blocks|key|1719255|text|不能，cp可以复制多个源，但只会复制到单个目标。您需要安排多次调用cp+-每个目的地一次-用于您想要做的事情；如您所说，使用循环或其他一些工具。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1719256|entityMap^0|3|2|X|2|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

No, <code>cp</code> can copy multiple sources but will only copy to a single destination. You need to arrange to invoke <code>cp</code> multiple times - once per destination - for what you want to do; using, as you say, a loop or some other tool.

blocks|key|1719323|text|您不能单独使用cp完成此操作，但您可以将cp与xargs结合使用|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1719324|echo+dir1+dir2+dir3+%7C+xargs+-n+1+cp+file1|code-block|syntax|javascript|1719325|将file1复制到dir1、dir2和dir3。xargs将调用cp+3次来执行此操作，有关详细信息，请参阅xargs的手册页。|1719326|entityMap^0|7|2|K|2|N|5|0|0|1|5|9|4|E|4|J|4|O|5|W|2|1I|5|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]|$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|N|$]]

You can't do this with <code>cp</code> alone but you can combine <code>cp</code> with <code>xargs</code>:

<pre><code>echo dir1 dir2 dir3 | xargs -n 1 cp file1
</code></pre>

Will copy <code>file1</code> to <code>dir1</code>, <code>dir2</code>, and <code>dir3</code>. <code>xargs</code> will call <code>cp</code> 3 times to do this, see the man page for <code>xargs</code> for details.

blocks|key|1504411|text|通配符也适用于Roberts代码|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1504412|echo+./fs*/*+%7C+xargs+-n+1+cp+test+|code-block|syntax|javascript|1504413|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Wildcards also work with Roberts code

<pre><code>echo ./fs*/* | xargs -n 1 cp test 
</code></pre>

blocks|key|1504535|text|根据我在https://superuser.com/questions/32630/parallel-file-copy-from-single-source-to-multiple-targets上看到的答案，我会使用cat和tee，而不是cp。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1504536|例如：|1504537|cat+inputfile+%7C+tee+outfile1+outfile2+>+/dev/null|code-block|syntax|javascript|1504538|entityMap|0|LINK|mutability|MUTABLE|url|https://superuser.com/questions/32630/parallel-file-copy-from-single-source-to-multiple-targets^0|32|3|36|3|3D|2|4|2N|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@$9|11|A|12|1|13]]|E|$]]|$1|F|3|G|5|6|7|14|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|15|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

I would use <code>cat</code> and <code>tee</code> based on the answers I saw at <a href="https://superuser.com/questions/32630/parallel-file-copy-from-single-source-to-multiple-targets">https://superuser.com/questions/32630/parallel-file-copy-from-single-source-to-multiple-targets</a> instead of <code>cp</code>.

For example:

<pre><code>cat inputfile | tee outfile1 outfile2 &gt; /dev/null
</code></pre>

blocks|key|1504487|text|据我所知，您可以使用以下内容：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1504488|ls+%7C+xargs+-n+1+cp+-i+file.dat|code-block|syntax|javascript|1504489|cp命令的-i选项意味着系统会询问您是否使用file.dat覆盖当前目录中的文件。虽然这不是一个完全自动的解决方案，但它对我来说是可行的。|offset|length|style|CODE|1504490|entityMap^0|0|0|0|2|5|2|M|8|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]|$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|X|8|@]|9|@]|A|$]]]|N|$]]

As far as I can see it you can use the following:

<pre><code>ls | xargs -n 1 cp -i file.dat
</code></pre>

The <code>-i</code> option of <code>cp</code> command means that you will be asked whether to overwrite a file in the current directory with the <code>file.dat</code>. Though it is not a completely automatic solution it worked out for me.

blocks|key|1720420|text|这些答案看起来都比显而易见的复杂：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1720421|for+i+in+/foo+/bar;+do+cp+"$file1"+"$i";+done|code-block|syntax|javascript|1720422|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

These answers all seem more complicated than the obvious:

<pre><code>for i in /foo /bar; do cp "$file1" "$i"; done
</code></pre>

blocks|key|1505252|text|不是使用cp本身，但是...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1505253|对于我来说，这是在将一张(慢速)SD卡上的大量Gopro素材复制到三个(慢速)+USB驱动器的背景下出现的。我只想读取数据一次，因为这需要永远的时间。我希望它是递归的。|1505254|$+tar+cf+-+src+%7C+tee+>(+cd+dest1+;+tar+xf+-+)+>(+cd+dest2+;+tar+xf+-+)+%7C+(+cd+dest3+;+tar+xf+-+)|code-block|syntax|javascript|1505255|(如果想要更多的输出，可以添加更多的>()部分。)|1505256|我没有对此进行基准测试，但它肯定比cp-in-a-loop+(或者一堆并行的cp调用)快得多。|1505257|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Not using cp per se, but...

This came up for me in the context of copying lots of Gopro footage off of a (slow) SD card to three (slow) USB drives. I wanted to read the data only once, because it took forever. And I wanted it recursive.

