ios - 当我以格式字符串调用NSLog并进一步使用该时间时，如何获得时间？

Question

当我以格式字符串调用NSLog并进一步使用该时间时，如何获得时间？
我需要几微秒的时间？

2017-09-28 20:19:39.383263+0300 NIR[5677:1383138] а 

Answer 1

这是您可以使用的方法。它返回带有当前时间戳的(NSString *)。有点混乱，因为没有格式化程序字符串来获取包括毫秒在内的秒数:

- (NSString*)dateTimeWithMS
{
    CFAbsoluteTime timeInSeconds = CFAbsoluteTimeGetCurrent();
    NSDate *d = [NSDate dateWithTimeIntervalSinceReferenceDate:timeInSeconds];
    NSDateFormatter *df = [[NSDateFormatter alloc] init];
    df.dateFormat = @"yyyy-MM-dd H:mm:ss";
    NSString *s = [df stringFromDate:d];
    int ms = (int) round(1000 * (timeInSeconds - floor(timeInSeconds)));

    NSString *result = [NSString stringWithFormat:@"%@\.%000d", s, ms];

    return(result);
}

基本上，我们得到时间，使用标准格式，然后附加毫秒。

您可以使用以下方法进行测试:

    for(int i = 0; i < 10; ++i)
        NSLog(@"Time #%d with MS = %@", i, [self dateTimeWithMS]);

您将获得类似以下的输出-您可以看到Xcode的时间戳与我们的时间戳匹配:

2017-09-28 18:28:10.781 TimeMillisecondTest[2497:463749] Time #0 with MS = 2017-09-28 18:28:10.780
2017-09-28 18:28:10.781 TimeMillisecondTest[2497:463749] Time #1 with MS = 2017-09-28 18:28:10.781
2017-09-28 18:28:10.781 TimeMillisecondTest[2497:463749] Time #2 with MS = 2017-09-28 18:28:10.782
2017-09-28 18:28:10.781 TimeMillisecondTest[2497:463749] Time #3 with MS = 2017-09-28 18:28:10.782
2017-09-28 18:28:10.782 TimeMillisecondTest[2497:463749] Time #4 with MS = 2017-09-28 18:28:10.782
2017-09-28 18:28:10.782 TimeMillisecondTest[2497:463749] Time #5 with MS = 2017-09-28 18:28:10.782
2017-09-28 18:28:10.782 TimeMillisecondTest[2497:463749] Time #6 with MS = 2017-09-28 18:28:10.782
2017-09-28 18:28:10.782 TimeMillisecondTest[2497:463749] Time #7 with MS = 2017-09-28 18:28:10.783
2017-09-28 18:28:10.782 TimeMillisecondTest[2497:463749] Time #8 with MS = 2017-09-28 18:28:10.783
2017-09-28 18:28:10.783 TimeMillisecondTest[2497:463749] Time #9 with MS = 2017-09-28 18:28:10.783