Question

我要回复：

./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

从中删除./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

这只是返回最后一个文件夹，我想返回/ swatches，/ product_shots / high_res等......

Answer 1

使用sed，如下所示，

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

一种稍微简单的方法：

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

注意：这不是很完美;当$BASE内的任何目录以连字符开头时，它将失败。只有当你能保证不是这种情况时才应该使用它。

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2 个答案: