Example:
Repeat every 2nd week on Mondays every year between January 15th and May 10th.
I don't think so. This is the closest you can get IMO:
DTSTART;VALUE=DATE:20160115
RRULE:FREQ=YEARLY;BYWEEKNO=3,5,7,9,11,13,15,17,19;BYDAY=MO
Unfortunately, this will expand January 13th or 14th in certain years, which also means that all other instances in that year will be off by one week (see the first 100 instances). I think there are a few dates for which this would be correct though (because certain days in the year are always in the same calendar week).
The only way to do that is, if start and end date of the range match the start and end of a month.
I.e., in order to expand an event to every 2nd Monday between January 1st and May 31st each year you could do this:
DTSTART;VALUE=DATE:20160101
RRULE:FREQ=YEARLY;BYMONTH=1,2,3,4,5;BYDAY=MO;BYSETPOS=1,3,5,7,9,11,13,15,17,19,21,23,25,27
Here are the first 100 instances of that rule. Though some RRULE implementations might not support this properly.
i'm trying to use the Datediff function to give me the years off a user but this is more complicated than i thought.
SELECT DATEDIFF( DD,'1963-07-22','2016-07-23')
This will give me 19360 Days i think that is because 2016 have a Leap Year and that is fine.
what i would like to do is get the YEAR and not the days.
if i change the interval from DD to YY(YYYY) it only calculates the year.
In my experience it does work best to use the number of days between the two dates and then divide that amount by 365.25 to be exact, then round off to even years. This would give you the most precise age in years I think.
The correct answer to calculate someone's age, or the difference in truncated years between two dates is
year(#today)-year(#birthDate)+floor((month(#today)-month(#birthdate)+floor((day(#today)-day(#birthdate))/31))/12);
This will work regardless of leap years. And correct for whether the person was born on a later month or even a later day in the same month. This will also ignore hours and minutes, as should be when calculating someone's age.
When using "yyyy" in DateDiff only the year parts of the dates are used to calculate the difference. The month and day are omitted. This will produce results that are sometimes correct and sometimes incorrect by one year.
Try using this instead.
SELECT Int((#2016-07-23#-#1963-07-22#)/365.25);
I am trying to write an algorithm that takes an input of a date ("2000-01-01") and also "y|m|d", where y is the number of years to add to the original date and m and d are the months and days. This algorithm needs to take into account leap years as well. Sorry I haven't posted any code, I haven't really got anything working yet.
The algorithm also needs to work like this: if you add three months to 30th November, you would get 28th February the next year, or 29th if it is a leap year. So if the month you are coming from as a length higher than the final resultant month, the last day of the final resultant month will be taken.
Could someone please give me some pointers on how to write it or link to any useful resources?
I hope that this makes sense, any questions let me know. Thanks.
You seem to have the problem quite well figured out. Here is the algorithm for finding a leap year:
if year is not divisible by 4 then common year
else if year is not divisible by 100 then leap year
else if year is not divisible by 400 then common year
else leap year
Most, if not all of what you need can be found in the <ctime> library (http://www.cplusplus.com/reference/ctime/).
In particular look up the time_t and struct tm types and how to convert between the two (localtime(), gmttime(), ...). Note that you can add days, hours, months, etc to the fields in a struct tm and they are handled properly, e.g., the "32nd of January" becomes "Feb. 1st".
The <ctime> library also handles leap years.
Your way of adding 3 months seems non conventinal, so you might have to check this by hand. Still easy enough by using <ctime>.
EDIT: Before somebody complains that <ctime> is c++, in plain c this library is in <time.h>.
I m trying to write a code in C on caclulating days until next thursday. I need help, using this paragrah.
http://en.wikipedia.org/wiki/Doomsday_rule#Finding_the_day_of_the_week_of_a_given_calendar_date
any ideas on how i should apporach it, and rough guideline
Separate your problem in smaller problems.
When you feel like a smaller problem is small enough to solve, solve that problem.
For instance: your master problem is "using the technique described in the wikipedia article calculate how many days are there till next Thursday".
Problems small enough to make into a function:
calculate (or lookup) a year's doomsday (Monday for 2011) yeardoomsday(2011)
calculate (or lookup) a month doomsday (7 for March) monthdoomsday(2011, 3)
calls yeardoomsday(2011) somewhere inside the function
...
I am stuck on the following problem from my C programming class:
Write a program that prompts the user to input a year, and then outputs the calendar(for the entire year).
I have no idea how to approach this problem. I can usually start my homework problems (this is an optional challenge problem), but I am really lost. We've worked through chapters 1-10 of Deitel & Deitel (loops, arrays, pointers, I/O, etc), but I don't know how to approach this at all. Any hints or suggestions would be appreciated.
