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jpegcropbatch-processinglossless

Lossless jpeg batch crop on Linux

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?

Solution

  • 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.