Get current time in seconds since the Epoch on Linux, Bash

This should work:

date +%s

Just to add.

Get the seconds since epoch(Jan 1 1970) for any given date(e.g Oct 21 1973).

date -d "Oct 21 1973" +%s

Convert the number of seconds back to date

date --date @120024000

The command date is pretty versatile. Another cool thing you can do with date(shamelessly copied from date --help). Show the local time for 9AM next Friday on the west coast of the US

date --date='TZ="America/Los_Angeles" 09:00 next Fri'

Better yet, take some time to read the man page http://man7.org/linux/man-pages/man1/date.1.html

So far, all the answers use the external program date.

Since Bash 4.2, printf has a new modifier %(dateformat)T that, when used with argument -1 outputs the current date with format given by dateformat, handled by strftime(3) (man 3 strftime for informations about the formats).

So, for a pure Bash solution:

printf '%(%s)T\n' -1

or if you need to store the result in a variable var:

printf -v var '%(%s)T' -1

No external programs and no subshells!

Since Bash 4.3, it's even possible to not specify the -1:

printf -v var '%(%s)T'

(but it might be wiser to always give the argument -1 nonetheless).

If you use -2 as argument instead of -1, Bash will use the time the shell was started instead of the current date. This can be used to compute elapsed times

$ printf -v beg '%(%s)T\n' -2$ printf -v now '%(%s)T\n' -1$ echo beg=$beg now=$now elapsed=$((now-beg))beg=1583949610 now=1583953032 elapsed=3422