Command Line – Get Today’s Date and Use It in Filename

You can use rename with a real Perl expression

$ rename 's/(\d+\.jpg)/use File::stat; sprintf("%s-%s", stat($&)->mtime, $1)/e' * -v
0000.jpg renamed as 1408057200-0000.jpg
0001.jpg renamed as 1408060861-0001.jpg
0002.jpg renamed as 1408064522-0002.jpg

That leaves the files with a time-since-epoch. Find for sorting, poor for reading.

The only problem here is you need to match the whole filename (so that you can stat $& —aka the entire match— inside the substitution). That could get a little tiresome with really complicated names.

We can build on that to have a more conventional date formatting:

$ rename 's/(\d+\.jpg)/use File::stat; use POSIX; sprintf("%s-%s", strftime("%Y%m%d-%H%M%S", localtime(stat($&)->mtime)), $1)/e' * -vn
0000.jpg renamed as 20140815-000000-0000.jpg
0001.jpg renamed as 20140815-010101-0001.jpg
0002.jpg renamed as 20140815-020202-0002.jpg

Here's one way to do it:

time=$(($(date -d"1776-07-04 00:00:00" +%s) - $(date +%s)))

sets a time variable you can use like so:

echo $time
-7595723059

The +%s tells date to format to seconds since 1970-01-01 00:00:00 UTC which helps give us a starting point for doing the math in seconds.

Now it can also be added to a script with the same variables.

:~$ time=$(($(date -d"1776-07-04 00:00:00" +%s) - $(date +%s)))
:~$ printf '%dh:%dm:%ds\n' $(($time/3600)) $(($time%3600/60)) $(($time%60))
-2109923h:-12m:-55s

If you want Years:Days:Hours:Minutes:Seconds since, it would be the following:

printf '%dy:%dd:%dh:%dm:%ds\n' $(($time/60/60/24/365)) $(($time/60/60/24%365)) $(($time/3600%24)) $(($time%3600/60)) $(($time%60))

Which would give you output like the following:

:~$ time=$(($(date -d"1776-07-04 00:00:00" +%s) - $(date +%s)))
:~$ printf '%dy:%dd:%dh:%dm:%ds\n' $(($time/60/60/24/365)) $(($time/60/60/24%365)) $(($time/3600%24)) $(($time%3600/60)) $(($time%60))
-242y:-11d:-7h:-46m:-37s

Hope this helps!