再加上一个完整的循环，检索每个文件夹的路径，使用Albert的答案和Gordans的组合。这应该很有用。

for i in $(ls -d /pathto/parent/folder/*/); do echo ${i%%/}; done

输出:

/pathto/parent/folder/childfolder1/
/pathto/parent/folder/childfolder2/
/pathto/parent/folder/childfolder3/
/pathto/parent/folder/childfolder4/
/pathto/parent/folder/childfolder5/
/pathto/parent/folder/childfolder6/
/pathto/parent/folder/childfolder7/
/pathto/parent/folder/childfolder8/

这是我正在用的

ls -d1 /Directory/Path/*;

试试这个。它适用于所有Linux发行版。

ls -ltr | grep drw

我用以下方法部分解决了这个问题:

cd "/path/to/pricipal/folder"

for i in $(ls -d .*/); do sudo ln -s "$PWD"/${i%%/} /home/inukaze/${i%%/}; done

 

    ln: «/home/inukaze/./.»: can't overwrite a directory
    ln: «/home/inukaze/../..»: can't overwrite a directory
    ln: accesing to «/home/inukaze/.config»: too much symbolics links levels
    ln: accesing to «/home/inukaze/.disruptive»: too much symbolics links levels
    ln: accesing to «/home/inukaze/innovations»: too much symbolics links levels
    ln: accesing to «/home/inukaze/sarl»: too much symbolics links levels
    ln: accesing to «/home/inukaze/.e_old»: too much symbolics links levels
    ln: accesing to «/home/inukaze/.gnome2_private»: too much symbolics links levels
    ln: accesing to «/home/inukaze/.gvfs»: too much symbolics links levels
    ln: accesing to «/home/inukaze/.kde»: too much symbolics links levels
    ln: accesing to «/home/inukaze/.local»: too much symbolics links levels
    ln: accesing to «/home/inukaze/.xVideoServiceThief»: too much symbolics links levels

好吧，这对我来说是主要的部分:)

如果不需要列出隐藏目录，我提供:

ls -l | grep "^d" | awk -F" " '{print $9}'

如果需要列出隐藏目录，请使用:

ls -Al | grep "^d" | awk -F" " '{print $9}'

Or

find -maxdepth 1 -type d | awk -F"./" '{print $2}'

Ls和awk(不含grep)

No need to use grep since awk can perform regularexpressino check so it is enough to do this:

ls -l | awk '/^d/ {print $9}'

ls -l列出有权限的文件 Awk滤波器输出 '/^d/'正则表达式，只搜索以字母d开头的行(作为目录)，并查看第一行-权限 {print}将打印所有列 {print $9}将只打印ls -l输出中的第9列(name)

非常简单明了