java.util.Date format SSSSSS: if not microseconds what are the last 3 digits? -

just tested code on both windows (8) workstation , aix:

    public static void main(string[] args) {         system.out.println(new simpledateformat("yyyy-mm-dd hh:mm:ss.ssssss").format(new date()));         system.out.println(new simpledateformat("yyyy-mm-dd hh:mm:ss.ssssss").format(new date()));     } 

and got similar result:

2013-10-07 12:53:26.000905 2013-10-07 12:53:26.000906 

can please explain me last digits, if not microseconds?

note: interact db2 database in chronological data stored using timed columns timestamp 6 digits after seconds i.e. microseconds (imo). "timestamps" created requesting following query:

select current timestamp currenttimestamp table ( values (1)) temp 

i wonder if, given above results, couldn't use in code new date() instead of selecting current timestamp database.

thanks.

ps: searched found no relevant (answered) questions, like: current time in microseconds in java or get time hour, minute, second, millisecond, microsecond

from documentation of simpledateformat:

letter     date or time component     presentation     examples   s          millisecond                number           978  

so milliseconds, or 1/1000th of second. format on 6 digits, add 3 leading zeroes...

you can check way:

    date d =new date();     system.out.println(new simpledateformat("yyyy-mm-dd hh:mm:ss.s").format(d));     system.out.println(new simpledateformat("yyyy-mm-dd hh:mm:ss.ss").format(d));     system.out.println(new simpledateformat("yyyy-mm-dd hh:mm:ss.sss").format(d));     system.out.println(new simpledateformat("yyyy-mm-dd hh:mm:ss.ssss").format(d));     system.out.println(new simpledateformat("yyyy-mm-dd hh:mm:ss.sssss").format(d));     system.out.println(new simpledateformat("yyyy-mm-dd hh:mm:ss.ssssss").format(d)); 

output:

2013-10-07 12:13:27.132 2013-10-07 12:13:27.132 2013-10-07 12:13:27.132 2013-10-07 12:13:27.0132 2013-10-07 12:13:27.00132 2013-10-07 12:13:27.000132 

(ideone fiddle)