[[NSDate date] timeIntervalSince1970];

它以双精度返回自epoch以来的秒数。我几乎可以肯定你可以从小数部分访问毫秒。

这基本上和@TristanLorach发布的答案是一样的，只是为Swift 3重新编码:

   /// Method to get Unix-style time (Java variant), i.e., time since 1970 in milliseconds. This 
   /// copied from here: http://stackoverflow.com/a/24655601/253938 and here:
   /// http://stackoverflow.com/a/7885923/253938
   /// (This should give good performance according to this: 
   ///  http://stackoverflow.com/a/12020300/253938 )
   ///
   /// Note that it is possible that multiple calls to this method and computing the difference may 
   /// occasionally give problematic results, like an apparently negative interval or a major jump 
   /// forward in time. This is because system time occasionally gets updated due to synchronization 
   /// with a time source on the network (maybe "leap second"), or user setting the clock.
   public static func currentTimeMillis() -> Int64 {
      var darwinTime : timeval = timeval(tv_sec: 0, tv_usec: 0)
      gettimeofday(&darwinTime, nil)
      return (Int64(darwinTime.tv_sec) * 1000) + Int64(darwinTime.tv_usec / 1000)
   }

[[NSDate date] timeIntervalSince1970];

它以双精度返回自epoch以来的秒数。我几乎可以肯定你可以从小数部分访问毫秒。

这是我给斯威夫特用的

var date = NSDate()
let currentTime = Int64(date.timeIntervalSince1970 * 1000)

print("Time in milliseconds is \(currentTime)")

使用本网站验证准确性http://currentmillis.com/

let timeInMiliSecDate = Date()
let timeInMiliSec = Int (timeInMiliSecDate.timeIntervalSince1970 * 1000)
print(timeInMiliSec)

NSTimeInterval time = ([[NSDate date] timeIntervalSince1970]); //double
long digits = (long)time; //first 10 digits        
int decimalDigits = (int)(fmod(time, 1) * 1000); //3 missing digits
/*** long ***/
long timestamp = (digits * 1000) + decimalDigits;
/*** string ***/
NSString *timestampString = [NSString stringWithFormat:@"%ld%03d",digits ,decimalDigits];