Lossless jpeg batch crop

问题

I need to crop a bunch of jpegs by 20 pixels on the right side losslessly. Do you know about any software that can do that? 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. Any ideas?

回答1:

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.