我想检查 2 NSDate
秒它们是否在同一天(时间可以不同)。我目前拥有的是这个:
- (NSPredicate*) predicateWithDate:(NSDate *)date {
NSCalendar *calendar = [NSCalendar currentCalendar];
NSDateComponents *components = [calendar components:(NSYearCalendarUnit | NSMonthCalendarUnit ) fromDate:date];
//create a date with these components
NSDate *startDate = [calendar dateFromComponents:components];
[components setMonth:0];
[components setDay:1];
[components setYear:0];
NSDate *endDate = [calendar dateByAddingComponents:components toDate:startDate options:0];
return [NSPredicate predicateWithFormat:@"((ANY notes.date >= %@) AND (ANY notes.date < %@))",startDate,endDate];
}
你的开始日期,结束日期计算看起来几乎正确,你只是忘记了NSDayCalendarUnit
:
NSDateComponents *components = [calendar components:(NSYearCalendarUnit | NSMonthCalendarUnit | NSDayCalendarUnit) fromDate:date];
有多种方法可以计算当天的开始和结束日期,我更喜欢以下稍短的代码:
NSCalendar *calendar = [NSCalendar currentCalendar];
NSDate *startDate;
NSTimeInterval interval;
[calendar rangeOfUnit:NSDayCalendarUnit startDate:&startDate interval:&interval forDate:date];
NSDate *endDate = [startDate dateByAddingTimeInterval:interval];
您的谓词将找到所有带有date >= startDate
,以及任何(可能不同)带有date < endDate
的注释.
如果要查找在给定日期至少有一个注释的对象,则需要一个 SUBQUERY:
[NSPredicate predicateWithFormat:@"SUBQUERY(notes, $n, $n.date >= %@ AND $n.date < %@).@count > 0",
startDate,endDate];