shell script to get collect pixel size of image

Question

I'm trying to create shell script that returns the largest picture in total dimensions pixel size?

For example:

I have many directory over 7000+, each directory have images:

dir_1/
picture_1.png = 800x600
picture_2.png = 80x100
picture_3.png = 80x640
picture_4.png = 500x630

dir_2/
p_1.png = 800x600
p_2.jpeg = 800x1000
p_3.png = 180x1640
p_4.gif = 500x30

So the result expected is:

 the largest one in dir_1 is: picture_1.png 
 the largest one is dir_2 is: p_2.png 

So I was thinking is best way is to find out the total of dimensions after collect the figures .. therefor I try to create bash script with sips command that can collect the figures

Here example :

 for f in *;
 do
 far=$( cd $f/OEBPS/image/ | ls * | egrep 'jpg|png|jpeg')

 W=$( sips -g pixelWidth $far | cut -f 2 -d":" )
 H=$( sips -g pixelHeight $far | cut -f 2 -d":" )

 coll=$(expr $W + $H)
 echo $f total is: $coll
 cd -
 done

But got error on result.

Any idea or better way?

Answer 1

here's a way to get the height and width in one step:

IFS=x read w h < <(identify "$file" | grep -oP '\d+x\d+(?=\+)')

identify is part of the ImageMagick package.

Your "$far" is surely not what you want:

for dir in */OEBPS/image/; do
    for image in "$dir"/*.{jpg,png,jpeg}; do
        IFS=x read w h < <(identify "$image" | grep -oP '\d+x\d+(?=\+)')
        echo $((w*h)) "$image"
    done | sort -n | tail -1 | {
        read size file
        echo "largest in $dir is $file"
    }
done

---

Actually, identify can take several files, so a more efficient technique:

for dir in */OEBPS/image/; do
    identify "$dir"/*.{jpg,png,jpeg} |
    awk '{split($(NF-6), a, /x/); split($0, b, /[[]/); print a[1]*a[2], b[1]}' |
    sort -n | tail -1 | {
        read size file
        echo "largest in $dir is $file"
    }
done

The awk command is a bit complicated because I want to handle image names that may contain spaces

Answer 2

#!/bin/bash
file=""
max=0
for f in /OEBPS/image/*{png,jpg,jpeg}
do
  id=$(identify "$f") 
  size=$(echo ${id} | sed -r 's/.* (JPEG|PNG) ([0-9]+)x([0-9]+) .*/\2*\3/')   
  area=$(($size))
  if (( area >= max ))
  then
    max=$area
    file="$f"
  fi
done
echo $max $file

• Don't use ls for iterating over files in scripts - blanks in filenames will brake them.
• Don't cd around.
• We don't need sorting to find the biggest file by area.
• For doing arithmetic in bash, you don't need expr, but $((a*b)).

Identify needs ImageMagick.

The sed-command will fail if your filename looks too close like the output of identify.