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rng 12-08-2015 10:53 AM
Excluding existing folders in bash
 
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 } '


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