如何使用bash返回设置点后的所有路径？

Question

我要把这个还回去：

./nas/cdn/catalog/swatches
./nas/cdn/catalog/product_shots
./nas/cdn/catalog/product_shots/high_res
./nas/cdn/catalog/product_shots/high_res/back
./nas/cdn/catalog/product_shots/high_res/front
./nas/cdn/catalog/product_shots/low_res
./nas/cdn/catalog/product_shots/low_res/back
./nas/cdn/catalog/product_shots/low_res/front
./nas/cdn/catalog/product_shots/thumbs
./nas/cdn/catalog/full_length
./nas/cdn/catalog/full_length/high_res
./nas/cdn/catalog/full_length/low_res
./nas/cdn/catalog/cropped
./nas/cdn/catalog/drawings

从这里删除./nas/cdn/catalog/的正确方法是什么？

到目前为止，这是我的代码。

BASE='./nas/cdn/catalog'
echo $BASE
for d in $(find . -type d -regex "${BASE}/[^.]*")
do
    echo $(basename $d)

done

bit --这只是返回最后一个文件夹，我喜欢返回/swatches、/product/high_res等.

Answer 1

像下面这样使用sed，

BASE='./nas/cdn/catalog'
echo $BASE
for d in $(find . -type d -regex "${BASE}/[^.]*")
do
    sed 's~^\([^/]*/\)\{4\}~~' <<< "$d"
done

示例：

$ var="./nas/cdn/catalog/drawings"
$ sed 's~^\([^/]*/\)\{4\}~~' <<< "$var"
drawings

Answer 2

一种稍微简单一些的方法：

BASE='./nas/cdn/catalog'
echo "$BASE"
( cd "$BASE" ; find */ -type d )

注意:这不是完全健壮的；当$BASE中的任何目录都以连字符开头时，它就会失败。只有当您能够保证不存在这种情况时，才应该使用它。