Passing value with xargs to generate dynamic output filename

Question

I have some files in a folder. By running the following command I can create a list with the filtered filenames ( test1, test2,test3) and pass it using xargs to a grep filtering command filter file in the command contains a few values to be filtered out.

ls -Art1 | grep test | xargs -L1 grep -vf filter > ouput 

However when run this command output file contain the filtered result of test1, test2 and test3. All in one.

I would like to have separate file for each test1 > test1_output, test2 >test2_output,... so taking the xargs value and just add an extra string to it "_output" to generate an output filename

Answer 1

There's no need to pipe the output of ls to search for a name match. Just use a shell glob. This also allows you to define an output file for each filter attempt

for file in *test* do [ -f "./$file" ] && grep -vf filter "./$file" >"${file}_output" done 

Technically this is potentially a slightly different set of file matches to your ls -A as your code considers dot files whereas mine does not. Fixable if relevant.

In your comment you mention you are performing two actions on each file. If I have understood you correctly, then for such a situation you can modify the code like this:

for file in *test* do if [ -f "./$file" ] then grep -f filter "./$file" >"${file}_delta" grep -vf filter "./$file" >"${file}_output" fi done