0; i < 15; i++)
{
var from = DateTime.Now.AddDays(i);
Console.WriteLine("{0} - {1}", from.ToShortDateString(), GetSoonestDrawDate(from, drawDates).ToShortDateString());
}
prints ( from - next ):
11/18/2013 - 11/18/2013
11/19/2013 - 11/20/2013
11/20/2013 - 11/20/2013
11/21/2013 - 11/23/2013
11/22/2013 - 11/23/2013
11/23/2013 - 11/23/2013
11/24/2013 - 11/25/2013
11/25/2013 - 11/25/2013
11/26/2013 - 11/27/2013
11/27/2013 - 11/27/2013
11/28/2013 - 11/30/2013
11/29/2013 - 11/30/2013
11/30/2013 - 11/30/2013
12/1/2013 - 12/2/2013
12/2/2013 - 12/2/2013
If you wanted a simple hardcoded list of draw days, you could do this:
DateTime GetNextDate(DateTime from)
{
DayOfWeek target;
switch (from.DayOfWeek)
{
case DayOfWeek.Friday:
case DayOfWeek.Saturday:
case DayOfWeek.Sunday:
target = DayOfWeek.Sunday;
break;
case DayOfWeek.Monday:
target = DayOfWeek.Monday;
break;
case DayOfWeek.Tuesday:
case DayOfWeek.Wednesday:
case DayOfWeek.Thursday:
target = DayOfWeek.Thursday;
break;
default:
throw new ArgumentException("from");
}
while (from.DayOfWeek != target)
from = from.AddDays(1);
return from;
}
A simple method to find the nearest day in the future would be the following:
DayOfWeek[] draw_days = { DayOfWeek.Sunday, DayOfWeek.Monday, DayOfWeek.Thursday };
Console.WriteLine(
"Nearest Draw Date: {0}",
draw_days.Min(d => (d <= DateTime.Now.DayOfWeek) ? (d + 7) : d)
);
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.