Ios 填充了时间[hh:mm:ss]的数组数组

如何创建格式为[hh:mm:ss]的时间数组

数组数组将包括一周中的每一天，然后是一周中每一天的特定时间

当最后我要比较当前时间[hh:mm:ss]与数组中的时间[hh:mm:ss]以找出下一个最近时间之间的时差时，最好的解决方案是什么

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谢谢

下面是一个示例，说明如何使用包含时间（从技术上讲是日期，其中包含时间成分）的NSArray字典。稍后我将演示如何在比较时间时忽略日期部分

基本上，您将为一周中的每一天构建一个NSArray，其中包含您希望在该天使用的所有时间。然后，您将这7个数组中的每一个都添加到NSDictionary中，并为一周中的某一天添加一个键（如果只添加了几天，但它会让您了解：

NSDateFormatter *dateFormat = [[NSDateFormatter alloc] init];
[dateFormat setDateFormat:@"yyyy-MM-dd HH:mm:ss Z"];

// Here are just some dummy times.  The date parts will be ignored in the compare function below.
NSDate *firstTime = [dateFormat dateFromString:@"1970-01-01 07:00:00 +0100"];
NSDate *secondTime   = [dateFormat dateFromString:@"1970-01-01 10:19:00 +0100"];
NSDate *thirdTime = [dateFormat dateFromString:@"1970-01-01 15:00:00 +0100"];

// Build our day of the weeek arrays of times.
NSArray *mondayTimes = @[firstTime, secondTime];
NSArray *tuesdayTimes = @[firstTime, thirdTime];
NSArray *wednesdayTimes = @[firstTime, secondTime, thirdTime];

NSDictionary *weekdayTimes = @{@"Monday": mondayTimes, @"Tuesday" : tuesdayTimes, @"Wednesday" : wednesdayTimes };

拥有NSDictionary后，可以通过简单的键查找从字典中检索时间数组，如下所示：

NSArray *tuesdayTimesFromDictionary = [weekdayTimes objectForKey:@"Tuesday"];

要将所有时间与当前时间进行比较，只需在NSDATE数组中循环，找到自现在以来时间间隔最短的时间：

NSDate *now = [NSDate date];
double minTimeSinceNow = MAXFLOAT;
NSDate *closestTime;
NSDateFormatter *timeFormat = [[NSDateFormatter alloc] init];
[timeFormat setDateFormat:@"HH:mm:ss Z"];

for( NSDate *time in tuesdayTimesFromDictionary)
{
    double timeDiff = [self timeDifferenceBetween:now andDate:time];
    NSLog( [NSString stringWithFormat:@"Time compare between to now(%@) and %@ is %f seconds away", [timeFormat stringFromDate:now], [timeFormat stringFromDate:time], timeDiff]);

    if( abs(timeDiff) < minTimeSinceNow)
    {
        minTimeSinceNow = abs(timeDiff);
        closestTime = time;
    }
}

NSLog( [NSString stringWithFormat:@"Closest time to now (%@) is: %@, which is %f seconds away", [timeFormat stringFromDate:now], [timeFormat stringFromDate:closestTime], minTimeSinceNow]);

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我想你会想要一本NSArrays字典，字典中的每个条目都有一周中某一天的键。对于时间数组，我只需要使用NSDate，它可以让你轻松地计算数组中两个时间之间的差异。非常感谢！这真的很有帮助，我将在这个周末尝试使用它看看我想到了什么！

- (double)timeDifferenceBetween:(NSDate *)date1 andDate:(NSDate *)date2
{

    unsigned int flags = NSHourCalendarUnit | NSMinuteCalendarUnit | NSSecondCalendarUnit;
    NSCalendar* calendar = [NSCalendar currentCalendar];

    NSDateComponents* components1 = [calendar components:flags fromDate:date1];
    NSDateComponents* components2 = [calendar components:flags fromDate:date2];

    NSDate* timeOnly1 = [calendar dateFromComponents:components1];
    NSDate* timeOnly2 = [calendar dateFromComponents:components2];

    return [timeOnly1 timeIntervalSince1970] - [timeOnly2 timeIntervalSince1970];

}