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:

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

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 *;
 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 -

But got error on result.

Any idea or better way?

  • 1
    What error do you see? Apr 25, 2015 at 15:12
  • 1
    For the "largest" image, do you want to consider the perimeter or the area? Apr 25, 2015 at 15:18
  • Don't do "cd" in scripts. You can iterate from your current directory over the files too. And don't do "ls" in scripts. In bash, you can do: for f in $f/OEBPS/image/*{png,jpeg,jpg}. A ls will bite you if you have blanks and such in fillenames. Apr 25, 2015 at 17:18
  • Also, what if two or more images in the same directory have the same size ? Apr 25, 2015 at 22:00
  • I've fix by add identify "$image" 2> /dev/null ... anyhow thanks for your effort ... last think .. i want to the result print with urlencode for example file: tadlil kalamy fmt.png to be tadlil%20kalamy%20fmt.png I'm use MAC OS X Apr 25, 2015 at 22:46

2 Answers 2


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"

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"

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

  • first script seems is good but need little fix .. i got the result but in same-time i got error like: /OEBPS/image/*.png': No such file or directory @ error/blob.c/OpenBlob/2675 /OEBPS/image/*.jpeg': No such file or directory @ error/blob.c/OpenBlob/2675 seems the script try to find all dir with jpg,jpeg,png and print it to me just need to hide that Apr 25, 2015 at 22:25
  • add shopt -s nullglob before the for loop Apr 25, 2015 at 23:09
  • you could also discard identity's stderr: identity "..." 2>/dev/null Apr 25, 2015 at 23:13
for f in /OEBPS/image/*{png,jpg,jpeg}
  id=$(identify "$f") 
  size=$(echo ${id} | sed -r 's/.* (JPEG|PNG) ([0-9]+)x([0-9]+) .*/\2*\3/')   
  if (( area >= max ))
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.

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