如何递归查找目录中文件的数量及其目录名称linux

Question

I have following structure in directory on linux:

-dirA
    -dir1
        -dirX
        -dirY
    -dir2
        -dirX
        -dirY
    -dir2
        -dirX
        -dirY
    -dir2
        -dirX
        -dirY
    -dir2
        -dirX
        -dirY
    -dir2
        -dirX
        -dirY
    .
    .
    .
    .
    .
    .
    .
    .
    .
    -dirn
        -dirX
        -dirY

I want to get details for count of files as follows or any format:

dir1
dirX = count
dirY = count

dir2
dirX = count
dirY = count

dir3
dirX = count
dirY = count

......
......

dirn
dirX = count
dirY = count

note here folder name of dirX and dirY is same inside subdirectory dir1,dir2.....dirn

i used following command but it gives accumulated result:

find $DIR -exec stat -c '%F' {} \; | sort | uniq -c | sort -rn

and this

find . -type f | wc -l

Answer 1

A simple while loop should be enough

find "$DIR" -type d -print0 |
    while IFS= read -r -d '' path; do
        echo -e "$(ls -A -- "$path" | wc -l)\t$path"
    done

If you want them sorted by file count, you can append a | sort -n | sort -n at the end of the command (right after the done keyword).