I need to filter search results based on values that were added yesterday. I have seen plenty on finding yesterday using:
NSCalendar *cal = [NSCalendar currentCalendar];
NSDateComponents *components = [cal components:( NSHourCalendarUnit | NSMinuteCalendarUnit | NSSecondCalendarUnit ) fromDate:[[NSDate alloc] init]];
[components setHour:-24];
[components setMinute:0];
[components setSecond:0];
NSDate *yesterday = [cal dateByAddingComponents:components toDate:[NSDate date] options:0];
predicate = [NSPredicate predicateWithFormat:#"created_at >= %#", yesterday];
But this finds 24 hours since this exact moment in time. I need to filter yesterday as 12:01am-12:00pm. So the actual 24 hour period that was yesterday.
I'm guessing that I need to do something along the lines of:
1. Take the current date
2. Find the time from the current date to 12:01am of the same day
3. Then subtract 24 hours from that date
I feel confident I can do #3 (and #1 of course), but I'm not sure how to go about #2. I maybe over thinking it but I can't seem to grasp how to say: "Ok, it's 8:03am, I need to remove 8 hours and 2 minutes which will put me at 12:01am".
Start with some date of today, for example "now":
NSCalendar *cal = [NSCalendar currentCalendar];
NSDate *now = [NSDate date];
Subtract one day to get some date of yesterday:
NSDateComponents *minusOneDay = [[NSDateComponents alloc] init];
[oneDay setDay:-1];
NSDate *nowMinusOneDay = [cal dateByAddingComponents:minusOneDay toDate:now options:0];
Compute start and end date of the "day calendar unit" that contains yesterday's date:
NSDate *startOfYesterday;
NSTimeInterval lengthOfYesterday;
[cal rangeOfUnit:NSDayCalendarUnit startDate:&startOfYesterday interval:&lengthOfYesterday forDate:nowMinusOneDay];
NSDate *endOfYesterday = [startOfYesterday dateByAddingTimeInterval:lengthOfYesterday];
This should work even if a daylight savings time transition occurs between today and yesterday.
Generally one should avoid to use explicit time intervals such as "24 hours", because not every day has that length.
I am using the following NSCalendar method to get the number of days in a calendar year:
NSRange range = [gregorian rangeOfUnit:NSDayCalendarUnit
inUnit:NSYearCalendarUnit
forDate:date];
I am expecting a range.length return value of type NSRange which is 365 or 366 (days).
However, this code returns a range.length of 31 for a date value of 2005-06-06.
What is wrong with my code?
This is the full code snippet:
NSCalendar *gregorian = [[NSCalendar alloc]
initWithCalendarIdentifier:NSGregorianCalendar];
[subArray enumerateObjectsUsingBlock:
^(NSDate *date, NSUInteger idx, BOOL *stop) {
NSUInteger numberOfDays = [gregorian ordinalityOfUnit:NSDayCalendarUnit
inUnit:NSYearCalendarUnit forDate:date];
}];
This calculates the number of days of a year of a given date:
NSDate *someDate = [NSDate date];
NSDate *beginningOfYear;
NSTimeInterval lengthOfYear;
NSCalendar *gregorian = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
[gregorian rangeOfUnit:NSYearCalendarUnit
startDate:&beginningOfYear
interval:&lengthOfYear
forDate:someDate];
NSDate *nextYear = [beginningOfYear dateByAddingTimeInterval:lengthOfYear];
NSInteger startDay = [gregorian ordinalityOfUnit:NSDayCalendarUnit
inUnit:NSEraCalendarUnit
forDate:beginningOfYear];
NSInteger endDay = [gregorian ordinalityOfUnit:NSDayCalendarUnit
inUnit:NSEraCalendarUnit
forDate:nextYear];
NSInteger daysInYear = endDay - startDay;
Sad caveat: This does not work correctly for year 1582.
The year 1582, the year when Gregor introduced the currently widespread used calendar, needed a fix to align solar with calendar years. So they went with the pragmatic solution: They just dropped October 5-14. (They were not crazy enough to change weekdays, too). As a result the year 1582 only has 355 days.
Addendum: The code above only works correctly for years after 1582. It returns 365 days for the year 1500, for example, even though this year was a leap year in the then used julian calendar. The gregorian calendar starts at October 15, 1582. Computations made on the gregorian calendar are just not defined before that date. So in this way Apple's implementation is correct. I'm not aware of a correct implementation for years before 1583 on Cocoa.
