Q: Delete file by date in its name I have a list of files that looks like this:
...
live-2014-04-28.tar.bz2
live-2014-04-29.tar.bz2
live-2014-04-30.tar.bz2
live-2014-05-01.tar.bz2
live-2014-05-01.tar.bz2
live-2014-05-02.tar.bz2
...

... and trying to delete files older then a week with bash script:
c=0
for i in `echo "ls /filebackup/daily" | sftp somepath.your-backup.de`
do
        c=`expr $c + 1`
        [ $c -le 3 ] && continue
        d=`echo $i | sed -r 's/[^0-9]*([0-9]+-[0-9]+-[0-9]+).*/\1/'`
        d=`date -d $d +'%s'`
        echo $c
        if [ `expr $dc - 691200` -ge $d ]
        then
                echo 'here file will be deleted'
        fi
done

I can't get inside echo 'here file will be deleted', was trying to debug – not quite sure what $dc part does (I'm quite with coding in bash).
I'm using this code from this article (see The File-Backup-Script part), and trying to understand why it doesn't work on my side.
Thank you for any help.

A: Here is one way of doing it. 
#!/bin/bash

# Get the current and store in a variable
currDate=$(date +'%Y%m%d')  

# set this shell option to prevent literal match when no files exists
shopt -s nullglob

# Iterate over your directory
for file in *.bz2; do
    f=${file%%.*}                        # Strip the trailing extensions => live-2014-04-28
    f=${f#*-}                            # Strip the leading hyphen => 2014-04-28
    f=${f//-/}                           # Substitute all hyphens in date => 20140428
    if (( f <= (currDate - 7) )); then   # If the date is less than week old
        echo rm $file ...                # delete it
    fi
done

I have put echo before rm command so that you can test the output. If you are satisfied with the result then you can remove the echo. 

A: 
rying to delete files older then a week with bash script.

What if you ignore the names of the files and just use the timestamp?
$ find . -name "live-*.tar.bz2" -mtime +7 -delete

It's not exactly what you want, but it's simple.
