On Jul 1, 2008, at 5:20 PM, Jason Wiggins wrote:
It look like what I thought was going to be something simple, has
turned out to be less so and fraught with danger programatically.
I was just thinking though, would it be easier just to covert the
weekday value into seconds (weekday * 60 * 60 * 24) and subtract
that from the current date? This would save you from the issue of
year and month boundaries. Following code is untested (written in
Mail):
I'm sorry if I've obfuscated Chris's reply. I'd thought it would be
possible to save a few steps by reusing the date components object you
get from decomposing the start date, but Chris indicates that the
short-cut is not guaranteed to work in all situations. His robust
solution is to do the following:
NSDate *today = [NSDate date];
[NSCalendar *gregorian = [[NSCalendar alloc]
initWithCalendarIdentifier:NSGregorianCalendar];
// Get the weekday component of the current date
NSDateComponents *weekdayComponents = [gregorian
components:NSWeekdayCalendarUnit fromDate:today];
/*
Create a date components to represent the number of days to subtract
from the current date.
The weekday value for Sunday in the Gregorian calendar is 1, so
subtract 1 from the number
of days to subtract from the date in question. (If today's Sunday,
subtract 0 days.)
*/
NSDateComponents *componentsToSubtract = [[NSDateComponents alloc]
init];
[componentsToSubtract setDay: - ([weekdayComponents weekday] - 1)];
NSDate *beginningOfWeek = [gregorian
dateByAddingComponents:componentsToSubtract toDate:today options:0];
/*
Optional step:
beginningOfWeek now has the same hour, minute, and second as the
original date (today).
To normalize to midnight, extract the year, month, and day components
and create a new date from those components.
*/
NSDateComponents *components = [gregorian components:
(NSYearCalendarUnit | NSMonthCalendarUnit | NSDayCalendarUnit)
fromDate: beginningOfWeek];
beginningOfWeek = [gregorian dateFromComponents: components];
mmalc
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