如何将匹配模式的文件复制到名称中包含模式的子目录
我有以下目录结构：
folder-1.2.3/subA/subB/file.txt
folder-1.2.3/subA/subB/other_sub/file2.txt
folder-1.2.4/subA/subB/file.txt
folder-1.2.4/subA/subB/other_sub/file2.txt
我想匹配以 folder-*/subA/subB 开头的所有路径并将其内容复制到 /new_dst/prefix-* 其中 * 是匹配的字符串（在我的例子中是 1.2.3 或 1.2.4）。 结果是：
new_dst/prefix-1.2.3/file.txt
new_dst/prefix-1.2.3/other_sub/file2.txt
new_dst/prefix-1.2.4/file.txt
new_dst/prefix-1.2.4/other_sub/file2.txt
我可以使用任何标准的 Linux 命令（cp、find、grep、aws、sed ..）。我最初的想法是使用 find (find . -path "folder-*/subA/subB") 但我无法直接访问仅匹配整个路径的模式。这就是为什么我尝试通过 sed 解析它（例如 .. -exec sed -i "s/folder-//g"），但它开始看起来比应有的更复杂。
I have following directory structure:
folder-1.2.3/subA/subB/file.txt
folder-1.2.3/subA/subB/other_sub/file2.txt
folder-1.2.4/subA/subB/file.txt
folder-1.2.4/subA/subB/other_sub/file2.txt
I want to match all paths starting with folder-*/subA/subB and copy its content to /new_dst/prefix-* where * is matched string (1.2.3 or 1.2.4 in my case).
The outcome would be:
new_dst/prefix-1.2.3/file.txt
new_dst/prefix-1.2.3/other_sub/file2.txt
new_dst/prefix-1.2.4/file.txt
new_dst/prefix-1.2.4/other_sub/file2.txt
I can use any standard linux command (cp, find, grep, aws, sed..). My initial idea was to use find (find . -path "folder-*/subA/subB") but I don't have direct access what pattern was matched only the whole path. That's why I tried to parse it through sed (for example .. -exec sed -i "s/folder-//g") but it's starting to look more complicated than it should be.
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建议首先使用
find命令和-path选项找到所需的文件。找到您的文件后。将找到的文件输入 cp 命令。
Suggesting first to locate the desired files using
findcommand and-pathoption.Once your files are located. Feed the found files into
cpcommand.