Lossless jpeg batch crop on Linux

Question

I need to crop a number of images in jpeg format by 20 pixels on the right side losslessly on Linux.

I checked jpegtran, but it needs the file size in pixels before cropping, and I don't know how to build a batch file with that.

How can I losslessly crop 20 pixels from the right side of images programmatically?

If you're happy with jpegtran and just need the pixels, you can get that with ImageMagick identify. you can also automate it for a batch with some shell commands. — acfrancis, Nov 07 '13 at 22:55
Linux or Windows. So the question should maybe run more like: How do I do this in Linux or Windows shell. I'll take a look at ImageMagick. — user2708058, Nov 07 '13 at 22:59
Not necessarily, certain operations can be performed without reencoding the rest of the image. Like cropping. — user2708058, Nov 08 '13 at 00:03

acfrancis · Accepted Answer · 2013-11-08T00:46:54.107

2

My shell scripting is a little rusty so please make a backup of your images before trying this script.

#!/bin/bash
FILES=/path/to/*.jpg

for f in $FILES
do
    identify $f | awk '{ split($3, f, "x"); f[1] -= 20; cl = sprintf("jpegtran -crop %dx%d+0+0 %s > new_%s", f[1], f[2], $1, $1); system(cl); }'
done

Points to note:

Adjust the path to the correct value
Do you need *.jpeg?
identify is an ImageMagick command
awk will grab the pixel dimensions from identify to use as a parameter (with the width reduced by 20px) for jpegtran to crop the image
The new image is saved as new_[old_name].jpg
jpegtran might adjust the cropping region so that it can perform losslessly. Check that the resulting images are the correct size and not slightly larger.

edited Nov 08 '13 at 00:46

answered Nov 08 '13 at 00:28

acfrancis

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identify prints out the image dimensions which is what you need (adjusted) as a parameter for jpegtran – acfrancis Nov 08 '13 at 00:40
Yeah your explanation gave me the idea that something else was changing the image size *before* passing it to jpegtran, but reading the commands clarified that. – Havenard Nov 08 '13 at 00:40
Coincidently this week I came across a situation where I was searching for backgrounds to a new prototype website, I found one that was perfect if I flipped it upside down but had the frustration of seeing the file multiply in size. This program did the trick and even reduced the file size, pretty neat thing. – Havenard Nov 08 '13 at 00:50

score 2 · Answer 2 · answered Nov 17 '20 at 09:37

Very similar to the accepted answer, the following would also work with file names containing spaces. And it is arguably simpler, using identify's built-in -format option instead of parsing the output with awk.

#!/bin/bash

X=0; Y=0   # offset from top left corner

for f in /path/to/*.jpg; do
    read -r W H < <(identify -format '%w %h' "$f") # get width and height
    (( W -= 20 ))                                  # substract 20 from width
    out="${f%%.jpg}-crop-20.jpg"                   # add "-crop-20" to filename
    jpegtran -crop ${W}x$H+$X+$Y "$f" > "$out"     # crop
done

Lossless jpeg batch crop on Linux

2 Answers2

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