I want to find a subdirectory of the current directory, which (that is the subdirectory) contains 2 or more regular files.
I am not interested in directories containing less than 2 files, neither in directories which contain only subdirectories.
I want to find a subdirectory of the current directory, which (that is the subdirectory) contains 2 or more regular files.
I am not interested in directories containing less than 2 files, neither in directories which contain only subdirectories.
Here is a completely different approach based on GNU find and uniq. This is much faster and much CPU-friendly than answers based on executing a shell command that counts files for each directory found.
find . -type f -printf '%h\n' | sort | uniq -d
The find command prints the directory of all files in the hierarchy and uniq only displays the directories that appear at least twice.
With the help of Gilles's answer on SU and its reverse and some modification, here what you need.
find . -type d -exec sh -c 'set -- "$1"/*;X=0;
for args; do [ -f "$args" ] && X=$((X+1)) ;done; [ "$X" -gt 1 ] ' _ {} \; -print
Directory tree.
.
├── test
│ ├── dir1
│ │ ├── a
│ │ ├── b
│ │ └── c
│ ├── dir2
│ │ ├── dira
│ │ │ └── a file\012with\012multiple\012line
│ │ ├── dirb
│ │ │ ├── file-1
│ │ │ └── file-2
│ │ └── dirc
│ ├── diraa
│ ├── dirbb
│ ├── dircc
│ └── x
│ └── x1
│ └── x2
└── test2
├── dir3
└── dir4
Result:
./test
./test/dir1
./test/dir2/dirb
find . -type d \
-exec sh -c 'c=0; for n in "$1"/*; do [ -f "$n" ] && [ ! -h "$n" ] && c=$(( c + 1 )); done; [ "$c" -ge 2 ]' sh {} ';' \
-print
This will find all names in or under the current directory and then filter out all names that are not names of directories.
The remaining directory names will be given to this short script:
c=0
for n in "$1"/*; do
[ -f "$n" ] && [ ! -h "$n" ] && c=$(( c + 1 ))
done
[ "$c" -ge 2 ]
This script will count the number of regular files (skipping symbolic links) in the directory given as the first command line argument (from find). The last command in the script is a test to see if the count was 2 or greater. The result of this test is the return value (exit status) of the script.
If the test succeeded, -print will cause find to print out the path to the directory.
To also consider hidden files (files whose names begins with a dot), change the sh -c script from saying
for n in "$1"/*; do
to
for n in "$1"/* "$1"/.*; do
Testing:
$ tree
.
`-- test
|-- a
|-- dir1
| |-- a
| |-- b
| `-- c
`-- dir2
|-- dira
|-- dirb
| |-- file-1
| `-- file-2
`-- dirc
6 directories, 6 files
$ find . -type d -exec sh -c 'c=0; for n in "$1"/*; do [ -f "$n" ] && [ ! -h "$n" ] && c=$(( c + 1 )); done; [ "$c" -ge 2 ]' sh {} ';' -print
./test/dir1
./test/dir2/dirb
Another find + wc approach:
find path/currdir -maxdepth 1 -type d ! -empty ! -path "path/currdir" \
-exec sh -c 'count=$(find "$1" -maxdepth 1 -type f | wc -l); [ $count -ge 2 ]' _ {} \; -print
path/currdir - path to your current directory
-maxdepth 1 - consider only direct child subfolders
! -empty - ignore empty subfolders
! -path "path/currdir" - ignore the current directory path
count=$(find "$1" -maxdepth 1 -type f | wc -l) - count is assigned with the number of files for each found subfolder
[ $count -ge 2 ] ... -print - print subfolder name/path containing 2 or more regular files