Bash命令删除几个嵌套目录

Question

我犯了一个错误，创建了嵌套目录，如：

dir1/dir2_A/out/dir2_A
file1
file2 
out
dir1/dir2_B/out/dir2_B
file1
file2 
out
dir1/dir2_C/out/dir2_C
file1
file2 
out
...

如何删除所有中间目录：

dir1/dir2_A/out/dir2_A/ → dir1/dir2_A/
dir1/dir2_B/out/dir2_B/ → dir1/dir2_B/
dir1/dir2_C/out/dir2_C/ → dir1/dir2_C/

等等。，在dir1中得到这个？

dir1/dir2_A
file1
file2 
out
dir1/dir2_B
file1
file2 
out
dir1/dir2_C
file1
file2 
out

Answer 1

下降到文件，将所有内容向上移动
接下来删除空目录。使用括号作为子shell，这样您将从d1返回而不使用cd ../..。

for d1 in dir1/*/*/*; do
test -d "$d1" || continue
(cd "$d1" && mv * ../.. )
done
# Cleanup without `rm -r` avoiding errors when something went wrong
rmdir dir1/*/out/* 
rmdir dir1/*/out