How to convert timestamps to dates in bash?

Question

I need a shell command or script that converts a UNIX timestamp to a date. The input can come either from the first parameter or from stdin, allowing for the following usage patterns:

ts2date 1267619929

and

echo 1267619929 | ts2date

Both commands should output "Wed Mar 3 13:38:49 2010".

Answer 1

I have written a script that does this myself:

#!/bin/bash
LANG=C
if [ -z "$1" ]; then
    if [  "$(tty)" = "not a tty" ]; then
            p=`cat`;
    else
            echo "No timestamp given."
            exit
    fi
else
    p=$1
fi
echo $p | gawk '{ print strftime("%c", $0); }'

Answer 2

In PHP

$unix_time = 1256571985;

echo date("Y-m-d H:i:s",$unix_time)

Answer 3

You can use this simple awk script:

#!/bin/gawk -f   
{ print strftime("%c", $0); }

Sample usage:

$ echo '1098181096' | ./a.awk 
Tue 19 Oct 2004 03:18:16 AM PDT
$

Answer 4

On later versions of common Linux distributions you can use:

date -d @1267619929

Answer 5

you can use GNU date eg

$ sec=1267619929
$ date -d "UTC 1970-01-01 $sec secs"

or

$ date -ud @$sec

Answer 6

This version is similar to chiborg's answer, but it eliminates the need for the external tty and cat. It uses date, but could just as easily use gawk. You can change the shebang and replace the double square brackets with single ones and this will also run in sh.

#!/bin/bash
LANG=C
if [[ -z "$1" ]]
then
    if [[ -p /dev/stdin ]]    # input from a pipe
    then
        read -r p
    else
        echo "No timestamp given." >&2
        exit
    fi
else
    p=$1
fi
date -d @$p +%c