Bash / python - 用条件递归删除文件夹

Question

我想在bash / python中编写一个脚本，它能够递归删除文件夹，里面没有'.txt'个文件。

例如：

| Folder1/          --> keep
|--SubFolderA/      --> keep
|----file.txt

| Folder2/          --> delete

| Folder3/          --> keep
|--subFolderA/      --> keep
|----file.txt

| Folder4/          --> delete
|--SubFolderA/      --> delete
|----SubFolderB/    --> delete
|------file.py

| Folder5/          --> keep
|--SubFolderA/      --> keep, because .txt file is in SubFolderB
|----file.sh
|----SubFolderB/    --> keep
|------file.txt

| Folder6/          --> keep
|--SubFolderA/      --> keep
|----file.txt
|----SubFolderB/    --> delete
|------file.py

Answer 1

绝对使用find。像这样：

find /path -type f -name "*.txt" -exec rm -rf {} \;

我不打算为你写if语句，但那应该让你开始......其实不论之处：

path=the/path/to/the/directory/
myvar=$(find $path -type f -name "*.txt" -exec rm -rf {} \;)
if [ -z $myvar ]
then
rmdir $path
else
#don't do anything...keep it
fi