Get Formatted Date From Timestamp With Rounded Milliseconds Bash Shell Script

Question

I need to get a date in a specific format but can't work out how to do it.

Here is how I'm getting the date at the moment.

date -r "$timestamp" +'%Y-%m-%dT%H:%M:%S.s'

However the issue is the milliseconds has too many digits for the format I need. I need the milliseconds to be limited to 3 digits.

Any idea how I can do such a thing?

Current Workaround

Not accurate but it works. I calculate the milliseconds afterwards and then just take the first three characters of the string. Obviously this doesn't take into account round up.

date_string_one=`date -r "$timestamp" +'%Y-%m-%dT%H:%M:%S.'`
date_string_milli=`date -r "$timestamp" +'%s'`
date_string="$date_string_one"`printf "%.3s" "$date_string_milli"`

Answer 1

You may simply use %3N to truncate the nanoseconds to the 3 most significant digits:

$ date +"%Y-%m-%d %H:%M:%S,%3N"
2014-01-08 16:00:12,746

or

$ date +"%F %T,%3N"
2014-01-08 16:00:12,746

testet with »GNU bash, Version 4.2.25(1)-release (i686-pc-linux-gnu)«

But be aware, that %N may not implemented depending on your target system or bash version. Tested on an embedded system »GNU bash, version 4.2.37(2)-release (arm-buildroot-linux-gnueabi)« there was no %N:

date +"%F %T,%N"
2014-01-08 16:44:47,%N

Answer 2

Million ways to do this.. one simple way is to remove the trailing numbers...

date +'%Y-%m-%d %H:%M:%S.%N' | sed 's/[0-9][0-9][0-9][0-9][0-9][0-9]$//g'

or

date +'%Y-%m-%d %H:%M:%S.%N' | sed 's/......$//g'