How to print only part of output

Question

I'm kinda new to shell scripts and I'm struggling to get only the information I need from a command.

I'm executing this script:

#!/bin/bash
myfunction(){
   find . -name '*' -exec file {} \; | grep "image"
}
myfunction

Basically, what I need is the list of the names of the images on my directory and the data about their resolution, but I get a lot of data that are useless to me:

./image222s.jpeg: JPEG image data, JFIF standard 1.01, aspect ratio, density 1x1, segment length 16, baseline, precision 8, 554x554, components 3
./fruta-png.png: PNG image data, 1400 x 1501, 8-bit/color RGBA, non-interlaced
./ddd.jpeg: JPEG image data, JFIF standard 1.01, aspect ratio, density 1x1, segment length 16, baseline, precision 8, 540x568, components 3
./images.jpeg: JPEG image data, JFIF standard 1.01, aspect ratio, density 1x1, segment length 16, baseline, precision 8, 452x678, components 3

I'd like to know how can I reduce the output to only the file name and its resolution. Something like:

image.jpeg: 452x678
image222.jpeg: 554x554

Thanks!

Answer 1

If you can use identify tool from imagemagick, you can pass a -format string to only print the needed information:

Install imagemagick if necessary:

sudo apt-get install imagemagick

Then run:

find . -type f -exec identify -format "%i: %P\n" {} \; 2>/dev/null

(or + instead of \; which will be faster but will print at the end only).

---

If you have a lot of non-image files matching your find, you might want to filter these first because file will be faster than identify:

find . -type f -exec sh -c 'file "$1" | grep -q image' find-sh {} \; -exec identify -format '%i: %P\n' {} \;

Note, that file might filter some file formats it doesn't know about, that identify could do, e.g. RAW images.