Shell – Run a command on every file, with a flag argument that depends on the filename

findshell

Suppose I have a folder containing files with names like

file1.txt
file2.txt
file2.txt

etc. I would like to run a command on each of them, like so:

mycommand file1.txt -o file1-processed.txt
mycommand file2.txt -o file2-processed.txt
mycommand file3.txt -o file3-processed.txt

etc.

There are several similar questions on this site – the difference is that I want to insert the -processed test into the middle of the file name, before the extension.

It seems like find should be the tool for the job. If it wasn't for the -o flag I could do

find *.txt -exec mycommand "{}" ";"

However, the {} syntax gives the whole file name, e.g. file1.txt etc., so I can't add the "-processed" in between the filename and its extension. A similar problem exists with using a simple bash for loop.

Is there a simple way to accomplish this task, using find or otherwise?

Best Answer

If all the files to be processed are in the same folder, you don't need to use find, and can make do with native shell globbing.

for foo in *.txt ; do
  mycommand "${foo}" -o "${foo%.txt}-processed.txt"
done

The shell idiom ${foo%bar} removes the smallest suffix string matching the pattern bar from the value of foo, in this case the .txt extension, so we can replace it with the suffix you want.

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