I'm writing a GUI for a legacy data object where a date is saved in the integer properties day of the year and year (real "genius" design...)
How can I convert this data into day and month given the year?
I'm currently trying:
int theYear = 2016;
int theDayOfTheYear = 222;
NSCalendar* gregorian = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
NSDateComponents *components = [[NSDateComponents alloc] init];
[components setYear:theYear];
[components setDay:theDayOfTheYear];
NSDate* theDate = [gregorian dateFromComponents:components];
This seems a bit too hacky to me. It could be just coincidence that it works with days >31 - haven't found any mention of this in the NSDateComponents class reference.
When I debug, po theDate gives me 2016-08-08 22:00:00 +0000
--> close, but it should be 2016-08-09 (!)
To perform the reverse operation, i.e. to convert a normal DD-MM-YYYY date into day of the year, I'm using this code:
NSDate *fullDate = [NSDate date];
NSCalendar *gregorian = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
NSUInteger dayOfYear = [gregorian ordinalityOfUnit:NSDayCalendarUnit
inUnit:NSYearCalendarUnit
forDate:fullDate];
But I can't seem to find the proper methods to do the opposite...
Sometimes a test is all takes :)
Don't forget that NSDate are GMT, so you might want to add [[NSTimeZone localTimeZone] secondsFromGMT] seconds to it.
It seems that you have it working :)
- (void) testDate
{
int startYear = 2016;
int startMonth = 1;
int startDay = 1;
// Create the startDate
NSDateComponents * startDateComp = [NSDateComponents new];
startDateComp.day = startDay;
startDateComp.month = startMonth;
startDateComp.year = startYear;
NSDate * startDate = [[NSCalendar currentCalendar] dateFromComponents:startDateComp];
for(int i=0 ; i < 365; i++)
{
// add i day to startDate
NSDate * testDate = [startDate dateByAddingTimeInterval:i * 24 * 60 * 60];
NSUInteger dayOfYear = [[NSCalendar currentCalendar] ordinalityOfUnit:NSDayCalendarUnit
inUnit:NSYearCalendarUnit
forDate:testDate];
XCTAssert(dayOfYear == (i + 1), #"the day of the year should be i + 1");
// Create a date using day of year
NSDateComponents *components = [[NSDateComponents alloc] init];
[components setYear:startYear];
[components setDay:dayOfYear];
NSDate* resultDate = [[NSCalendar currentCalendar] dateFromComponents:components];
// testing result date against test date
NSInteger test, result;
test = [[NSCalendar currentCalendar] component:NSCalendarUnitDay fromDate:testDate];
result = [[NSCalendar currentCalendar] component:NSCalendarUnitDay fromDate:resultDate];
XCTAssert(test == result);
test = [[NSCalendar currentCalendar] component:NSCalendarUnitMonth fromDate:testDate];
result = [[NSCalendar currentCalendar] component:NSCalendarUnitMonth fromDate:resultDate];
XCTAssert(test == result);
test = [[NSCalendar currentCalendar] component:NSCalendarUnitYear fromDate:testDate];
result = [[NSCalendar currentCalendar] component:NSCalendarUnitYear fromDate:resultDate];
XCTAssert(test == result);
}
Please try this. Its the larger calendar unit NSCalendarUnitEra :)
NSDate *fullDate = [NSDate date];
NSCalendar *gregorian = [NSCalendar currentCalendar];
gregorian.timeZone = [NSTimeZone systemTimeZone];
NSUInteger dayOfYear = [gregorian ordinalityOfUnit:NSCalendarUnitDay
inUnit:NSCalendarUnitEra
forDate:fullDate];
NSDateComponents *dateCom = [[NSDateComponents alloc] init];
dateCom.day = dayOfYear;
NSDate *expectedDate = [gregorian dateFromComponents:dateCom];
NSLog(#"%#",expectedDate);
Related
I need Objective C method for converting Gregorian date to Julian days same as this PHP method (GregorianToJD).
Precision: Incorporating time of day in Julian Date conversions
These Julian date conversion methods yield results identical to the U.S. Naval Observatory Online Julian Date Converter, which is more precise than NSDateFormatter's Julian Date conversion. Specifically, the functions below incorporate time-of-day (e.g. hour, minute and seconds), whereas NSDateFormatter rounds to noon GMT.
