Half OT: Convenient way to a copy with path of single files

Half OT: Convenient way to a copy with path of single files

[meino.cramer]

Hi,

I am looking for a convenient way to copy a handful
of files as follows ("mycp" stands for what ever I
have to use to accomplish the copy):

    mycp /home/user/A/B/C/D/files.txt /home/A/B/other/.

results in :

   /home/A/B/other/C/D/files.txt

where 'files.txt' is a placeholder for everything which my
be matched by regular expressions inclusive furthere directories,
symlinks etc...

Is there a alreadu a command in the treasure of zsh or do I have
to script something?

Thank you very much in advance for any help!
Best regards,
mcc

Re: Half OT: Convenient way to a copy with path of single files

[Bart Schaefer]

On Sep 24,  7:09am, meino.cramer@gmx.de wrote:
}
}     mycp /home/user/A/B/C/D/files.txt /home/A/B/other/.
} 
} results in :
} 
}    /home/A/B/other/C/D/files.txt

Something along these lines should work:

autoload -Uz zmv
zmv -C '/(home)/user/(A)/(B)/(C)/(D)/(files.txt)' '/$1/$2/$3/other/$4/$5/$6'

Unfortunately that doesn't translate well to replacing "files.txt" with
an arbitrary subtree pattern, because parenthesized glob patterns can
only refer to files within a single directory.

However, the special case of a recursive glob will work if you make use
of the -w option:

zmv -wC '/(home)/user/(A)/(B)/(C)/(D)/**/*(.txt)' '/$1/$2/$3/other/$4/$5$6$7'

Here $5 contains everything found by **/ _including_ the trailing slash,
and $7 contains the .txt extension; hence no slashes in $5$6$7.