Here's a relatively clean version. It's a category on NSCalendar.
- (NSInteger) daysInYear:(NSInteger) year {
NSDateComponents *dateComponents = [[NSDateComponents alloc] init];
dateComponents.year = year;
dateComponents.month = dateComponents.day = 1;
NSDate *firstOfYear = [self dateFromComponents:dateComponents];
dateComponents.year = year + 1;
NSDate *firstOfFollowingYear = [self dateFromComponents:dateComponents];
return [[self components:NSDayCalendarUnit
fromDate:firstOfYear
toDate:firstOfFollowingYear
options:0] day];
}
Thanks to AKV for pointing out that -components:fromDate:toDate:options will work in this case.
This doesn't seem to work for 1582 or prior, which per Nikolai Ruhe's answer is because Gregorian calendar calculations simply aren't defined prior to then.
What about finding number of days in a given year as: Although this is not the relative solution for the question.
-(NSInteger)daysBetweenTwoDates:(NSDate *)fromDateTime andDate:(NSDate *)toDateTime{
NSDate *fromDate;
NSDate *toDate;
NSCalendar *calendar = [NSCalendar currentCalendar];
[calendar rangeOfUnit:NSDayCalendarUnit startDate:&fromDate interval:NULL forDate:fromDateTime];
[calendar rangeOfUnit:NSDayCalendarUnit startDate:&toDate interval:NULL forDate:toDateTime];
NSDateComponents *difference = [calendar components:NSDayCalendarUnit fromDate:fromDate toDate:toDate options:0];
return [difference day];
}
-(void)someMethod {
NSString *yourYear=#"2012";
NSDate *date1 = [NSDate dateWithString:[NSString stringWithFormat:#"%#-01-01 00:00:00 +0000",yourYear]];
NSDate *date2 = [NSDate dateWithString:[NSString stringWithFormat:#"%#-12-31 00:00:00 +0000",yourYear]];
NSInteger numberOfDays=[self daysBetweenTwoDates:date1 andDate:date2];
NSLog(#"There are %ld days in between the two dates.", numberOfDays);
}
Edit:
As dateWithString: is deprecated, you can use
NSDateFormatter *dateFormatter=[[NSDateFormatter alloc] init];
[dateFormatter dateFormat]=#"dd-MM-yyyy";
NSDate *date=[dateFormatter dateFromString:dateString];
to form date1 and date2
You are getting the number of days in the month of your date.
You could use the same function and just pass in a date with February as the month. If range.length returns 28, the Total days in the year will be 365, if it returns 29, then the number of days will be 366.
Does anyone know if there is a way to set the first day of the week on a NSCalendar, or is there a calendar that already has Monday as the first day of the week, instead of Sunday.
I'm currently working on an app that is based around a week's worth of work, and it needs to start on Monday, not Sunday. I can most likely do some work to work around this, but there will be a lot of corner cases. I'd prefer the platform do it for me.
Thanks in advance
Here's some the code that I'm using. it's saturday now, so what I would hope is that weekday would be 6, instead of 7. that would mean that Sunday would be 7 instead of rolling over to 0
NSCalendar *gregorian = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
[gregorian setFirstWeekday:0];
unsigned unitFlags = NSYearCalendarUnit | NSMonthCalendarUnit | NSDayCalendarUnit | NSWeekCalendarUnit | NSWeekdayCalendarUnit;
NSDateComponents *todaysDate = [gregorian components:unitFlags fromDate:[NSDate date]];
int dayOfWeek = todaysDate.weekday;
Edit: This does not check the edge case where the beginning of the week starts in the prior month. Some updated code to cover this: https://stackoverflow.com/a/14688780/308315
In case anyone is still paying attention to this, you need to use
ordinalityOfUnit:inUnit:forDate:
and set firstWeekday to 2. (1 == Sunday and 7 == Saturday)
Here's the code:
NSCalendar *gregorian = [[[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar] autorelease];
[gregorian setFirstWeekday:2]; // Sunday == 1, Saturday == 7
NSUInteger adjustedWeekdayOrdinal = [gregorian ordinalityOfUnit:NSWeekdayCalendarUnit inUnit:NSWeekCalendarUnit forDate:[NSDate date]];
NSLog(#"Adjusted weekday ordinal: %d", adjustedWeekdayOrdinal);
Remember, the ordinals for weekdays start at 1 for the first day of the week, not zero.
