如果父文件夹与bash同名，则移动子文件夹的内容

Question

When doing tar extract operations sometimes the contents extract directly to the parent folder which is awful since they get all disordered. By example:

tar -xzf foo1.tar.gz

Extracts:

./file1
./file2
./file3

The solution to that is to create an specific folder for the tar.gz By example:

tar -C foo -xzf foo1.tar.gz

Extracts:

./foo1/file1
./foo1/file2
./foo1/file3

So if I have many .tar.gz to extract I will just do:

find -name \*.tar.gz | xargs -n 1 basename -s .tar.gz | xargs -I {} tar -C {} -xzf {}.tar.gz

To extract them safely, but with that I'll endup with things like this:

./foo3/foo3/file1
./foo3/foo3/file2
./foo3/foo3/file3

Is there an automatic way to remove the duplicated child folder with bash for those cases that need it?

Answer 1

计算目录中条目的数量（提示： dotglob ， (*)和${#var[@]} ），如果单个目录条目具有相同的名称，则复制属性并移动内容（请注意，在该目录中也有相同的名称），然后删除现在为空的目录。

Answer 2

This is a possible solution

To move files from the child to the toplevel directory:

find ! -name . -maxdepth 1 -type d | xargs -I {} sh -c "find {}/{} -maxdepth 1 | xargs -I [] echo \"mv [] {}\""

This works because if the child folder doesnt exists find would do nothing:

find foo4/foo4 
find: 'foo4/foo4': No such file or directory

To remove reapted child directory

find ! -name . -maxdepth 1 -type d | xargs -I {} find {}/{} -type d -maxdepth 1 | xargs rmdir

I think it did what it was intended for. Disadvantage:

It shows many harmless errors.