It might help you to understand the mathematics of the calendar. If the fabulous book Calendrical Calculations is not in your university library, they may be able to get you a reprint of the article by the same authors in Software—Practice & Experience. And ask your prof to request the book for the library.
In general, when you have a big problem like this one, you want to break it down into little problems that are easier to solve.
Here's one possible little problem to start with: if you know how many days there are in a month, and what weekday the first of the month falls on, could you output a calendar for that month?
The hardest part is determining which day of the week the year starts on.
http://en.wikipedia.org/wiki/Calculating_the_day_of_the_week
But even without that knowledge, when I first implemented this, I used a reference date (for example, you know that today, January 11, 2010 is a Monday) and counted days from there. (Just keep in mind that leap years have an extra day, and that leap years are every 4 years except every 100 years except every 400 years.)
http://en.wikipedia.org/wiki/Leap_year
Does this code qualify? :-)
char command[]="cal 2010";
sprintf(command,"cal %d",argv[1]);
system(command);
It assumes a Unix machine with cal in the path.
you need to find out first the day on 1st of january
and then print the dates.
GO to https://sourceforge.net/projects/c-cpp-calender/
go through the code and you will understand it
A good start may be the localtime(3) and mktime(3) functions. Alternatively, you can implement the relevant date arithmetic from scratch. Then, simply, generate the first line of the calendar (find the weekday that corresponds to January 1st, then print taht in the right place, followed by the rest of the week), then print all but the last lines, then print the last line.
Depending on if you want a calendar paginated by month or not, this MAY be better done ona per-month instead of a per-year basis.
Well, first figure out the algorithmic part of your problem - given a year, find what day Jan 1st is.
After this, just note the number of days in each month (store it in an array, say num_days[]), and the note the number of months in a year and an array of strings for the months.
For e.g. the outermost loop iterates over the months. Say, iterate for(i=0;i<NumMonths;++i). Then, for each month, print the string, e.g. month[i], then a newline.
Then, with simple tabs, print Sun Mon Tue ..., and another newline.
Then using the day Jan 1st corresponds to (call it FirstDay), insert spaces, and start with that day. Keep printing the dates and newlines till you hit max_month[i] which is 31 (for January). Store the name of day of the last day of the previous month and just reiterate treating that day as FirstDay.
You need a couple pieces to start. First, you need a formula that computes the day of the week for January 1 of whatever year is entered. You'll also need a formula to determine if the year is a leap year. Both of these formulae are easily found with a simple Google search. The third item you need is a simple array containing the number of days in each of the 12 months for a non-leap year.
Once you have these things, its trivial to determine the week day for each month of the year. Make sure to account for February 29 in a leap year. From there, you just need to create a function that primts out the monthly calendar in a form that looks similar to the calendar hanging on the wall. Try sketching out the desired layout on paper first and use that as a template for creating the appropriate format statements.
Might check out the doomsday algorithm. This would get you certain "dooms days" like Jan 31 is a doomsday, for 2008, that was a saturday. You can work backwords from there
Basically, there are two approaches:
The easy / pragmatical way: Solve the task, and forget about everything else. Here you may check documentation for mktime() (you'll find an example based on mktime below..).
The scientific / engineering way: Learn to know how it works! You can start at the great wikipedia article about the gregorian calendar. Read it, understand it, and write code that implements the underlying algorithms (which are known, no rocket science, it is possible). This will improve your skills a lot (in fact, you really should do such a thing, maybe not the calender but another topic, it will give you a big leap in understanding all things).
Now some pragmatic code to start with. mktime() has a great feature: it knows the calender details, and it accepts e.g. a date "2010-01-60" and will convert it into February 29th, 2010. But, this will only work for dates after 1970. It will not work for earlier dates (though I am not 100% sure, but it shouldn't work, because unix time starts at January 1st, 1970, but try with other dates, maybe mktime() is not restricted to unix time).
Pseudo-Code, this prints each day on a single line (YYYY-MM-DD):
void print_cal( int year ) {
static char weekdays[] = { "Sun", "Mon", "Tue", "Wed", "Thu", "Fri", "Sat" };
struct tm tm;
for( int day=0; day<365; ++day ) {
memset( &tm, 0, sizeof(tm) );
tm.tm_year = year - 1900;
tm.tm_mday = day;
mktime( &tm ); // modifies tm
printf( "%04d-%02d-%02d, %s\n", tm.tm_year, tm.tm_mon+1, tm.tm_mday, weekdays[tm.tm_wday] );
}
}
This code ignores leap years. You still have to adjust it to be correct for leap years! Also the result is not very pretty yet, just one line per day.
EDIT: added output of weekdays.