Swift examples:
func jdFromDate(date : NSDate) -> Double {
let JD_JAN_1_1970_0000GMT = 2440587.5
return JD_JAN_1_1970_0000GMT + date.timeIntervalSince1970 / 86400
}
func dateFromJd(jd : Double) -> NSDate {
let JD_JAN_1_1970_0000GMT = 2440587.5
return NSDate(timeIntervalSince1970: (jd - JD_JAN_1_1970_0000GMT) * 86400)
}
Objective-C examples:
double jdFromDate(NSDate *date) {
double JD_JAN_1_1970_0000GMT = 2440587.5;
return JD_JAN_1_1970_0000GMT + date.timeIntervalSince1970 / 86400;
}
NSDate dataFromJd(double jd) {
double JD_JAN_1_1970_0000GMT = 2440587.5;
return [[NSDate alloc] initWithTimeIntervalSince1970: (jd - JD_JAN_1_1970_0000GMT) * 86400)];
}
Note: Research confirms that the accepted answer rounds the date to a 24-hour interval because it uses the g format-specifier of NSDateFormatter, which returns the Modified Julian Day, according to the UNICODE standard's Date Format Patterns that Apple's date formatting APIs adhere to (according to the Date Formatting Guide).
According to http://en.wikipedia.org/wiki/Julian_day, the Julian day number for January 1, 2000, was 2,451,545. So you can compute the number of days between your date and this
reference date. For example (Jan 1, 2014):
NSUInteger julianDayFor01012000 = 2451545;
NSCalendar *cal = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
[cal setTimeZone:[NSTimeZone timeZoneForSecondsFromGMT:0]];
NSDateComponents *comp = [[NSDateComponents alloc] init];
comp.year = 2014;
comp.month = 1;
comp.day = 1;
NSDate *date = [cal dateFromComponents:comp];
comp.year = 2000;
comp.month = 1;
comp.day = 1;
NSDate *ref = [cal dateFromComponents:comp];
NSDateComponents *diff = [cal components:NSDayCalendarUnit fromDate:ref toDate:date options:0];
NSInteger julianDays = diff.day + julianDayFor01012000;
NSLog(#"%ld", (long)julianDays);
// Output: 2456659
This gives the same result as http://www.php.net/manual/en/function.gregoriantojd.php:
<?php
$jd = GregorianToJD(1, 1, 2014);
echo "$jd\n";
?>
Inverse direction (Julian days to Gregorian year/month/day):
NSInteger julianDays = 2456659; // From above example
NSUInteger julianDayFor01012000 = 2451545;
NSCalendar *cal = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
[cal setTimeZone:[NSTimeZone timeZoneForSecondsFromGMT:0]];
NSDateComponents *comp = [[NSDateComponents alloc] init];
comp.year = 2000;
comp.month = 1;
comp.day = 1;
NSDate *ref = [cal dateFromComponents:comp];
NSDateComponents *diff = [[NSDateComponents alloc] init];
diff.day = julianDays - julianDayFor01012000;
NSDate *date = [cal dateByAddingComponents:diff toDate:ref options:0];
comp = [cal components:NSDayCalendarUnit|NSMonthCalendarUnit|NSYearCalendarUnit fromDate:date];
NSLog(#"%04ld-%02ld-%02ld", (long)comp.year, (long)comp.month, (long)comp.day);
// Output: 2014-01-01
UPDATE: As Hot Licks correctly stated in a comment, it is easier to use a date
formatter with the "g" format:
NSCalendar *cal = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
NSDateComponents *comp = [[NSDateComponents alloc] init];
comp.year = 2014;
comp.month = 1;
comp.day = 1;
NSDate *date = [cal dateFromComponents:comp];
NSDateFormatter *fmt = [[NSDateFormatter alloc] init];
[fmt setDateFormat:#"g"];
NSInteger julianDays = [[fmt stringFromDate:date] integerValue];
NSLog(#"%ld", (long)julianDays);
// Output: 2456659
And for the inverse direction:
NSInteger julianDays = 2456659;
NSDateFormatter *fmt = [[NSDateFormatter alloc] init];
[fmt setDateFormat:#"g"];
NSDate *date = [fmt dateFromString:[#(julianDays) stringValue]];
NSCalendar *cal = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
NSDateComponents *comp = [cal components:NSDayCalendarUnit|NSMonthCalendarUnit|NSYearCalendarUnit fromDate:date];
NSLog(#"%04ld-%02ld-%02ld", (long)comp.year, (long)comp.month, (long)comp.day);
// Output: 2014-01-01
let date = Date() // now
let cal = Calendar.current
var day = 0
day = cal.ordinality(of: .day, in: .year, for: date) ?? 0
I want to compare two NSDates with NOW ([NSDate date]).
