如果在 bash 中包含特定文件名，则仅列出目录

Question

Hel lo everyone, I need help with bash commands.

Here is the deal :

I have several directories :

such as

path1/A/path2/
 file1
 file2
path1/B/path2/
 file1
path1/C/path2/
 file1
 file2
path1/D/path2/
 file1
 file2

and a /path_to_this_file/file.txt :

A
B
C
D

that I use such as :

cat /path_to_this_file/file.txt | while read line; do ls path1/$line/path2/

then I can list all content in the paths BUT I would like to do a ls only for the path2 that does not have a file2 into their directory..

Here only the path1/B/path2/ should be listed

Does someone have a code for that ?

Answer 1

Added an if statement to your code:

cat /path_to_this_file/file.txt |
    while read line
    do
        if [ ! -f "path1/$line/path2/file2" ]; then
            ls path1/$line/path2/
        fi
    done

Alternatively:

xargs -I {} bash -c "[ ! -f "path1/{}/path2/file2" ] && ls path1/{}/path2" < /path_to_this_file/file.txt

Answer 2

This will do the job

while read line; do
    ls "path1/$line/path2/file2" &> /dev/null || ls "path1/$line/path2"
done < /path_to_this_file/file.txt

Answer 3

Using mapfile aka readarray bash4+ only and a for loop

#!/usr/bin/env bash

mapfile -t var < path_to_this_file/file.txt

for i in "${var[@]}"; do
  if [[ ! -e path1/$i/path2/file2 ]]; then
    ls "path1/$i/path2/"
  fi
done

The ls in the above code

 ls "path1/$i/path2/" 

Output is

file1

If the you want to print just the PATH change the ls "path1/$i/path2/" to

echo "path1/$i/path2/" 

Output is

path1/B/path2/

If you want to print both PATH and file use

echo "path1/$i/path2/"*

Output is

path1/B/path2/file1