Bash command to delete all but last 5 directories [duplicate]

Question

Possible Duplicate:
Delete all but the most recent X files in bash

I have a script to create incremental backups daily and I need to delete all backups but last 5.

For example, I have this folders:

 drwxr-xr-x  4 root root 4096 Oct 29 01:10 2010-10-29 drwxr-xr-x  4 root root 4096 Oct 30 01:10 2010-10-30 drwxr-xr-x  4 root root 4096 Oct 31 01:10 2010-10-31 drwxr-xr-x  4 root root 4096 Nov  1 01:10 2010-11-01 drwxr-xr-x  4 root root 4096 Nov  2 01:10 2010-11-02 drwxr-xr-x  4 root root 4096 Nov  3 01:10 2010-11-03 drwxr-xr-x  4 root root 4096 Nov  4 01:10 2010-11-04 drwxr-xr-x  4 root root 4096 Nov  5 01:10 2010-11-05 drwxr-xr-x  4 root root 4096 Nov  6 01:10 2010-11-06 drwxr-xr-x  4 root root 4096 Nov  7 01:10 2010-11-07 drwxr-xr-x  4 root root 4096 Nov  8 01:10 2010-11-08 

And I need to maintain only the last 5 directories and delete the others. After command execute, I need to have only this:

 drwxr-xr-x  4 root root 4096 Nov  4 01:10 2010-11-04 drwxr-xr-x  4 root root 4096 Nov  5 01:10 2010-11-05 drwxr-xr-x  4 root root 4096 Nov  6 01:10 2010-11-06 drwxr-xr-x  4 root root 4096 Nov  7 01:10 2010-11-07 drwxr-xr-x  4 root root 4096 Nov  8 01:10 2010-11-08 

I don't need to delete previous to 5 days, I need to delete all except 5 last directories :)

Now I'm using:

find /backup/increment -maxdepth 1 -type d -mtime +5 -exec rm -rf {} \;

But I need to improved not based in time :)

EDIT: This is an example for a server that do backups all days, but I need an script that delete all folders previous to last 5 because my computer do backups at 00:10 at night, but not all nights the backup is done it, because my computer isn't working all days, and I need to have always the last 5 backups :)

Answer 1

use the tail command to print lines starting with the n th line (Option -n +N):

rm `ls -t | tail -n +6` 

ls -t outputs the current directory sorted by time. tail -n +6 takes al lines starting with the 6th line. Quoting with backticks feeds the result of the pipe into the rm command.

OLD SOLUTION, not correct ...

use the head command, which prints the first n lines of some output:

rm `ls -t1 | head -n 5` 

ls -t outputs the current directory sorted by time. head -n 5 takes the first five entries of the previous output. Quoting with backticks feeds the result of the pipe into the rm command.

Please try out first before applying to live data :) ...

Answer 2

The first thing that came to my mind. It's not elegant:

a=0; for i in `ls -t`; do     a=`expr $a + 1`;     if [ $a  -gt 5 ]; then           echo "removing $i";           rm -rf $i     fi; done