Documentation link.
This code constructs a date that is set to Monday of the current week:
NSCalendar *gregorian = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
NSDate *today = [NSDate date];
NSDate *beginningOfWeek = nil;
BOOL ok = [gregorian rangeOfUnit:NSWeekCalendarUnit startDate:&beginningOfWeek
interval:NULL forDate: today];
setFirstWeekday: on the NSCalendar object.
Sets the index of the first weekday for the receiver.
- (void)setFirstWeekday:(NSUInteger)weekday
Should do the trick.
In my opinion this settings should be dynamic according to the user locale.
Therefore one should use:
NSCalendar *gregorian = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
[gregorian setLocale:[NSLocale currentLocale]];
This will cause the calendar to set the first week day according to the user locale automatically. Unless you are developing your app for a specific purpose/user locale (or prefer to allow the user to choose this day).
I've done it like this.
NSCalendar *gregorian = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
NSDate *today = [NSDate date];
NSDateComponents *compForWeekday = [gregorian components:(NSWeekdayCalendarUnit) fromDate:today];
NSInteger weekDayAsNumber = [compForWeekday weekday]; // The week day as number but with sunday starting as 1
weekDayAsNumber = ((weekDayAsNumber + 5) % 7) + 1; // Transforming so that monday = 1 and sunday = 7
I had trouble with a lot of the answers here. . maybe it was just me. .
Here's an answer that works for me:
- (NSDate*)firstDayOfWeek
{
NSCalendar* cal = [[NSCalendar currentCalendar] copy];
[cal setFirstWeekday:2]; //Override locale to make week start on Monday
NSDate* startOfTheWeek;
NSTimeInterval interval;
[cal rangeOfUnit:NSWeekCalendarUnit startDate:&startOfTheWeek interval:&interval forDate:self];
return startOfTheWeek;
}
- (NSDate*)lastDayOfWeek
{
NSCalendar* cal = [[NSCalendar currentCalendar] copy];
[cal setFirstWeekday:2]; //Override locale to make week start on Monday
NSDate* startOfTheWeek;
NSTimeInterval interval;
[cal rangeOfUnit:NSWeekCalendarUnit startDate:&startOfTheWeek interval:&interval forDate:self];
return [startOfTheWeek dateByAddingTimeInterval:interval - 1];
}
Update:
As pointed out (elsewhere) by #vikingosegundo, in general its best to let the local determine which day is the start of the week, however in this case the OP was asking for the start of the week to occur on Monday, hence we copy the system calendar, and override the firstWeekDay.
The problem with Kris' answer is the edge case where the beginning of the week starts in the prior month. Here's some easier code and it also checks the edge case:
// Finds the date for the first day of the week
- (NSDate *)getFirstDayOfTheWeekFromDate:(NSDate *)givenDate
{
NSCalendar *calendar = [NSCalendar currentCalendar];
// Edge case where beginning of week starts in the prior month
NSDateComponents *edgeCase = [[NSDateComponents alloc] init];
[edgeCase setMonth:2];
[edgeCase setDay:1];
[edgeCase setYear:2013];
NSDate *edgeCaseDate = [calendar dateFromComponents:edgeCase];
NSDateComponents *components = [calendar components:NSYearCalendarUnit|NSMonthCalendarUnit|NSWeekCalendarUnit|NSWeekdayCalendarUnit fromDate:edgeCaseDate];
[components setWeekday:1]; // 1 == Sunday, 7 == Saturday
[components setWeek:[components week]];
NSLog(#"Edge case date is %# and beginning of that week is %#", edgeCaseDate , [calendar dateFromComponents:components]);
// Find Sunday for the given date
components = [calendar components:NSYearCalendarUnit|NSMonthCalendarUnit|NSWeekCalendarUnit|NSWeekdayCalendarUnit fromDate:givenDate];
[components setWeekday:1]; // 1 == Sunday, 7 == Saturday
[components setWeek:[components week]];
NSLog(#"Original date is %# and beginning of week is %#", givenDate , [calendar dateFromComponents:components]);
