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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).

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whats the problem with the date syntax? you don't like the formatting options? I think a date -d @TIMESTAMP is really simple... – echox Oct 11 '10 at 13:57
@echox I was completely not seeing the @TIMESTAMP in the docs. – xenoterracide Oct 11 '10 at 13:59
ah, ok, that explains it :-) – echox Oct 11 '10 at 14:00

11 Answers 11

up vote 108 down vote accepted

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
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+1 for the comment about the lack of portability (why doesn't the POSIX spec include a way to do this? grr) – Richard Hansen Feb 1 '12 at 22:50
What does the @ mean in date -d @1234567890? man date made no reference to that... – Chris Markle Jan 14 '13 at 21:10
@ChrisMarkle GNU man pages are often woefully incomplete. “The date string format is more complex than is easily documented here but is fully described in the info documentation.” To wit: gnu.org/software/coreutils/manual/html_node/… – Gilles Jan 14 '13 at 21:56
The info date is quite complete. The entry at 28.9 Seconds since the Epoch explains in detail about the @timestamp. – BinaryZebra Feb 10 at 8:31

date -d @1190000000 Replace 1190000000 with your epoch

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Assuming GNU date, that is. – Gilles Oct 11 '10 at 18:14
$ 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)
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With bash-4.2 or above:

printf '%(%F %T)T\n' 1234567890

(where %F %T is the strftime()-type format)

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 '%(%F %T)T\n' '#1234567890'

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

$ TZ=Europe/London bash -c 'printf "%(%c)T\n" 0'
Thu 01 Jan 1970 01:00:00 BST
$ TZ=Europe/London ksh93 -c 'printf "%(%c)T\n" "#0"'
Thu Jan  1 00:00:00 1970
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The two I frequently use are:

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


$ tclsh
% clock format 1234567890
Sa Feb 14 00:31:30 CET 2009
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With zsh you could use the strftime builtin:

strftime format epochtime

      Output the date denoted by epochtime in the format specified.


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).

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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: http://stackoverflow.com/questions/3249827/convert-from-unixtime-at-command-line

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Only works for GNU date and GNU awk. Neither awk nor nawk support strftime. – DarkHeart Jul 28 '14 at 3:56

PowerShell – Convert Unix or Epoch time to readable format


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Hi! When posting a link, it's best to include the most important parts of information from it, so that it is still available here on U+L, even when the link eventually breaks. It would be great if you could edit your answer to improve it. – dhag Dec 14 '15 at 16:48
Powershell being a windows only utility is quite off topic on unix.stackexchange.com – jlliagre Apr 12 at 22:44

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=%Y\nMON=%m\nDAY=%d\nHOUR=%H\nMIN=%M\nSEC=%S";
    size_t formatSize = strlen(formatStr) + 2;
    char * output     = (char *)malloc(sizeof(char)*formatSize);

    strftime(output, formatSize, formatStr, &input_tm);

    return 0;


clang -o epoch2datetime main.c

eval `./epoch2datetime 1450196411`
#2015 12 16 00 20 11
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Wouldn't be a real solution without a little node.js:

    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

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>.

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If your epoch time is in milliseconds instead of seconds, remove the last three digits before passing it to date -d:

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

This gives incorrect data. 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.
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What utility prints included milliseconds (without a dot) ? – BinaryZebra Feb 10 at 8:35
I have seen that WebLogic Application server mostly returns data time values with milliseconds and no dots when using scripting tool. e.g., lastSuccessfulConnectionUse=1455086371603 – KnockTurnAl Feb 10 at 8:49
I see. This page confirms Raw Time Value The timestamp in milliseconds. Thanks. – BinaryZebra Feb 10 at 10:11

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