bash 如何压缩多个文件夹，每个文件夹都有自己的 zip 存档？
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How to compress multiple folders, each into its own zip archive?
提问by Villi Magg
So I want to create a script that enables me to compresses each folder into its own zip archive and not into a one big zip file.
所以我想创建一个脚本，使我能够将每个文件夹压缩到它自己的 zip 存档中，而不是压缩到一个大的 zip 文件中。
As an example, I've got a directory:
例如，我有一个目录：
+ MyDirectory/
| |
| + Folder_01/
| |
| + Folder_02/
| |
| + Folder_03/
|
When I'm done running the script under MyDirectorythen I would have a zip file of each Folder which is inside MyDirectory: Folder_01.zip, Folder_02.zipand Folder_03.zip.
I know some BASH but this is something I can't figure out.
当我做下运行的脚本，MyDirectory然后我会每个文件夹的zip文件是内部MyDirectory：Folder_01.zip，Folder_02.zip和Folder_03.zip。我知道一些 BASH，但这是我无法弄清楚的。
How can this be done?
如何才能做到这一点？
Kind regards.
亲切的问候。
回答by Igor Chubin
for i in *
do
[ -d "$i" ] && zip -r "$i.zip" "$i"
done
You walk through all the directories and create zip for each of them.
您遍历所有目录并为每个目录创建 zip。
Or even more concise:
或者更简洁：
for i in */; do zip -r "${i%/}.zip" "$i"; done
(thanks to damienfrancois for suggestion).
（感谢 damienfrancois 的建议）。
回答by ahmed zeaad
here
这里
for i in */; do tar -czvf "${i%/}.tar.gz" "$i"; done
回答by Gajus
Consider using xzalgorithm if you require smaller output (at the cost of longer process):
xz如果您需要较小的输出（以较长的过程为代价），请考虑使用算法：
for d in */ ; do
outName=$d;
outName=${outName// /\-};
outName=${outName//[!0-9a-z-]};
dirName=$d;
dirName=${dirName//\/}
tar -c "$dirName" | xz -e > $outName.tar.xz
done
This code will sanitise folder names and produce .tar.xzfor every folder in the current directory.
此代码将清理文件夹名称并.tar.xz为当前目录中的每个文件夹生成。