NSDate *date1 = [NSDate dateWithString:#"1982-02-12 07:00:00 +0100"];
NSDate *now = [NSDate dateWithString:#"2012-01-25 10:19:00 +0100"]; //example
NSDate *date2 = [NSDate dateWithString:#"1989-02-12 15:00:00 +0100"];
I would like to check if now is between date1 and date2. In the example above this is the case. The date component should be completely ignored, so only the time component should be compared. How could I accomplish this?
Thanks in advance!
unsigned int flags = NSCalendarUnitHour | NSCalendarUnitMinute | NSCalendarUnitSecond;
NSCalendar* calendar = [NSCalendar currentCalendar];
NSDateComponents* components = [calendar components:flags fromDate:date1];
NSDate* timeOnly = [calendar dateFromComponents:components];
This will give you a date object where everything but the hours/minutes/seconds have been reset to some common value. Then you can use the standard NSDate compare functions on them.
For reference, here is the opposite question to yours: Comparing two NSDates and ignoring the time component
You can create a date representing the start of today and add the time as components to it to get the boundary dates.
NSDate *now = [NSDate date];
NSDate *startOfToday;
[[NSCalendar currentCalendar] rangeOfUnit:NSDayCalendarUnit startDate:&startOfToday interval:NULL forDate:now];
NSDateComponents *startComps = [[NSDateComponents alloc] init];
startComps.hour = 7;
startComps.minute = 30;
NSDateComponents *endComps = [[NSDateComponents alloc] init];
endComps.hour = 20;
NSDate *startDate = [[NSCalendar currentCalendar] dateByAddingComponents:startComps toDate:startOfToday options:0];
NSDate *endDate = [[NSCalendar currentCalendar] dateByAddingComponents:endComps toDate:startOfToday options:0];
if ([startDate timeIntervalSince1970] < [now timeIntervalSince1970] && [now timeIntervalSince1970] < [endDate timeIntervalSince1970]) {
NSLog(#"good");
}
NSDateFormatter* formatterDate = [[[NSDateFormatter alloc] init] autorelease];
formatterDate.dateStyle = NSDateFormatterMediumStyle; // whatever format you like
NSDate *first_Date = [formatterDate dateFromString:#""];
NSDate *second_Date = [formatterDate dateFromString:#""];
NSDate *todaysDate = [NSDate date];
NSTimeInterval timeIntFormFirstDate = [todaysDate timeIntervalSinceDate:First_Date];
NSTimeInterval timeIntFronLastDate = [second_Date timeIntervalSinceDate:todaysDate];
int interval1 = timeIntFormFirstDate/60;
int interval2 = timeIntFronLastDate/60;
if (interval1 >0 && interval2 >0)
{
NSLog(#"Today's date is between first and second date");
}
I want to set the NSDate time with my desired hours:minutes:seconds
currently im working with NSDate component but it is not giving the desired result
[comps setHour: -hours];
[comps setMinute:0];
[comps setSecond:0];
NSDate *minDate = [calendar_c dateFromComponents:comps];
This works great as an NSDate category.
/** Returns a new NSDate object with the time set to the indicated hour,
* minute, and second.
* #param hour The hour to use in the new date.
* #param minute The number of minutes to use in the new date.
* #param second The number of seconds to use in the new date.
*/
-(NSDate *) dateWithHour:(NSInteger)hour
minute:(NSInteger)minute
second:(NSInteger)second
{
NSCalendar *calendar = [NSCalendar currentCalendar];
NSDateComponents *components = [calendar components: NSYearCalendarUnit|
NSMonthCalendarUnit|
NSDayCalendarUnit
fromDate:self];
[components setHour:hour];
[components setMinute:minute];
[components setSecond:second];
NSDate *newDate = [calendar dateFromComponents:components];
return newDate;
}
With the above category, if you have an existing date you want to change the time on, you do so like this:
NSDate *newDate = [someDate dateWithHour:10 minute:30 second:00];
If, however, you are trying to add or subtract hours from an existing date, a category method to do that is also straightforward:
/** Returns a new date with the given number of hours added or subtracted.
* #param hours The number of hours to add or subtract from the date.