<pre><code>$ tar cf - src | tee &gt;( cd dest1 ; tar xf - ) &gt;( cd dest2 ; tar xf - ) | ( cd dest3 ; tar xf - )
</code></pre>

(And you can add more of those >() sections if you want more outputs.)

I haven't benchmarked that, but it's definitely a lot faster than cp-in-a-loop (or a bunch of parallel cp invocations).

blocks|key|1505055|text|如果您不想使用forked命令：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1505056|tee+<inputfile+file2+file3+file4+...+>/dev/null|offset|length|style|CODE|1505057|entityMap^0|0|0|1B|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

If you want to do it without a forked command:

<code>tee &lt;inputfile file2 file3 file4 ... &gt;/dev/null</code>

blocks|key|1720242|text|本质上等同于xargs答案，但如果您想要并行执行：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1720243|parallel+-q+cp+file1+:::+/foo/+/bar/|code-block|syntax|javascript|1720244|例如，要将file1复制到当前文件夹的所有子目录中(包括递归)：|1720245|parallel+-q+cp+file1+:::+`find+-mindepth+1+-type+d`|1720246|注意:这可能只在非常特定的用例中传递了任何明显的速度提升，例如，如果每个目标目录是不同的磁盘。|1720247|它在功能上也类似于xargs的'-P‘参数。|1720248|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Essentially equivalent to the xargs answer, but in case you want parallel execution:

<pre><code>parallel -q cp file1 ::: /foo/ /bar/
</code></pre>

So, for example, to copy file1 into all subdirectories of current folder (including recursion):

<pre><code>parallel -q cp file1 ::: `find -mindepth 1 -type d`
</code></pre>

N.B.: This probably only conveys any noticeable speed gains for very specific use cases, e.g. if each target directory is a distinct disk. 

It is also functionally similar to the '-P' argument for xargs.

blocks|key|1504305|text|另一种方法是使用cat和tee，如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1504306|cat+<source+file>+%7C+tee+<destination+file+1>+%7C+tee+<destination+file+2>+[...]+>+<last+destination+file>|code-block|syntax|javascript|1504307|我认为这将是非常低效的，因为作业将被分成几个进程(每个目的地一个)，并且硬盘驱动器将一次在盘片的不同部分写入几个文件。但是，如果您想将一个文件写出到几个不同的驱动器，这种方法可能会非常有效(因为所有副本都可以并发发生)。|1504308|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Another way is to use cat and tee as follows:

<pre><code>cat &lt;source file&gt; | tee &lt;destination file 1&gt; | tee &lt;destination file 2&gt; [...] &gt; &lt;last destination file&gt;
</code></pre>

I think this would be pretty inefficient though, since the job would be split among several processes (one per destination) and the hard drive would be writing several files at once over different parts of the platter. However if you wanted to write a file out to several different drives, this method would probably be pretty efficient (as all copies could happen concurrently).

blocks|key|1719935|text|不，你不能。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1719936|我多次发现我可以使用这个功能，所以我做了自己的工具来做这件事。|1719937|http://github.com/ddavison/branch|offset|length|1719938|很简单-|1719939|branch+myfile+dir1+dir2+dir3|style|CODE|1719940|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|0|X|0|0|0|0|S|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|W|8|@]|9|@$F|X|G|Y|1|Z]]|A|$]]|$1|H|3|I|5|6|7|10|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|11|8|@$F|12|G|13|L|M]]|9|@]|A|$]]|$1|N|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|O|$P|$5|Q|R|S|A|$T|E]]]]

No - you cannot.

I've found on multiple occasions that I could use this functionality so I've made my own tool to do this for me.

<a href="http://github.com/ddavison/branch" rel="nofollow">http://github.com/ddavison/branch</a>

pretty simple - 
<code>branch myfile dir1 dir2 dir3</code>

blocks|key|1505213|text|ls+-d+*/+%7C+xargs+-iA+cp+file.txt+A|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1505214|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>ls -d */ | xargs -iA cp file.txt A
</code></pre>

blocks|key|1505323|text|使用bash脚本|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1505324|DESTINATIONPATH[0]="xxx/yyy"
DESTINATIONPATH[1]="aaa/bbb"
++++++++++++++++..
DESTINATIONPATH[5]="MainLine/USER"
NumberOfDestinations=6

for+((+i=0;+i<NumberOfDestinations;+i%2B%2B))
++++do
++++++++cp++SourcePath/fileName.ext+${DESTINATIONPATH[$i]}

++++done
exit|code-block|syntax|javascript|1505325|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Using a bash script

<pre><code>DESTINATIONPATH[0]="xxx/yyy"
DESTINATIONPATH[1]="aaa/bbb"
 ..
DESTINATIONPATH[5]="MainLine/USER"
NumberOfDestinations=6

for (( i=0; i&lt;NumberOfDestinations; i++))
 do
 cp SourcePath/fileName.ext ${DESTINATIONPATH[$i]}

 done
exit
</code></pre>

blocks|key|1720023|text|如果你想把多个文件夹复制到多个文件夹，你可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1720024|echo+dir1+dir2+dir3+%7C+xargs+-n+1+cp+-r+/-r/-n+dir3/{subdir1，subdir2，subdir3}|1720025|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

if you want to copy multiple folders to multiple folders one can do something like this:

echo dir1 dir2 dir3 | xargs -n 1 cp -r /path/toyourdir/{subdir1,subdir2,subdir3}

blocks|key|1505382|text|如果您的所有目标目录都与路径表达式匹配-就像它们都是path/to的子目录-那么只需将find与cp结合使用，如下所示：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1505383|find+./path/to/*+-type+d+-exec+cp+[file+name]+{}+\;|code-block|syntax|javascript|1505384|就这样。|1505385|entityMap^0|Q|7|17|4|1C|2|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

