How to set date with Epoch format

date

I want to set date from a seconds since epoch value, for instance I want to set date with input value 1452053571.

I read through date -help but not found anything.

Is there any parameters to do it?

Best Answer

With GNU date, you can use the same date string format for both -d and -s options.

To convert from seconds since epoch to human readable format:

date -d '@2147483647'

To set it:

date -s '@2147483647'

With *BSD date:

# Convert seconds since epoch
$ date -r 2147483647 
Tue Jan 19 03:14:07 UTC 2038

# Set date by seconds since epoch
$ date "$(date -r 2147483647 +'%y%m%d%H%M.%S')"
Tue Jan 19 03:14:07 UTC 2038

Related Solutions

Shell – Leap seconds and date

I didn't find a simple solution to my question, so I've written a small Bash script to solve it. You need to download the leap seconds file given in the link below and put it with the script or change the path to it. I didn't write utc2unix.sh yet, but it's very easy to adapt. Do not hesitate to comment/give suggestions...

unix2utc.sh:

#!/bin/bash

# Convert a Unix timestamp to the real number of seconds
# elapsed since the epoch.

# Note: this script only manage additional leap seconds

# Download leap-seconds.list from
# https://github.com/eggert/tz/blob/master/leap-seconds.list

# Get current timestamp if nothing is given as first param
if [ -z $1 ]; then
    posix_time=$(date --utc +%s)
else
    posix_time=$1
fi

# Get the time at which leap seconds were added
seconds_list=$(grep -v "^#" leap-seconds.list | cut -f 1 -d ' ')

# Find the last leap second (see the content of leap-seconds.list)
# 2208988800 seconds between 01-01-1900 and 01-01-1970:
leap_seconds=$(echo $seconds_list | \
               awk -v posix_time="$posix_time" \
               '{for (i=NF;i>0;i--)
                   if (($i-2208988800) < posix_time) {
                    print i-1; exit
                    }
                } END {if (($(i+1)-2208988800) == posix_time) 
                    print "Warning: POSIX time ambiguity:",
                            posix_time,
                          "matches 2 values in UTC time!",
                          "The smallest value is given." | "cat 1>&2"
                }')
# echo $leap_seconds

# Add the leap seconds to the timestamp
seconds_since_epoch=$(($posix_time + $leap_seconds))

echo $seconds_since_epoch

Just some tests:

date --utc +%s && ./unix2utc.sh -> today and at least until June 2015 the difference is 25 sec.
./unix2utc.sh 78796799 -> 78796799
./unix2utc.sh 78796801 -> 78796802
./unix2utc.sh 78796800 -> 78796800 + on stderr: Warning: POSIX time ambiguity: 78796800 matches 2 consecutive values in UTC time! Only the smallest value is given.

Linux – How to Set Time with UNIX Timestamp Using BusyBox Date Command

Try

date @`date +%s`

I don't think it's got anything to do with bash.

Busybox's date command is a lightweight version of the more classic GNU/FSF date

Best Answer

Related Solutions

Shell – Leap seconds and date

Linux – How to Set Time with UNIX Timestamp Using BusyBox Date Command

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