Command line to automatically crop an image?

Question

By using Gimp's menu, you can automatically crop the image (removing white borders). I have a lot of images with white borders of different sizes. I want to remove them using Gimp in command line but I cannot figure out what the command is.

Anyone has an idea?

Maybe by using ImageMagick?

Answer 1

By using ImageMagick:

convert -trim image.jpg image.jpg

To trim/autocrop the whole directory:

for a in *.jpg; do convert -trim "$a" "$a"; done

Or using find:

find -name "*.jpg" -exec convert -trim "{}" "{}" \;

Answer 2

I haven't used this in a while but hopefully it will help. Make a gimp batch script (I call mine crop-png.scm), and put it in ~/.gimp-2.6/scripts/).

(define (crop-png filename)
  (let* 
    (
    (image (car (gimp-file-load RUN-NONINTERACTIVE filename filename)))
    (drawable (car (gimp-image-get-active-layer image)))
    )

  ; crop the image
  (plug-in-autocrop RUN-NONINTERACTIVE image drawable)

  ; save in original png format
  (file-png-save RUN-NONINTERACTIVE image drawable filename filename
       0 6 0 0 0 1 1)

  ; clean up the image
  (gimp-image-delete image)
  )
)

Then save this shell scrip (e.g., pngcrop.sh) and call it on the png files like this: 'pngcrop.sh *.png'

#!/bin/bash

if [ $# -le 0 ]; then
    echo
    echo "Usage: $(basename $0) file1.png [file2.png ...]"
    echo
    echo "  This script uses gimp to autocrop PNG files and"
    echo "  save them to PNG format.  You must have"
    echo "  crop-png.scm installed in your gimp "
    echo "  scripts directory."
    echo
    exit 1
fi

# set the filelist
files=$*

# # set the base command
# CMD="gimp -i -b "

# loop and add each file
for i in ${files[*]} ; do
  # #echo $i
  # ARGS="\"(crop-png \\\"$i\\\")\""
  # CMD="$CMD $ARGS"

  gimp -i -b "(crop-png \"$i\")" -b "(gimp-quit 0)"
done

# # add the end to quit
# TAIL="-b \"(gimp-quit 0)\""
# CMD="$CMD $TAIL"
# 
# #echo $CMD
# eval $CMD