Shell – How to set a file’s mtime in epoch format

On *BSD:

date -r 1234567890

On Linux (specifically, with GNU coreutils ≥5.3):

date -d @1234567890

With older versions of GNU date, you can calculate the relative difference to the UTC epoch:

date -d '1970-01-01 UTC + 1234567890 seconds'

If you need portability, you're out of luck. The only time you can format with a POSIX shell command (without doing the calculation yourself) line is the current time. In practice, Perl is often available:

perl -le 'print scalar localtime $ARGV[0]' 1234567890

date +%s.%N will give you, eg., 1364391019.877418748. The %N is the number of nanoseconds elapsed in the current second. Notice it is 9 digits, and by default date will pad this with zeros if it is less than 100000000. This is actually a problem if we want to do math with the number, because bash treats numbers with a leading zero as octal. This padding can be disabled by using a hyphen in the field spec, so:

echo $((`date +%s`*1000+`date +%-N`/1000000))

would naively give you milliseconds since the epoch.

However, as Stephane Chazelas points out in comment below, that's two different date calls which will yield two slightly different times. If the second has rolled over in between them, the calculation will be an entire second off. So:

echo $(($(date +'%s * 1000 + %-N / 1000000')))

Or optimized (thanks to comments below, though this should have been obvious):

echo $(( $(date '+%s%N') / 1000000));