删除linux目录中的所有文件夹和文件，除了一个文件夹和该文件夹中的所有内容

Question

I have a directory structure as :-

/usr/testing/member/
   ---> public--->folder1--->file1
           \----> file2
   ---> folder3:- contains files and folders
   ---> folder4:- contains files and folders
   ---> several files

I want to keep the public folder and all its contents (further folders and files within it) but want to delete everything else under the directory /usr/testing/member/. But that also means member folder is not deleted. Is there any shell script or command that can be used to achieve this exactly as i stated.

Answer 1

Here's one way to do it:

(cd /usr/testing/member; find . -maxdepth 1 \( ! -name . -a ! -name public \) -exec echo rm -fr {} +)

That is: cd into /usr/testing/member , find all files and directories there, without going further below, and exclude the current directory (".") and any file or directory named "public", and execute a command for the found files.

This will print what would be deleted. Verify it looks good, and then drop the echo .

Answer 2

I think below will do the work,

$ cd  /usr/testing/member/
$ rm -rf $(ls | grep -v "public")

explanation:

we are passing everything inside /usr/testing/member/ but public to rm by making use of -v(exclude) option of grep