[SOLVED] Excluding existing folders in bash

rng · 12-08-2015, 10:53 AM

I have a list of directories in a file:

Code:

/root
/boot
/sys
/a_dir
/b_dir

I want to remove already existing directories from this list:

Code:

/a_dir
/b_dir

How can I do this? I know xargs can deal with one line at a time and [ -d dirname ] will show if the dirname is present. So I tried:

Code:

cat dirlist | xargs -I {} [ -d {} ]

but it does not work. Thanks for your help.

HMW · 12-08-2015, 01:48 PM

I don't understand the question. Why don't you simply delete the unwanted entries using sed?

Code:

$ echo "/root
/boot
/sys
/a_dir
/b_dir" | sed '/\/[a-b]_dir/d'
/root
/boot
/sys

But I am probably missing something...

Best regards,
HMW

rng · 12-08-2015, 06:54 PM

The filter should check if that directory exists. If it does not exist, only then its name should be printed, else not. It could be something like:

Code:

cat dirlist | awk 'if( [ ! -d $1 ] ) print ($1)'

How can this awk code be corrected or any other method can be used?

onlyesterday16 · 12-08-2015, 08:13 PM

I think you need use a loop:

Code:

for dir in $(cat dirlist); do
        if [ ! -d $dir ]; then
                do smt
        fi
done

rng · 12-08-2015, 09:59 PM

Thanks. I also managed using awk as follows:

Code:

cat dirlist | awk ' { if ( system(" [ -d " $1 " ] ")  ) print $1 } '