How do I convert an epoch timestamp to a human readable format on the cli?

How do I convert an epoch timestamp to a human readable format on the cli? I think there’s a way to do it with date but the syntax eludes me (other ways welcome).

Asked By: xenoterracide

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date -d @1190000000 Replace 1190000000 with your epoch

Answered By: fschmitt
$ echo 1190000000 | perl -pe 's/(d+)/localtime($1)/e' 
Sun Sep 16 20:33:20 2007

This can come in handy for those applications which use epoch time in the logfiles:

$ tail -f /var/log/nagios/nagios.log | perl -pe 's/(d+)/localtime($1)/e'
[Thu May 13 10:15:46 2010] EXTERNAL COMMAND: PROCESS_SERVICE_CHECK_RESULT;HOSTA;check_raid;0;check_raid.pl: OK (Unit 0 on Controller 0 is OK)
Answered By: Stefan Lasiewski

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

The two I frequently use are:

$ perl -leprint scalar localtime 1234567890
Sat Feb 14 00:31:30 2009

and

$ tclsh
% clock format 1234567890
Sa Feb 14 00:31:30 CET 2009
Answered By: al.

Custom format with GNU date:

date -d @1234567890 +'%Y-%m-%d %H:%M:%S'

Or with GNU awk:

awk 'BEGIN { print strftime("%Y-%m-%d %H:%M:%S", 1234567890); }'

Linked SO question: https://stackoverflow.com/questions/3249827/convert-from-unixtime-at-command-line

Answered By: Ivan Chau

With zsh you could use the strftime builtin:

strftime format epochtime

      Output the date denoted by epochtime in the format specified.

e.g.

zmodload zsh/datetime
strftime '%A, %d %b %Y' 1234567890
Friday, 13 Feb 2009

There’s also dateconv from dateutils:

dateconv -i '%s' -f '%A, %d %b %Y' 1234567890
Friday, 13 Feb 2009

keep in mind dateutils tools default to UTC (add -z your/timezone if needed).

Answered By: don_crissti

With bash-4.2 or above:

$ printf '%(%FT%T%z)Tn' 1234567890
2009-02-13T23:31:30+0000

(where %FT%T%z is the strftime()-type format, here using standard unambiguous format which includes the UTC offset (%z))

That syntax is inspired from ksh93.

In ksh93 however, the argument is taken as a date expression where various and hardly documented formats are supported.

For a Unix epoch time, the syntax in ksh93 is:

printf '%(%FT%T%z)Tn' '#1234567890'

ksh93 however seems to use its own algorithm for the timezone and can get it wrong. For instance, in mainland Britain, it was summer time all year in 1970, but:

$ TZ=Europe/London bash -c 'printf "%(%c)Tn" 0'
Thu 01 Jan 1970 01:00:00 BST
$ TZ=Europe/London ksh93 -c 'printf "%(%c)Tn" "#0"'
Thu Jan  1 00:00:00 1970

ksh93 (and zsh’s strftime builtin) support subsecond, not bash yet:

$ ksh -c 'printf "%(%FT%T.%6N%z)Tn" 1234567890.123456789'
2009-02-13T23:31:30.123456-0000
$ zsh -c 'zmodload zsh/datetime; strftime %FT%T.%6.%z 1234567890 123456780'
2009-02-13T23:31:30.123457+0000
Answered By: Stéphane Chazelas

You could also use a little C program for printing the datetime in the format that can be directly parsed by shell

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>

int main(int argc, char * argv[]) {

    if (argc==1) { 
        return 1;
    }

    struct tm input_tm;
    char * formatStr  = "YEAR=%YnMON=%mnDAY=%dnHOUR=%HnMIN=%MnSEC=%S";
    size_t formatSize = strlen(formatStr) + 2;
    char * output     = (char *)malloc(sizeof(char)*formatSize);

    strptime(argv[1],"%s",&input_tm);
    strftime(output, formatSize, formatStr, &input_tm);

    printf("%sn",output);
    free(output);
    return 0;
}

usage:

#compile
clang -o epoch2datetime main.c

#invoke
eval `./epoch2datetime 1450196411`
echo $YEAR $MON $DAY $HOUR $MIN $SEC
#output
#2015 12 16 00 20 11
Answered By: Meow

Wouldn’t be a real solution without a little node.js:

epoch2date(){
    node -p "new Date($1)"
}

add that to ~/.bash_aliases and make sure its sourced in ~/.bashrc with . ~/.bash_aliases

if [ -f ~/.bash_aliases ]; then
    . ~/.bash_aliases
fi

To get node on your system goto http://nvm.sh and run the curl command. It’ll install node version manager (nvm) which allows you to switch versions of node.

Just type nvm ls-remote and pick a version to nvm install <version>.

Answered By: chovy

If your epoch time is in milliseconds instead of seconds, either put a dot before last three digits (as hinted in comments by user79743), or remove the last three digits before passing it to date -d:

Entered directly, this gives incorrect result :

$ date -d @1455086371603
Tue Nov  7 02:46:43 PST 48079     #Incorrect

Put a dot before last three digits:

$ date -d @1455086371.603
Tue Feb  9 22:39:32 PST 2016     #Correct

Or, remove the last three digits:

$ date -d @1455086371
Tue Feb  9 22:39:31 PST 2016      #Correct after removing the last three digits. You may remove and round off the last digit too.
Answered By: KnockTurnAl

In PowerShell:

(([System.DateTimeOffset]::FromUnixTimeMilliSeconds($unixtime)).DateTime).ToString("s")
Answered By: Kulwinder Singh

If UTC is your preference (for the sample epoch timestamp 1666666666),

$ # long options; Linux
$ date --date=@1666666666 --utc
Tue 14 Nov 22:13:20 UTC 2023

$ # short options; Linux
$ date -d @1666666666 -u
Tue 14 Nov 22:13:20 UTC 2023
Answered By: Andrew Richards
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