Number of Tuesdays in the current month in objective c - objective-c

I'm looking for a way to calculate the number of Tuesdays in the current month in Objective C.
For example if the code ran today (July 16, 2012) it would ouput 5 because there are 5 Tuesdays in the month of July in 2012 (the 3rd, 10th, 17th, 24th, and 31st).
I have seen solutions online for doing it in Excel, but I am struggling translating it to Objective C.
Thanks for your help! Long time, first time.

We're going to need a calendar:
NSCalendar *calendar = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
Now let's get the current month and year:
NSDate *now = [NSDate date];
NSDateComponents *monthAndYear = [calendar components:NSMonthCalendarUnit | NSYearCalendarUnit fromDate:now];
We can use that to get the first Tuesday of the current month and year:
NSDateComponents *firstTuesdayComponents = [monthAndYear copy];
firstTuesdayComponents.weekday = 3; // Sunday = 1
firstTuesdayComponents.weekdayOrdinal = 1; // First Tuesday
NSDate *firstTuesday = [calendar dateFromComponents:firstTuesdayComponents];
We can also use it to get the first day of next month:
NSDateComponents *firstOfNextMonthComponents = [monthAndYear copy];
firstOfNextMonthComponents.month += 1;
firstOfNextMonthComponents.day = 1;
NSDate *firstOfNextMonth = [calendar dateFromComponents:firstOfNextMonthComponents];
Now we can ask for the number of days between the two dates:
NSDateComponents *differenceComponents = [calendar components:NSDayCalendarUnit fromDate:firstTuesday toDate:firstOfNextMonth options:0];
Most weeks have seven days and a single Tuesday, so we should divide the number of days by 7. If there's a remainder, we should round it up because we started counting from a Tuesday.
int tuesdayCount = (differenceComponents.day + 6) / 7; // Adding 6 makes the integer division round up.
NSLog(#"There are %d Tuesdays in month %d of year %d.", tuesdayCount, (int)monthAndYear.month, (int)monthAndYear.year);
Now let's hop in the time machine to test it:
There are 5 Tuesdays in month 1 of year 2012.
There are 4 Tuesdays in month 2 of year 2012.
There are 4 Tuesdays in month 3 of year 2012.
There are 4 Tuesdays in month 4 of year 2012.
There are 5 Tuesdays in month 5 of year 2012.
There are 4 Tuesdays in month 6 of year 2012.
There are 5 Tuesdays in month 7 of year 2012.
There are 4 Tuesdays in month 8 of year 2012.
There are 4 Tuesdays in month 9 of year 2012.
There are 5 Tuesdays in month 10 of year 2012.
There are 4 Tuesdays in month 11 of year 2012.
There are 4 Tuesdays in month 12 of year 2012.

When looking at the outputs of cal 2012, cal 2011, and cal 2000, I've drawn a few conclusions:
There are either four or five Tuesdays in every month.
To find the months with five Tuesdays, note:
For months with 31 days, the first must fall on a Sunday, Monday, or Tuesday
For months with 30 days, the first must fall on a Monday or Tuesday
For months with 29 days, the first must fall on a Tuesday
Hopefully this is an easier problem to solve.

Using an algorithm based on the same observation sarnold made in his answer, this should do the trick:
- (NSUInteger)numberOfTuesdaysInMonthContainingDate:(NSDate *)date
{
NSCalendar *calendar = [NSCalendar currentCalendar];
NSUInteger numDaysInMonth = [calendar rangeOfUnit:NSDayCalendarUnit inUnit:NSMonthCalendarUnit forDate:date].length;
NSDate *firstDayOfMonth;
NSTimeInterval firstDayOfMonthLength;
if (![calendar rangeOfUnit:NSMonthCalendarUnit startDate:&firstDayOfMonth interval:&firstDayOfMonthLength forDate:date]) {
NSLog(#"Unable to calculate first day of the month for %#", date);
return 0;
}
NSUInteger firstDayOfTheMonthWeekday = [calendar ordinalityOfUnit:NSWeekdayCalendarUnit inUnit:NSWeekCalendarUnit forDate:firstDayOfMonth];
// The following is for Tuesday only, the math is more complex for an abitrary weekday
NSUInteger numTuesdays = 4;
NSUInteger minValidWeekday = 32 - numDaysInMonth;
if (firstDayOfTheMonthWeekday >= minValidWeekday && firstDayOfTheMonthWeekday <= 3) {
numTuesdays = 5;
}
return numTuesdays;
}

