两个NSDate之间的时差

在这里，我试图计算两个date之间的小时数。 当我运行应用程序，它崩溃。 你能告诉我这个代码中的错误吗？

NSString *lastViewedString = @"2012-04-25 06:13:21 +0000"; NSDateFormatter *dateFormatter = [[[NSDateFormatter alloc] init] autorelease]; [dateFormatter setDateFormat: @"yyyy-MM-dd HH:mm:ss zzz"]; NSDate *lastViewed = [[dateFormatter dateFromString:lastViewedString] retain]; NSDate *now = [NSDate date]; NSLog(@"lastViewed: %@", lastViewed); //2012-04-25 06:13:21 +0000 NSLog(@"now: %@", now); //2012-04-25 07:00:30 +0000 NSTimeInterval distanceBetweenDates = [now timeIntervalSinceDate:lastViewed]; double secondsInAnHour = 3600; NSInteger hoursBetweenDates = distanceBetweenDates / secondsInAnHour; NSLog(@"hoursBetweenDates: %@", hoursBetweenDates); 

Answer 1

我认为差异应该是int值…

 NSLog(@"hoursBetweenDates: %d", hoursBetweenDates); 

希望对你有帮助..

Answer 2

提到一个类似问题的答案，更好的和苹果认可的方式将使用NSCalendar方法，如下所示：

 - (NSInteger)hoursBetween:(NSDate *)firstDate and:(NSDate *)secondDate { NSUInteger unitFlags = NSCalendarUnitHour; NSCalendar *calendar = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar]; NSDateComponents *components = [calendar components:unitFlags fromDate:firstDate toDate:secondDate options:0]; return [components hour]+1; } 

如果您的目标是iOS 8或更高版本，请使用NSCalendarIdentifierGregorian而不要使用已弃用的NSGregorianCalendar 。

Answer 3

NSInteger无法使用显示

 NSLog(@"%@", hoursBetweenDates); 

改为使用：

 NSLog(@"%d", hoursBetweenDates); 

如果不确定在Apple开发人员文档中使用什么外观： http : //developer.apple.com/library/mac/#documentation/Cocoa/Conceptual/Strings/Articles/formatSpecifiers.html#//apple_ref/doc/uid/TP40004265