mozcjpeg - optimize original image

Question

I'm trying to optimize sample.jpg with mozcjpeg, but I want it to compress sample.jpg, and not creating another file.

I run this command:

mozcjpeg -quality 80 sample.jpg > sample.jpg

and I get

Empty input file

What's the right way to do it?

P.S. I have a bash script that runs once a day to optimize images created in the last 24 hours.

score 0 · Answer 1 · answered Apr 25 '17 at 12:43

0

Just use a different output file name, otherwise it gets immediately overwritten.

answered Apr 25 '17 at 12:43

Vadim Landa

Vadim, I need to overwrite original file, that's what the question is about. jpegtran can do that, so I think mozjpeg can do it too. – Grin Apr 25 '17 at 13:12

score 0 · Answer 2 · answered Feb 25 '19 at 01:20

0

You’re reading the file and overwriting it at the same time. Instead, make an intermediary file:

tmp_file="$(mktemp)"
mozcjpeg -quality 80 sample.jpg > "${tmp_file}"
mv "${tmp_file}" sample.jpg

answered Feb 25 '19 at 01:20

user137369

score 0 · Answer 3 · answered Aug 10 '20 at 20:44

anycmd foo.jpg > foo.jpg is understood as:

So, by the time the command is executed, the original file is already gone. How can you work around that?

Well, you can create a temporary file, as suggested in the other answer.

Alternatively, you can use sponge. It's part of moreutils package.

$ whatis sponge
sponge (1)           - soak up standard input and write to a file

This tool allows you to write:

mozcjpeg -quality 80 sample.jpg | sponge sample.jpg

It works because sponge will first read all of the stdin (in memory), and only after stdin reaches EOF it will open the file for writing.

3 Answers3