*/
-(NSDate*)dateByAddingHours:(NSInteger)hours
{
NSDateComponents *components = [[NSDateComponents alloc] init];
[components setHour:hours];
return [[NSCalendar currentCalendar]
dateByAddingComponents:components toDate:self options:0];
}
Your approach should work fine. I needed a solution for this type problem (setting the individual date components) and the following code works as expected for me. My situation: I wanted to create a date object that used the current date but had the time set to a value that was passed in as a string.
NSString *string = #"7:00";
NSLocale *locale = [[NSLocale alloc] initWithLocaleIdentifier:#"en_US_POSIX"];
NSDateFormatter *timeOnlyFormatter = [[NSDateFormatter alloc] init];
[timeOnlyFormatter setLocale:locale];
[timeOnlyFormatter setDateFormat:#"h:mm"];
NSCalendar *calendar = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
NSDate *today = [NSDate date];
NSDateComponents *todayComps = [calendar components:(NSDayCalendarUnit | NSMonthCalendarUnit | NSYearCalendarUnit) fromDate:today];
NSDateComponents *comps = [calendar components:(NSHourCalendarUnit | NSMinuteCalendarUnit) fromDate:[timeOnlyFormatter dateFromString:string]];
comps.day = todayComps.day;
comps.month = todayComps.month;
comps.year = todayComps.year;
NSDate *date = [calendar dateFromComponents:comps];
[calendar release];
[timeOnlyFormatter release];
[locale release];
One thing to note is that you really have to pay attention to time zones when you are judging whether a time appears to be accurate. For example, in my app, when you stop at a breakpoint, you will see the time in GMT (so it looks different than the input time, which is in my local time), but when the time is actually displayed on screen in the app, it is being formatted to display in the local timezone. You may need to take this into consideration to determine whether the result is actually different from what you would expect.
If this does not help, can you elaborate on "not giving the desired result"? What result is it giving and how does that compare to what you expected?
is Swift2
extension NSDate {
func dateWithHour (hour: Int, minute:Int, second:Int) ->NSDate?{
let calendar = NSCalendar.currentCalendar(),
components = calendar.components([.Day,.Month,.Year], fromDate: self)
components.hour = hour;
components.minute = minute;
components.second = second;
return calendar.dateFromComponents(components)
}
}
You can set 0 to hour, min, and second.
NSDateFormatter *tFmt = [[NSDateFormatter alloc] init];
tFmt.dateFormat = #"yyyy-MM-dd";
NSString *strNowDate = [NSString stringWithFormat:#"%# 00:00:00",[tFmt stringFromDate:[NSDate date]]];
NSDate *nowDate = [NSDate dateWithString:strNowDate formatString:#"yyyy-MM-dd HH:mm:ss"];
Swift 5 solution (based on #dattk answer) for those who fear Deprecation warnings :)
func date(withHour hour: Int, withMinute minute: Int, withSeconds second: Int) -> Date? {
let now = Date()
let calendar = NSCalendar.current
var components = calendar.dateComponents([.day,.month,.year], from: now)
components.hour = hour
components.minute = minute
components.second = second
return calendar.date(from: components)
}
NSCalendar *calendar = [[NSCalendar alloc] initWithCalendarIdentifier:NSCalendarIdentifierGregorian];
NSDateComponents *comps = [calendar components:(NSCalendarUnitHour | NSCalendarUnitMinute) fromDate:[NSDate date]];
comps.hour = 0;
comps.minute = 15;
NSDate *date = [calendar dateFromComponents:comps];
iOS 8
I need to change a NSDate object. What I am basically doing is changing the year value.
for example:
NSString *someYear = #"2093";
NSDate *date = [NSDate date]; // Gets the current date.
... Create a new date based upon 'date' but with specified year value.
So with 'date' returning 2011-03-06 22:17:50 +0000 from init, I would like to create a date with 2093-03-06 22:17:50 +0000.
However I would like this to be as culturally neutral as possible, so it will work whatever the timezone.
Thanks.
Here's my code for setting the UIDatePicker limits for a Date Of Birth selection. Max age allowed is 100yrs
_dateOfBirth.maximumDate = [NSDate date];
//To limit the datepicker year to current year -100
NSDate *currentDate = [NSDate date];
NSUInteger componentFlags = NSYearCalendarUnit;
NSDateComponents *components = [[NSCalendar currentCalendar] components:componentFlags fromDate:currentDate];
NSInteger year = [components year];
NSLog(#"year = %d",year);
[components setYear:-100];
NSDate *minDate = [[NSCalendar currentCalendar] dateByAddingComponents:components toDate:currentDate options:0];
_dateOfBirth.minimumDate = minDate;
Take a look at NSCalendar, especially components:fromDate: and dateFromComponents: methods.
