Selective remove awk-bash

Dedalus · #1 (**permalink**) 10-02-2009

Hi

I have a folder that contain up to 1000 subfolder. Genereally i need just 2 or 3 of them and i can know apriori the name of the subfolder i want to keep.
I'd like to erase all the unnecessary subfolders.

I've tried with:

Code:

ls -l | awk '$8 !~ /name_fold/{print "rm -r " $8}'| sh

and it works becouse i keep the name_fold1 ...but just for 1 folder, I'd like to do something like:

Code:

ls -l | awk '$8 !~ /name_fold1 || name_fold2 /{print "rm -r " $8}'| sh

in order to erase almost all and kepp the name_fold1 and name_fold2

Any idea?

thanks

D

Franklin52 · #2 (**permalink**) 10-02-2009

Something like this?

Code:

ls -l | awk '$8 !~ /name_fold1/ && $8 !~ /name_fold2/{print "rm -r " $8}'| sh

danmero · #3 (**permalink**) 10-02-2009

Shorter

Code:

ls -l | awk '$8 !~ /name_fold[12]/{print "rm -r " $8}'| sh

Dedalus · #4 (**permalink**) 10-02-2009

that's great!

thanks both

radoulov · #5 (**permalink**) 10-02-2009

Some shells support extended globbing:

ksh

Code:

rm -r -- !(name_fold[12])

bash

Code:

shopt -s extglob
rm -r -- !(name_fold[12])

zsh (setopt extendedglob if not enabled):

Code:

rm -r -- ^name_fold[12]

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