CODEWITHSUNDEEP
bashmozjpeg
Grin
Grin

Reputation: 577

mozcjpeg - optimize original image

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.

Upvotes: 2

Views: 783

Answers (3)

Denilson Sá Maia
Denilson Sá Maia

Reputation: 49527

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

  • The shell will open foo.jpg for write (overwriting any existing data).
  • The shell will fork and set this open file as the stdout.
  • The shell will exec anycmd foo.jpg.

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.

Upvotes: 2

user137369
user137369

Reputation: 5714

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

Upvotes: 1

Vadim Landa
Vadim Landa

Reputation: 2854

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

Upvotes: 0

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