return [calendar dateFromComponents:components];
}
I see misunderstanding in the other messages. The first weekday, whichever it is, has number 1 not 0. By default Sunday=1 as in the "Introduction to Date and Time Programming Guide for Cocoa: Calendrical Calculations":
"The weekday value for Sunday in the Gregorian calendar is 1"
For the Monday as a first workday the only remedy I have is brute force condition to fix the calculation
NSCalendar *cal=[[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
NSDateComponents *comps = [cal components:NSWeekdayCalendarUnit fromDate:[NSDate date]];
// set to 7 if it's Sunday otherwise decrease weekday number
NSInteger weekday=[comps weekday]==1?7:[comps weekday]-1;
Below also covers the edge case,
- (NSDate *)getFirstDayOfTheWeekFromDate:(NSDate *)givenDate
{
NSCalendar *calendar = [NSCalendar currentCalendar];
NSDateComponents *components = [calendar components:NSYearCalendarUnit|NSMonthCalendarUnit|NSWeekCalendarUnit|NSWeekdayCalendarUnit fromDate:givenDate];
[components setWeekday:2]; // 1 == Sunday, 7 == Saturday
if([[calendar dateFromComponents:components] compare: curDate] == NSOrderedDescending) // if start is later in time than end
{
[components setWeek:[components week]-1];
}
return [calendar dateFromComponents:components];
}
You can just change .firstWeekday of the calendar.
NSCalendar *calendar = [[NSCalendar alloc] initWithCalendarIdentifier:NSCalendarIdentifierGregorian];
calendar.firstWeekday = 2;
Then use rangeOfUnit:startDate:interval:forDate: to get the first day
NSDate *startOfWeek;
[calendar rangeOfUnit:NSCalendarUnitWeekOfYear startDate:&startOfWeek interval:nil forDate:[NSdate date]];
Try this:
NSCalendar *yourCal = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar]
[yourCal setFirstWeekday:0];
Iv found out the way to display any weekday name using nscalender..using the following code..
Just open your console from xcode menu bar to see the results.Copy Paste the following code in your viewDidLoad method to get the first day of the week
NSDate *today = [NSDate date];
NSDateFormatter *dateFormat = [[NSDateFormatter alloc] init];
[dateFormat setDateFormat:#"MM/dd/yyyy :EEEE"];
NSString *dateString = [dateFormat stringFromDate:today];
NSLog(#"date: %#", dateString);
[dateFormat release];
NSCalendar *gregorian = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
NSDateComponents *components = [gregorian components:NSWeekdayCalendarUnit | NSYearCalendarUnit | NSMonthCalendarUnit | NSDayCalendarUnit fromDate:today];
[components setDay:([components day]-([components weekday]-1))];
NSDate *beginningOfWeek = [gregorian dateFromComponents:components];
NSDateFormatter *dateFormat_first = [[NSDateFormatter alloc] init];
[dateFormat_first setDateFormat:#"MM/dd/yyyy :EEEE"];
NSString *dateString_first = [dateFormat_first stringFromDate:beginningOfWeek];
NSLog(#"First_date: %#", dateString_first);
The Output will be:
date: 02/11/2010 :Thursday
First_date: 02/07/2010 :Sunday
since i had run this program on 2/11/2010 u will get the desired output depending on the current date.
Similarly if u want to get the first working day of the week i.e Monday's date then just modify the code a bit:
CHANGE :[components setDay:([components day]-([components weekday]-1))];
TO
[components setDay:([components day]-([components weekday]-2))];
to get Mondays date for that week..
Similarly u can try to find the date of any of seven workdays by changing the integer -1,-2 and so on...
Hope u r question is answered..
Thanks,
Bonson Dias
The ISO 8601 calendar appears to have it's first weekday set to monday by default.
Using the Calendar nextWeekend (iOS 10 or later) and ordinality (thanks #kris-markel). I've gotten Monday as first of the week for the en_US calendar.
