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

Question

I need something simple like date, but in seconds since 1970 instead of the current date, hours, minutes, and seconds.

date doesn't seem to offer that option. Is there an easy way?

Answer 1

This should work:

date +%s

As recently corrected in the date manual:

%s = seconds since the Epoch (1970-01-01 00:00 UTC)

Answer 2

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

Answer 3

Pure bash solution

Since bash 5.0 (released on 7 Jan 2019) you can use the built-in variable EPOCHSECONDS.

$ echo $EPOCHSECONDS
1547624774

There is also EPOCHREALTIME which includes fractions of seconds.

$ echo $EPOCHREALTIME
1547624774.371210

EPOCHREALTIME can be converted to micro-seconds (μs) by removing the decimal point. This might be of interest when using bash's built-in arithmetic (( expression )) which can only handle integers.

$ echo ${EPOCHREALTIME/./}
1547624774371210

In all examples from above the printed time values are equal for better readability. In reality the time values would differ since each command takes a small amount of time to be executed.

Answer 4

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

Answer 5

With most Awk implementations:

awk 'BEGIN {srand(); print srand()}'

Answer 6

This is an extension to what @pellucide has done, but for Macs:

To determine the number of seconds since epoch (Jan 1 1970) for any given date (e.g. Oct 21 1973)

$ date -j -f "%b %d %Y %T" "Oct 21 1973 00:00:00" "+%s"
120034800

Please note, that for completeness, I have added the time part to the format. The reason being is that date will take whatever date part you gave it and add the current time to the value provided. For example, if you execute the above command at 4:19PM, without the '00:00:00' part, it will add the time automatically. Such that "Oct 21 1973" will be parsed as "Oct 21 1973 16:19:00". That may not be what you want.

To convert your timestamp back to a date:

$ date -j -r 120034800
Sun Oct 21 00:00:00 PDT 1973

Apple's man page for the date implementation: https://developer.apple.com/library/mac/documentation/Darwin/Reference/ManPages/man1/date.1.html

Answer 7

use this bash script (my ~/bin/epoch):

#!/bin/bash

# get seconds since epoch
test "x$1" == x && date +%s && exit 0

# or convert epoch seconds to date format (see "man date" for options)
EPOCH="$1"
shift
date -d @"$EPOCH" "$@"