Related

Given an NSDate, find the last day of fourth prior month

I am trying to calculate an NSDate object based on the current date. If the current date is April 1st, 2015, I need to generate the date, December 31, 2014. If the current date is April 30th, 2015, I STILL need to generate the date, December 31, 2014. If however, it is May 1st, 2015, I need to generate January 31st, 2015. In other words, whatever month I am in, I need the date of the end of the month, from four months ago, regardless of where I am in the current month.
The code I have thus far is:
NSCalendar *theCalendar = [NSCalendar currentCalendar];
NSDateComponents *dayComponent = [[NSDateComponents alloc] init];
[dayComponent setDay:-90];
NSDate *nextDate = [theCalendar dateByAddingComponents:dayComponent toDate:[NSDate date] options:0];
NSLog(#"The date I am getting is: %#", nextDate);
The above code gives me the date value of exactly 90 days prior to the current date, but I need the date to always be the end of the month that is 4 months earlier.
As you've already discovered, you need a starting date and a calendar:
NSDate *startingDate = [NSDate date];
NSCalendar *calendar = [NSCalendar autoupdatingCurrentCalendar];
You'll need the components of the current date but only down to the current month, because you don't care about the specific day within the month:
NSDateComponents *components = [calendar
components:NSCalendarUnitEra | NSCalendarUnitYear | NSCalendarUnitMonth
fromDate:startingDate];
You say you want the last day of the fourth prior month. Since months have different numbers of days, the last day varies depending on the month. But all months have first days, and those first days are always numbered 1. So it's easiest to compute “the last day of the fourth prior month” by first going back three months:
components.month -= 3;
Then, go one day prior to that month:
components.day = -1;
Finally, you need to get clear in your head that an NSDate represents an instant in time, but a day (like “April 1st, 2015”) is an interval of time, starting and ending at specific instants. If you're going to represent a whole day using an NSDate, you're going to be storing one instant within that interval. You don't want to store the first or last instant (which will both be midnights); that causes problems for some days in some time zones. Instead, use noon as your instant:
components.hour = 12;
Now you're ready to ask the calendar for a new NSDate:
NSDate *lastDayOfFourthPriorMonth = [calendar dateFromComponents:components];
You want to get familiar with NSCalendar and NSDateComponents. If you read those two API documents, you will be able to assemble the answer.
Conceptually, you want to use NSCalendar components:fromDate: to get the components (day, month, year) of the current date.
You will then walk that month value back 4 months. Now if that wraps past January, you know you need to determine how much, so that you adjust and stay within months [1..12]. Further, you'll need to decrement the year at that point too.
Knowing the month, you can find the last day of that month through several means; the crudest of which is to maintain your own enum...but there's probably a better way using NSCalendar that will also account for February in leap years.
At the end, you can build the resultant date from the components you've assembled, using the NSCalendar method dateFromComponents:.
See #rob mayoff's excellent answer which is a more concrete realization of this theory and IMO, the correct answer.

nsdate format to last digit in year, then day of year

The following code will output this (current day is Jan 15, 2015):
2015015
I need it to output this:
5015
I want it in the format of yddd, where y is the last number in the year, and ddd is the day of the year. so January 1st, 2000 is 0001 and December 31st, 2007 is 7365.
NSDate *julianLabel = [NDDate date];
NSDateFormatter *julianFormatter = [[NSDateFormatter alloc] init];
[julianFormatter setDateFormat:#"yddd"];
self.julian.text = [julianFormatter stringFromDate:julianLabel];
Surely you could just use substringFromIndex on the string, something that should work for the next eight thousand years or so (a). In other words, something like:
self.julian.text = [[julianFormatter stringFromDate:julianLabel] substringFromIndex:3];
(a) It'll work as per your specification until we reach the year 10,000 but keep in mind you'll start getting duplicates in 2025. I'm assuming that's not a problem due to the detail provided.

Programmatically getting the date "next Sunday at 5PM"