Here is an example of it with fallback to firstWeekday:
extension Calendar {
var firstWorkWeekday: Int {
guard #available(iOS 10.0, *) else{
return self.firstWeekday
}
guard let endOfWeekend = self.nextWeekend(startingAfter: Date())?.end else {
return self.firstWeekday
}
return self.ordinality(of: .weekday, in: .weekOfYear, for: endOfWeekend) ?? self.firstWeekday
}
}
The Swift solution (note, use .yearForWeekOfYear, not .year):
let now = Date()
let cal = Calendar.current
var weekComponents = cal.dateComponents([.yearForWeekOfYear, .weekOfYear,
.weekday], from: now)
//weekComponents.weekday = 1 // if your week starts on Sunday
weekComponents.weekday = 2 // if your week starts on Monday
cal.date(from: weekComponents) // returns date with first day of the week
… is there a calendar that already has Monday as the first day of the week, instead of Sunday.
Someday, there will be.
My simple way of doing this is to get Monday = 0, Sunday = 6:
NSDateComponents *dateComponents = [[NSCalendar currentCalendar] components:NSWeekdayCalendarUnit fromDate:[NSDate date]];
NSInteger dayNumStartingFromMonday = ([dateComponents weekday] - 2 + 7) % 7; //normal: Sunday is 1, Monday is 2
I'm making my custom calendar view for an app for the European market. In a function I should get number of day of week... I do, but it returns the number of the day of week starting with Sunday. How should I hardcode returning this number starting with Monday?
Thanks
Here is what I have so far:
-(int)getWeekDay:(NSDate*)date_
{
NSLocale *frLocale = [[NSLocale alloc] initWithLocaleIdentifier:#"fr_FR"];
NSCalendar *gregorian = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
[gregorian setLocale:frLocale];
NSDateComponents *comps = [gregorian components:NSWeekdayCalendarUnit fromDate:date_];
int weekday = [comps weekday];
NSLog(#"Week day is %i", weekday);
return weekday;
}
You should use [NSCalendar setFirstWeekday:] to do this.
The best way to do this would be to use [NSCalendar setFirstWeekday:], as Joshua said in his answer.
Otherwise, you can do integer arithmetic. Vova's method is straightforward:
if (weekday>1)
weekday--;
else
weekday=7;
This one below works too, although it's a bit confusing:
int europeanWeekday = ((weekday + 5) % 7) + 1;
I have the simplest of tasks that I can't figure out how to do correctly. With the snippet below I can find out the beginning of a month.
NSCalendar *cal = [NSCalendar currentCalendar];
NSDate *beginning = nil;
[cal rangeOfUnit:NSMonthCalendarUnit startDate:&beginning interval:NULL forDate:self];
return beginning;
Now I want to determine the end of that same month (the last second at 23:59:59 would suit my purposes). My first thought was to start at the first day of the next month and subtract a second. But I couldn't break down the date in an NSDateComponent instance to [dateComponents setMonth:[dateComponents month] + 1], because in the case of December that method wouldn't accept 12 + 1 = 13 to get to January, right?
Then how would I get to the last second of a month?
Do this to get the beginning of next month:
NSDateComponents *comps = [[NSDateComponents alloc] init];
[comps setMonth:1];
NSDate *beginningOfNextMonth = [cal dateByAddingComponents:comps toDate:beginning options:0];
[comps release];
Alright, I think I have it. I'm using the rangeOfUnit:inUnit:forDate: method to determine the length of the current month in days, and add the corresponding NSTimeInterval to the month's start date.
- (NSDate *)firstSecondOfTheMonth {
NSCalendar *cal = [NSCalendar currentCalendar];
NSDate *beginning = nil;
if ([cal rangeOfUnit:NSMonthCalendarUnit startDate:&beginning interval:NULL forDate:self])
return beginning;
return nil;
}
- (NSDate *)lastSecondOfTheMonth {
NSDate *date = [self firstSecondOfTheMonth];
NSCalendar *cal = [NSCalendar currentCalendar];
NSRange month = [cal rangeOfUnit:NSDayCalendarUnit inUnit:NSMonthCalendarUnit forDate:date];
return [date dateByAddingTimeInterval:month.length * 24 * 60 * 60 - 1];
}
Note: Although this code works as I intended it to work, the answer that #spacehunt gave more clearly communicates the purpose.
The last second of the month... Most of the time, that boils down to taking the first second of the next month, and subtracting one second.
But are you asking the right question? What's the meaning of that last second? My guess is that you want to express a time interval starting from (and including) the first second of month N, and ending right before the first second of month N+1.