识别（并删除）所有.gif文件在所有子目录中具有超过一帧

Question

I having the following linux command to identify all .gif files and list its frames in the active directory:

identify -format '%n %i\n' -- *.gif

Now I want to modify and expand the command to fullfil the following: 1. indentify just the .gif files with MORE than 1 frame 2. look also in all subdirectories (right now the command is just looking at the active directory) 3. a command to: a) list the identified files b) delete the identified files

I would really appreciate you guys help to achieve this...

Thank you a lot in advance!!

Best, Florian

Answer 1

Your on the right path. I would recommend leveraging find , sort and awk commands to generate a list of files to remove.

find /path/to/directory \
     -type f \
     -name "*.gif" \
     -exec identify -format '%n %i\n' {} \; \
     | sort -u \
     | awk '$1 > 1 {print $2}'

How this works

• find will scan all sub-directors for file ending with *.gif , and pass the results to ImageMagick.

• identify will print the filename + frame count (what your already doing)

• sort -u will remove duplicate items ( uniq utility will also work.)
• awk should print the second column (filename) if the first column (frame count) is greater than 1.

note: this will not remove the files, but provide a list that can be inspected, and passed to rm command