How to delete everything except 2 files in directories?

Hi

I have 1 directory in my server i will delete everything except 2 files (index.html and .htaccess) with cronjob how can do it?

 

rm -R *.* !(index.html*) !(.htaccess*)

 

From SSH:

rm -R *.* !(index.html*) !(.htaccess*)

Code (markup):

From FTP:

Just select and delete the files

Code (markup):

 

Good question I was also wondering how it is done

 

[root@test~]# mkdir z ; cd z ; touch 1 2 3 4 5 6
[root@test z]# ls
1  2  3  4  5  6
[root@test~]# find z/ -type f  \( ! -iname '2' -and ! -iname '3' \) -print
z/1
z/4
z/5
z/6
[root@test~]#

Code (markup):

You could use find with -delete instead of -print to remove the files.

 

This is actually what I was about to suggest; it's a much more efficient way of doing things.

 

i have a different angle on it: how about not keeping your auto generated files (logs, cache, tmp) in the same directory as your static files?

 

The below should work
find <directory here> ! -name index.html ! -name .htaccess -exec rm -frv {} \;