Edited 07/08/13: Apple has an excellent set of WWDC videos that really helped me understand the various date and time classes in Objective-C, and how to correctly perform time calculations/manipulations with them.
"Solutions to Common Date and Time Challenges" (HD video, SD video, slides (PDF)) (WWDC 2013)
"Performing Calendar Calculations" (SD video, slides (PDF)) (WWDC 2011)
Note: links require a free Apple Developer membership.
I'm writing an app for a friend's podcast. She broadcasts her show live every Sunday at 5PM, and I would like to write some code in my app to optionally schedule a local notification for that time, so that the user is reminded of when the next live show is. How would I go about getting an NSDate object that represents "the next Sunday, at 5 PM Pacific time." (obviously this would have to be converted into whatever timezone the user is using)
First get the current day of the week:
NSDate *now = [NSDate date];
NSCalendar *calendar = [[NSCalendar alloc] initWithCalendarIdentifier:NSCalendarIdentifierGregorian];
NSDateComponents *dateComponents = [calendar components:NSCalendarUnitWeekday | NSCalendarUnitHour fromDate:now];
NSInteger weekday = [dateComponents weekday];
The Apple docs define a weekday as:
Weekday units are the numbers 1 through n, where n is the number of
days in the week. For example, in the Gregorian calendar, n is 7 and
Sunday is represented by 1.
Next figure out how many days to add to get to the next sunday at 5:
NSDate *nextSunday = nil;
if (weekday == 1 && [dateComponents hour] < 5) {
// The next Sunday is today
nextSunday = now;
} else {
NSInteger daysTillNextSunday = 8 - weekday;
int secondsInDay = 86400; // 24 * 60 * 60
nextSunday = [now dateByAddingTimeInterval:secondsInDay * daysTillNextSunday];
}
To get it at 5:00 you can just change the hour and minute on nextSunday to 5:00. Take a look at get current date from [NSDate date] but set the time to 10:00 am

NSDateFormatter show wrong year

im using xcode 4.5(4G182) with iOS 6. NSDateFormatter show wrong year in iOS 6, how to solve?
NSDateFormatter *dateFormatter = [[[NSDateFormatter alloc] init] autorelease];
[dateFormatter setDateFormat:#"YYYY-MM-dd"];
NSString *str = [dateFormatter stringFromDate:[NSDate date]];
NSLog(#"%# == %#",str,[[dateFormatter dateFromString:str] description]);
it print out "2012-09-14 == 2011-09-13 16:00:00 +0000"
YYYY is not the same as yyyy.
According to this page which the iOS Date format page references;
`y`: Year
`Y`: Year (in "Week of Year" based calendars). This year designation is used in
ISO year-week calendar as defined by ISO 8601, but can be used in
non-Gregorian based calendar systems where week date processing is desired.
May not always be the same value as calendar year.
The operative sentence being the last one. Use yyyyinstead.
Further details on how and why the year values may deviate when using YYYY:
The ISO week-numbering year starts at the first day (Monday) of week
01 and ends at the Sunday before the new ISO year (hence without
overlap or gap). It consists of 52 or 53 full weeks. The ISO
week-numbering year number deviates from the number of the traditional
Gregorian calendar year on a Friday, Saturday, and Sunday, or a
Saturday and Sunday, or just a Sunday, at the start of the traditional
Gregorian calendar year (which are at the end of the previous ISO
week-numbering year) and a Monday, Tuesday and Wednesday, or a Monday
and Tuesday, or just a Monday, at the end of the traditional Gregorian
calendar year (which are in week 01 of the next ISO week-numbering
year). For Thursdays, the ISO week-numbering year number is always
equal to the traditional Gregorian calendar year number.
Examples:
Monday 29 December 2008 is written "2009-W01-1"
Sunday 3 January 2010 is written "2009-W53-7"
From https://en.wikipedia.org/wiki/ISO_8601#Week_dates
(bold styling added)

Pick a day in a cycle based on a starting date and the current date

An app allows a user to define an arbitrary set of days like so:
Day 1 - Pick flowers
Day 2 - Have coffee
Day 3 - Go swimming
This set of days may have only one entry or it may have many. Suppose that this cycle of days begins on March 23, 2010.
Is there a general algorithm that can determine which day in a cycle corresponds to a given calendar day?
On March 24th, the algorithm should return Day 2, March 25th should be Day 3, March 26th should be Day 1, and so on...
More specifically, I am writing this code for a Cocoa application. So, barring the existence of a more general technique can the Calendar classes help me?
NSCalendar *gregorian = ...;
NSDate *startDate = ...;
NSDate *currentDate = [NSDate date];
NSUInteger unitFlags = NSDayCalendarUnit;
NSDateComponents *components = [gregorian components:unitFlags fromDate:startDate toDate:currentDate options:0];
NSInteger days = [components day];
days now contains the number of days which have elapsed since the start date. This can be used as an index (zero-based) into your array of activities. You can use the modulus operator with the length of the cycle as the dividend to compute indices after the first transit around the cycle.
Calculate the Julian date (or a variant) of the start date and the current date (This part is left as an exercise to the reader). The current day of the cycle is then (currentDate - startDate) % lengthOfCycle + 1.
After typing this out I figured out the answer and it is straightforward.
Calculate the number of days between the starting date and the current date then take that number modulo the total number of days in the cycle and you have the current day. Easy peezy.