If all your target directories match a path expression — like they're all subdirectories of <code>path/to</code> — then just use <code>find</code> in combination with <code>cp</code> like this:

<pre><code>find ./path/to/* -type d -exec cp [file name] {} \;
</code></pre>

That's it.

blocks|key|1504971|text|如果您需要指定要将文件复制到哪些文件夹中，您可以将find与一个或多个grep结合使用。例如，要替换任何子文件夹中出现的任何favicon.ico，您可以使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1504972|find+.+%7C+grep+favicon\.ico+%7C+xargs+-n+1+cp+-f+/root/favicon.ico|code-block|syntax|javascript|1504973|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If you need to be specific on into which folders to copy the file you can combine find with one or more greps. For example to replace any occurences of favicon.ico in any subfolder you can use:

<pre><code>find . | grep favicon\.ico | xargs -n 1 cp -f /root/favicon.ico
</code></pre>

blocks|key|1719744|text|这将复制到立即子目录，如果您想要更深入，请调整-maxdepth参数。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1719745|find+.+-mindepth+1+-maxdepth+1+-type+d%7C+xargs+-n+1+cp+-i+index.html|code-block|syntax|javascript|1719746|如果你不想复制到所有目录，希望你可以过滤掉你不感兴趣的目录。复制到以a开头的所有文件夹的示例|1719747|find+.+-mindepth+1+-maxdepth+1+-type+d%7C+grep+\/a+%7Cxargs+-n+1+cp+-i+index.html|1719748|如果复制到任意/不相交的目录集，您将需要Robert+Gamble的建议。|1719749|entityMap^0|N|9|0|0|Y|1|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|10|8|@]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|R|$]]

This will copy to the immediate sub-directories, if you want to go deeper, adjust the <code>-maxdepth</code> parameter. 

<pre><code>find . -mindepth 1 -maxdepth 1 -type d| xargs -n 1 cp -i index.html
</code></pre>

If you don't want to copy to all directories, hopefully you can filter the directories you are not interested in. Example copying to all folders starting with <code>a</code>

<pre><code>find . -mindepth 1 -maxdepth 1 -type d| grep \/a |xargs -n 1 cp -i index.html
</code></pre>

If copying to a arbitrary/disjoint set of directories you'll need Robert Gamble's suggestion.

blocks|key|1720318|text|我喜欢将一个文件复制到多个目录中，比如：cp+file1+/foo/;+cp+file1+/bar/;+cp+file1+/foo2/;+cp+file1+/bar2/，然后将一个目录复制到其他目录中：cp+-r+dir1/+/foo/;+cp+-r+dir1/+/bar/;+cp+-r+dir1/+/foo2/;+cp+-r+dir1/+/bar2/|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1720319|我知道这就像发出几个命令，但当我想输入一行然后离开一段时间时，它对我来说很有效。|1720320|entityMap^0|K|1S|2T|24|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]|$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|O|8|@]|D|@]|E|$]]|$1|H|3|-4|5|6|7|P|8|@]|D|@]|E|$]]]|I|$]]

I like to copy a file into multiple directories as such:
<code>cp file1 /foo/; cp file1 /bar/; cp file1 /foo2/; cp file1 /bar2/</code>
And copying a directory into other directories:
<code>cp -r dir1/ /foo/; cp -r dir1/ /bar/; cp -r dir1/ /foo2/; cp -r dir1/ /bar2/</code>

I know it's like issuing several commands, but it works well for me when I want to type 1 line and walk away for a while.

blocks|key|1504755|text|例如，如果您在目标文件夹的父目录中，则可以执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1504756|for+I+in+$(ls)；do+cp+sourcefile+$i；done|1504757|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

For example if you are in the parent directory of you destination folders you can do:

for i in $(ls); do cp sourcefile $i; done

Is it possible to copy a single file to multiple directories using the cp command ?

I tried the following , which did not work: 

<pre><code>cp file1 /foo/ /bar/
cp file1 {/foo/,/bar}
</code></pre>

I know it's possible using a for loop, or find. But is it possible using the gnu cp command?

How to copy a file to multiple directories using the gnu cp command

是否可以使用cp命令将单个文件复制到多个目录？我尝试了以下方法，但都不起作用：cp file1 /foo/ /bar/cp file1 {/foo/,/bar}我知道可以使用for循环或find。但是可以使用gnu cp命令吗？

问如何使用gnu cp命令将文件复制到多个目录
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回答 19

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回答 19

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问如何使用gnu cp命令将文件复制到多个目录
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