gpt4 book ai didi

objective-c - 如何获得当前一周的天数?

转载 作者:行者123 更新时间:2023-12-01 17:59:54 25 4
gpt4 key购买 nike

如何获得当前一周中的几天?
例如,今天是saturday - 28 july
我需要带有'28 Saturday',27 Friday26 Thursday25 Wednesday ... 22 Sunday的数组?

[代码] NSCalendar * myCalendar = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];

NSDateComponents *currentComps = [myCalendar components:( NSYearCalendarUnit | NSMonthCalendarUnit | NSWeekOfYearCalendarUnit | NSWeekdayCalendarUnit | NSHourCalendarUnit | NSMinuteCalendarUnit) fromDate:weekDate];
int ff = currentComps.weekOfYear;
NSLog(@"1 %d", ff);

[currentComps setWeekday:1]; // 1: sunday
NSDate *firstDayOfTheWeek = [myCalendar dateFromComponents:currentComps];
[currentComps setWeekday:7]; // 7: saturday
NSDate *lastDayOfTheWeek = [myCalendar dateFromComponents:currentComps];

NSLog(@"first - %@ \nlast - %@", firstDayOfTheWeek, lastDayOfTheWeek); [code]

我做到了,但是我有“firstDayOfWeek”(2012-07-29)和“lastDayOfWeek”(2012-08-04)

最佳答案

这是您要寻找的内容-您只需要相应地格式化结果即可:

NSDate *weekDate = [NSDate date];
NSCalendar *myCalendar = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];

NSDateComponents *currentComps = [myCalendar components:( NSYearCalendarUnit | NSMonthCalendarUnit | NSWeekOfYearCalendarUnit | NSWeekdayCalendarUnit | NSHourCalendarUnit | NSMinuteCalendarUnit) fromDate:weekDate];
int ff = currentComps.weekOfYear;
NSLog(@"1 %d", ff);

[currentComps setWeekday:1]; // 1: sunday
NSDate *firstDayOfTheWeek = [myCalendar dateFromComponents:currentComps];
[currentComps setWeekday:7]; // 7: saturday
NSDate *lastDayOfTheWeek = [myCalendar dateFromComponents:currentComps];

NSDateFormatter *myDateFormatter = [[NSDateFormatter alloc] init];
myDateFormatter.dateFormat = @"dd EEEE";
NSString *firstStr = [myDateFormatter stringFromDate:firstDayOfTheWeek];
NSString *secondStr = [myDateFormatter stringFromDate:lastDayOfTheWeek];

NSLog(@"first - %@ \nlast - %@", firstStr, secondStr);

关于objective-c - 如何获得当前一周的天数?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/11700707/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com