combining sed with xargs to obtain a source and output file name

Question

I have a situation where I have some input files like this:

M2U0001.MPG
M2U0180.MPG

And I want to run a command (in a bash shell) on each similarly named file in the directory. I'd like the current file name to be given to this command as an input and a modified version of the filename to be given as an output file. Here's an example:

ffmpeg -i M2U0001.MPG M2U0001_fixed.MPG

I had the idea of using xargs and sed, but this is as far as I got:

ls -1 *.MPG | xargs -I{} ffmpeg -i {} `echo {} | sed -r 's/[0-9]{2,}/&_fixed/'`

But this results in the original filename being output in both positions. Am I totally going about this the wrong way?

I found that if I echo the filename directly to the embedded chunk like this it works:

echo M2U0001.MPG | sed -r 's/[0-9]{2,}/&_fixed/'

Answer 1

Or alternatively:

for i in *.MPG ; do ffmpeg -i $i `basename $i .MPG`_fixed.MPG ; done

Edit: Thank joshbaptiste for the hint.

Answer 2

find . -iname "*.mpg" -exec sh -c "ffmpeg -i {} `echo {} | sed -e 's/\./_fixed\./'`" \;

Answer 3

for i in $(ls) should not be used, ls(1) output should not be used for parsing via scripts etc.. due to word splitting and is a common mistake I see in bash scripts at my job.

In this case parameter expansion works fine and is not susceptible to word splitting errors.

for i in *.MPG; do ffmpeg -i "$i" "${i%%.*}"_fixed.MPG ; done

Reference: http://mywiki.wooledge.org/BashPitfalls