I managed to figure the answer with the pointer Hoha gave me.
NSNumber *newYear = [[NSNumber alloc] initWithInt:[message intValue]];
NSCalendar* gregorian = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
unsigned int unitFlags = NSYearCalendarUnit | NSDayCalendarUnit | NSMonthCalendarUnit;
NSDateComponents* dateComponents = [gregorian components:unitFlags fromDate:[NSDate date]];
[dateComponents setYear:[newYear intValue]];
NSDate *newDate = [gregorian dateFromComponents:dateComponents];
[newYear release];
Starting in iOS 8 you can set an specific date component. For example:
date = [calendar dateBySettingUnit:NSCalendarUnitYear value:year ofDate:date options:0];
I want to create a function which results the date of next Friday but I have no plan how to do it. Has anyone a good hint to me ?
E.g. Get current date using NSDate, then use 'components>fromDate:' from NSCalendar to get the NSDateComponents, then add the time difference to next Friday and create a new NSDate and Bob's is your uncle.
Here is my working solution for getting the next 5 Sundays in Gregorian calendar:
self.nextBeginDates = [NSMutableArray array];
NSDateComponents *weekdayComponents = [[NSCalendar currentCalendar] components:NSWeekdayCalendarUnit fromDate:[NSDate date]];
int currentWeekday = [weekdayComponents weekday]; //[1;7] ... 1 is sunday, 7 is saturday in gregorian calendar
NSDateComponents *comp = [[NSDateComponents alloc] init];
[comp setDay:8 - currentWeekday]; // add some days so it will become sunday
// build weeks array
int weeksCount = 5;
for (int i = 0; i < weeksCount; i++) {
[comp setWeek:i]; // add weeks
[nextBeginDates addObject:[[NSCalendar currentCalendar] dateByAddingComponents:comp toDate:[NSDate date] options:0]];
}
[comp release];
This should work
+ (NSDate *) dateForNextWeekday: (NSInteger)weekday {
NSDate *today = [[NSDate alloc] init];
NSCalendar *gregorian = [[NSCalendar alloc]
initWithCalendarIdentifier:NSGregorianCalendar];
// Get the weekday component of the current date
NSDateComponents *weekdayComponents = [gregorian components:NSWeekdayCalendarUnit
fromDate:today];
/*
Add components to get to the weekday we want
*/
NSDateComponents *componentsToSubtract = [[NSDateComponents alloc] init];
NSInteger dif = weekday-weekdayComponents.weekday;
if (dif<=0) dif += 7;
[componentsToSubtract setDay:dif];
NSDate *beginningOfWeek = [gregorian dateByAddingComponents:componentsToSubtract
toDate:today options:0];
return beginningOfWeek;
}
Keep it simple, safe, and readable! (....KISSAR?)
#define FRIDAY 6
NSCalendar *gregorian = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
NSDateComponents *dayComponent = [[NSDateComponents alloc] init];
dayComponent.day = 1;
NSDate *nextFriday = [NSDate date];
NSInteger iWeekday = [[gregorian components:NSWeekdayCalendarUnit fromDate:nextFriday] weekday];
while (iWeekday != FRIDAY) {
nextFriday = [gregorian dateByAddingComponents:dayComponent toDate:nextFriday options:0];
iWeekday = [[gregorian components:NSWeekdayCalendarUnit fromDate:nextFriday] weekday];
}
Now nextFriday has your date.
Hope this helps!
EDIT
Note that if the current date was already a Friday, it would return that instead of the next Friday. If that's undesirable just init your nextFriday to a day later (so if current date was a Friday, it would start on Saturday, forcing the next Friday. And if current date was a Thursday you'd automatically have your next Friday without needing the loop).
Here is my solution, and just to warn you, on a saturday is the friday before shown.
cheers to all
NSDate *today = [[NSDate alloc] init];
NSCalendar *gregorian = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
NSDateComponents *weekdayComponents = [gregorian components:NSWeekdayCalendarUnit fromDate:today];
int weekday = [weekdayComponents weekday];
int iOffsetToFryday = -weekday + 6;
weekdayComponents.weekday = iOffsetToFryday;
NSDate *nextFriday = [[NSCalendar currentCalendar] dateByAddingComponents:weekdayComponents toDate:today options:0];