Bash：递归复制命名文件，保留文件夹结构
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Bash: Copy named files recursively, preserving folder structure
提问by mahemoff
I was hoping:
我希望：
cp -R src/prog.js images/icon.jpg /tmp/package
would yield a symmetrical structure in the destination dir:
将在目标目录中产生对称结构：
/tmp
|
+-- package
|
+-- src
| |
| +-- prog.js
|
+-- images
|
+-- icon.jpg
but instead, both of the files are copied into /tmp/package. A flat copy. (This is on OSX).
但是，这两个文件都被复制到 /tmp/package 中。平面副本。（这是在 OSX 上）。
Is there a simple bash function I can use to copy all files, including files specified by wildcard (e.g. src/*.js) into their rightful place within the destination directory. A bit like "for each file, run mkdir -p $(dirname "$file"); cp "$file" $(dirname "$file")", but perhaps a single command.
是否有一个简单的 bash 函数可以用来将所有文件（包括通配符（例如 src/*.js）指定的文件）复制到目标目录中的正确位置。有点像“对于每个文件，运行mkdir -p $(dirname "$file"); cp "$file" $(dirname "$file")”，但也许是一个命令。
This is a relevant thread, which suggests it's not possible.The author's solution isn't so useful to me though, because I would like to simply provide a list of files, wildcard or not, and have all of them copied to the destination dir. IIRC MS-DOS xcopy does this, but there seems to be no equivalent for cp.
这是一个相关的线程，这表明这是不可能的。不过，作者的解决方案对我来说并不是那么有用，因为我只想提供一个文件列表，无论是否通配符，并将它们全部复制到目标目录。IIRC MS-DOS xcopy 执行此操作，但似乎没有 cp 的等效项。
回答by ustun
Have you tried using the --parents option? I don't know if OS X supports that, but that works on Linux.
您是否尝试过使用 --parents 选项？我不知道 OS X 是否支持它，但它适用于 Linux。
cp --parents src/prog.js images/icon.jpg /tmp/package
If that doesn't work on OS X, try
如果这在 OS X 上不起作用，请尝试
rsync -R src/prog.js images/icon.jpg /tmp/package
as aif suggested.
正如 aif 所建议的那样。
回答by Randy Proctor
One way:
单程：
tar cf - <files> | (cd /dest; tar xf -)
回答by Jonathan Leffler
Alternatively, if you're old-school, use cpio:
或者，如果您是老派，请使用 cpio：
cd /source;
find . -print | cpio -pvdmB /target
Clearly, you can filter the file list to your heart's content.
显然，您可以根据自己的喜好过滤文件列表。
The '-p' option is for 'pass-through' mode (as against '-i' for input or '-o' for output). The '-v' is verbose (list the files as they're processed). The '-m' preserves modification times. The '-B' means use 'big blocks' (where big blocks are 5120 bytes instead of 512 bytes); it is possible it has no effect these days.
'-p' 选项用于 'pass-through' 模式（与 '-i' 用于输入或 '-o' 用于输出）。'-v' 是冗长的（列出处理过的文件）。'-m' 保留修改时间。“-B”表示使用“大块”（其中大块是 5120 字节而不是 512 字节）；这几天可能没有效果。
回答by Ryan Bright
rsync's -R optionwill do what you expect. It's a very feature-rich file copier. For example:
rsync 的 -R 选项将满足您的期望。这是一个功能非常丰富的文件复制器。例如：
$ rsync -Rv src/prog.js images/icon.jpg /tmp/package/
images/
images/icon.jpg
src/
src/prog.js
sent 197 bytes received 76 bytes 546.00 bytes/sec
total size is 0 speedup is 0.00
Sample results:
示例结果：
$ find /tmp/package
/tmp/package
/tmp/package/images
/tmp/package/images/icon.jpg
/tmp/package/src
/tmp/package/src/prog.js
回答by EMPraptor
Try...
尝试...
for f in src/*.js; do cp $f /tmp/package/$f; done
so for what you were doing originally...
所以对于你最初在做什么......
for f in `echo "src/prog.js images/icon.jpg"`; do cp $f /tmp/package/$f; done
or
或者
v="src/prog.js images/icon.jpg"; for f in $v; do cp $f /tmp/package/$f; done
