仅列出具有特定名称的所有目录的直接子目录

Question

我有一个如下所示的目录结构：

.
??? Untitled Folder
?   ??? Untitled Folder
?   ?   ??? zing
?   ?       ??? first
|   |       |   ??? dum
?   ?       ??? second
?   ?       ??? third
?   ??? Untitled Folder 2
?   ??? Untitled Folder 3
??? Untitled Folder 2
?   ??? Untitled Folder
?   ??? Untitled Folder 2
?   ?   ??? zing
?   ?       ??? fifth
|   |       |   ??? dum
?   ?       ??? fourth
?   ?       ??? sixth
?   ??? Untitled Folder 3
??? Untitled Folder 3
    ??? Untitled Folder
    ??? Untitled Folder 2
    ??? Untitled Folder 3
        ??? zing
            ??? eighth
            ??? seventh
                ??? dum

我只想列出所有zing目录的直接子目录。

示例输出为：

first
second
third
fifth
fourth
sixth
eighth
seventh

我试过

% find . -type d | grep zing
./Untitled Folder 3/Untitled Folder 3/zing
./Untitled Folder 3/Untitled Folder 3/zing/eigthth
./Untitled Folder 3/Untitled Folder 3/zing/seventh
./Untitled Folder 3/Untitled Folder 3/zing/seventh/dum
./Untitled Folder 2/Untitled Folder 2/zing
./Untitled Folder 2/Untitled Folder 2/zing/fifth
./Untitled Folder 2/Untitled Folder 2/zing/fifth/dum
./Untitled Folder 2/Untitled Folder 2/zing/fourth
./Untitled Folder 2/Untitled Folder 2/zing/sixth
./Untitled Folder/Untitled Folder/zing
./Untitled Folder/Untitled Folder/zing/first
./Untitled Folder/Untitled Folder/zing/first/dum
./Untitled Folder/Untitled Folder/zing/third
./Untitled Folder/Untitled Folder/zing/second

它甚至显示了我不想要的zing（在这种情况下dum）的孩子的孩子。

我怎样才能得到我的预期输出？

Answer 1

• Bash、Globstar 已激活（使用shopt -s globstar）。

  printf '%s\n' **/zing/*/
  

  • 仅对于最后一个路径组件，

     for dir in **/zing/*/; do basename "$dir"; done
    

• POSIX 查找。

  find . -type d -path '*/zing/*' -prune
  

  -prune 避免在匹配的目录中降序。

  • 仅对于最后一个路径组件，

    find . -type d -path '*/zing/*' -prune -exec basename {} \;