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Accepted
kumar
Asked: 2015-03-19 01:25:58 +0800 CST

How to cp only into the latest directory name?

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mkdir test_dir test_dir1 test_dir2 test_dir3
touch file1 file2 file3
cp -v file* test_dir*/

This copies only to test_dir3 and not in test_dir, test_dir1, test_dir2. Why?

For that matter it copies only to latest directory say test_dir4 and not into any other direcotry (it is omitting them).

bash

2 Answers

  1. Best Answer

    You have these three commands

    mkdir test_dir test_dir1 test_dir2 test_dir3
touch file1 file2 file3
cp -v file* test_dir*/

    Assuming no other files or directories in . before the example is started, the wildcards in that last line get expanded thus:

    cp -v file1 file2 file3 test_dir/ test_dir1/ test_dir2/ test_dir3/

    (You can see this by changing cp to echo cp and observing the result.) What you didn't tell us is the diagnostic messages produced by cp -v, which show what it is trying to do, i.e. copy every item on the command line but the last into the last item, which must therefore be a directory:

    ‘file1’ -> ‘test_dir3/file1’
‘file2’ -> ‘test_dir3/file2’
‘file3’ -> ‘test_dir3/file3’
cp: omitting directory ‘test_dir/’
cp: omitting directory ‘test_dir1/’
cp: omitting directory ‘test_dir2/’
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  2. Because the cp commands works so. If it has more as 2 argument, the last argument must be a directory, and every argument before that will be copied into.

    I think you want a cp which behaves much more like the windows copy command, and yes, there are such tools, although they aren't widely used. Even I had to google for that.

    Your actual goal could be easily reached by a simple loop:

    for i in test_dir*; do cp -v test_file* $i; done

    It is much more simple as it seems on the first spot. Bash is actually a basic-like programming language, and doing loops or even more complex operations in single-line commands in a unix environment is quite common.

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