Error when extracting a filename to name other files

Question

I am trying to use FFmpeg to convert a video to images in Bash. I would like to use the videos' filenames to name corresponding images (followed by a integer number).

I was able to do this if I was exporting the files in the same directory:

for file in `find . -name "*.mp4"`; do ffmpeg -i $file -q 1 $file'_'%d.jpeg; done

However, if I were to export images into a desired directory, I got errors. Sees like the value of $file is not just the file name, but a directory.

what I was trying:

for file in `find . -name "*.mp4"`; do ffmpeg -i $file -q 1 ~/testfolder/$file'_'%d.jpeg; done

My question is: how could I properly extract just the filename to use here?

Answer 1

Do it like this, as it will work with files containing any special characters (including newlines, which are rarely used but valid in filenames).

destdir=~/testfolder/
while read -u 5 -r -d '' file
do
  name=$(basename "$file")
  ffmpeg -i "$file" -q 1 "$destdir/${file}_%d.jpeg" </dev/null
done 5< <(find . -name "*.mp4" -print0)

The basename command is an external program. You can achieve the same result with the following lines using only shell expansions :

name=${file%/}
name=${name##*/}

Answer 2

You don't need neither basename nor a loop for this; You can use -type f to avoid find reporting directories in it's result combined with -exec option to execute directly your command.

Something like this should work for all your mp4 files:

find . -type f -name '*.mp4' -exec bash -c 'ffmpeg -i $0 -q 1 $0'_'%d.jpeg' {} \; 

See a small demo test:

$ find . -type f -name 'a*.txt' -exec bash -c 'echo ffmpeg -i $0 ' {} \; 
ffmpeg -i ./a.txt
ffmpeg -i ./a spaced file.txt
ffmpeg -i ./aa.txt
ffmpeg -i ./cheatsheets/awk-cheat-sheet-gv.txt

As you can see second file has spaces in the file name, but is handled correctly by find.
On the contrary such a spaced file will break with a for loop.

If you insist to make this job with a loop , then this must be a while loop as advised by Fred.

As a modification to Fred solution you can avoid the use of basename using the -printf capabilities of find:

while read -r -d '' file;do 
  ffmpeg -i "$file" -q 1 "${file}_%d.jpeg"
done < <(find . -name "*.mp4" -printf %f\\0)

-printf %f according to man page of find prints the file name stripped, and appending \\0 (null char) after each file name we ensure correct filenames handling even if names contain spaces or other special chars.

Small test:

$ while read -r -d '' file;do echo "ffmpeg -i $file -q 1 ${file}_%d.jpeg";done < <(find . -name "a*.txt" -printf %f\\0)
ffmpeg -i a.txt -q 1 a.txt_%d.jpeg
ffmpeg -i a spaced file.txt -q 1 a spaced file.txt_%d.jpeg
ffmpeg -i aa.txt -q 1 aa.txt_%d.jpeg
ffmpeg -i awk-cheat-sheet-gv.txt -q 1 awk-cheat-sheet-gv.txt_%d.jpeg

Answer 3

The bash command basename can be used for this.

In this case the following should list only the files:

for file in $(find . -name "*.mp4"); do echo $(basename $file); done

You can use $(basename $file) in your original code to get the filename only.

for file in $(find . -name "*.mp4"); do 
  ffmpeg -i "${file}" -q 1 "~/testfolder/$(basename $file)'_'%d.jpeg"
done