I have a directory containing a large number of directories. Each directory contains some files and in some cases another directory.
parent_directory
sub_dir_1
sub_dir_1.txt
sub_dir_1_1.txt
sub_dir_2
sub_dir_2.txt
sub_dir_2_1.txt
sub_dir_3
sub_dir_3.txt
sub_dir_3_1.txt
sub_dir_4
sub_dir_4.txt
sub_dir_4_1.txt
sub_dir_5
sub_dir_5.txt
sub_dir_5_1.txt
I need to check that each sub_dir contains at least one file with the exact same name. I don' need to check any further down if there are sub directories within the sub_dirs.
I was thinking of using for d in ./*/ ; do (command here); done
for d in ./*/ ; do (command here); done
for d in ./*/ ; do (command here); done
but I dont know how to get access to the sub_dir name inside the for loop
for d in ./*/ ;
do
(if directory does not contain 1 file that is the same name as the directory then echo directory name );
done
What is the best way to do this or is there a simpler way?
from the parent directory
find -maxdepth 1 -type d -printf "%f\n" |
xargs -I {} find {} -maxdepth 1 -type f -name {}.txt
will give you the name/name.txt pair. Compare with the all dir names to find the missing ones.
UPDATE
this might be simpler, instead of scanning you can check whether file exists or not
for f in $(find -maxdepth 1 -type d -printf "%f\n");
do if [ ! -e "$f/$f.txt" ];
then echo "$f not found";
fi; done
Maybe not understand fully, but
find . -print | grep -P '/(.*?)/\1\.txt'
this will print any file which is inside of the same-named directory, eg:
./a/b/b.txt
./a/c/d/d.txt
etc...
Similarly
find . -print | sed -n '/\(.*\)\/\1\.txt/p'
this
find . -print | grep -P '/(.*?)/\1\.'
will list all files regardless of the extension in same-named dirs.
You can craft other regexes following the backreference
logic.
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.