How to display filename only as an output of cp command?

Question

I'm copying files from src folder to dest folder usign the command:

cp -nv /src/* /dest/ > copy_result.txt

The result of copied files is in the output file as shown below:

 « /src/test1.txt » -> « /dest/test1.txt »
 « /src/test2.txt » -> « /dest/test2.txt »

Otherwise , I want that the output file contains the filename only, without the whole path and without the file extension, like so:

test1
test2

Answer 1

Try the command

cp -nv /src/* /dest/ | awk '{print $6}' | rev | cut -d/ -f1 | rev | cut -d. -f1 > copy_result.txt

awk prints the sixth field. And rev reverses the string. cut splits the string at a delimiter specified by -d, extract portion specified by -f.

Edit:
As pointed out, this is not friendly with files/directories with spaces. But this will by extracting substrings between '-> « ' and ' »' :

cp -nv /src/* /dest/ | sed -e 's/.*-> « \(.*\) ».*/\1/' | rev | cut -d/ -f1 | rev | cut -d. -f1 > copy_result.txt

Answer 2

-v is not specified by POSIX. This means each cp implementation may handle it differently (or not at all). Where your cp says « foo » my says 'foo'. It's possible to craft a command that would transform your undesired output into the desired one, still I'd like to present a solution independent from cp implementation.

---

After /src/* in your command expands, cp copies the resulting entries one by one. This means you can turn cp -n /src/* /dest/ into

for file in /src/*; do cp -n "$file" /dest/; done

Now it's easy to do something with $file. This will generate the output you want:

for path in /src/*; do
   file="${path##*/}"
   dst="/dest/$file"
   [ ! -e "$dst" ] && cp "$path" "$dst" && printf '%s\n' "${file%.*}"
done

(To redirect to copy_result.txt use done > copy_result.txt instead of just done).

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Note if you used cp -r -v and at least one of the operands from the expansion of /src/* was a directory, cp would copy and print its contents, while the above code (after we add -r) would copy as much but print only the directory itself. -r may not be the only option that would introduce discrepancy.

So while the approach should be independent from cp implementation, it relies on some option(s) not being used. Another disadvantage is the code runs a separate cp process for every object in /src/, this is far from optimal.

There's also a race condition: if the $dst file appears between [ and cp, cp will do its job. You can use cp -n to avoid overwriting, but printf will still report the file.