Crop MP3 to last N seconds

Question

I want to generate a new (fully valid) MP3 file from an existing MP3 file. The new file should only contain the last N seconds of the track.

There's -t option in avconv for getting first N seconds:

-t duration (output)

Stop writing the output after its duration reaches duration. duration may be a number in seconds, or in "hh:mm:ss[.xxx]" form.

Is there an option to crop last N seconds? avconv is not requirement in my case, all other tools are acceptable.

Answer 1

You can simply use ffprobe to get the duration of the audio.

ffprobe -i input_audio -show_entries format=duration -v quiet -of csv="p=0"

Calculate nth_second and use it with ffmpeg

ffmpeg -i input_audio -ss nth_second output_file

Following is the shell script for the process.

#!/bin/bash
duration=$(ffprobe -i ${1} -show_entries format=duration -v quiet -of csv="p=0")
nth_second=$(echo "${duration} - ${2}"|bc)
ffmpeg -i ${1} -ss ${nth_second} ${3}

Answer 2

If you want to release the script, you should check to ensure mp3info is installed, and the argument count, etc. but this should get you started. Save it as a .sh and call it with mp3lastnsecs.sh input.mp3 output.mp3 numsecs

Also, you'll need to add the other ffmpeg flags for tagging, bitrate, etc.

#!/usr/bin/bash

SECS=$(mp3info -p "%S" $1)
if [ "$SECS" -ge "$3" ]; then
  START=$((SECS-$3))
else
  echo 'Error: requested length is longer than the song.'
  exit 1
fi

#echo Starting song $1 from $START seconds.
ffmpeg -i "$1" -ss $START "$2"

Sample output:

[  @  tmp] ./lastN.sh 01-glaube.mp3 output.mp3 20
Starting song 01-glaube.mp3 from 96 seconds.
ffmpeg -i 01-glaube.mp3 -ss 96 output.mp3 # I had an echo in there for testing, it now calls ffmpeg directly
[  @  tmp] ./lastN.sh 01-glaube.mp3 output.mp3 1234
Error: requested length is longer than the song.