Use a command over a set of files and save results

Question

I have a set of files in a directory

test1.in
test2.in
test3.in
test4.in
...

I have written a conversion tool to extract some data from those files and save it into another file.

I want to run that conversion tool for each file and save the results into different output files

./tool -i test1.in -i test1.out
./tool -i test2.in -o test2.out

I have seen a lot of answer to iterate on the files and run the desired command for each file, but I am not sure how to specify the output.

for file in dir/*.in; do
    ./tool -i file -o ???
done

How can I create a name for the output file?

Read up in the bash manual about all the nifty things you can with parameter expansion. — Shawn, Sep 13 '19 at 08:14
Your example looks odd! You mean you want to overwrite `test2.out` each time? — Mark Setchell, Sep 13 '19 at 08:23
ups, that's a typo. Ill correct it. Thanks for pointing it out — jjcasmar, Sep 13 '19 at 08:23

pii_ke · Accepted Answer · 2019-09-13T11:37:45.193

1

for file in dir/*.in; do
    ./tool -i "$file" -o "${file%.*}".out
done

edited Sep 13 '19 at 11:37

answered Sep 13 '19 at 08:18

pii_ke

Are you sure this is correct? I have added `echo` to see the output file and its always outputing in.out – jjcasmar Sep 13 '19 at 09:24
No Wait. I'll test and update.. – pii_ke Sep 13 '19 at 09:28
Done. I think this works now.. – pii_ke Sep 13 '19 at 09:37

1 Answers1