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I want to concatenate multiple WAV files into a single WAV file using FFMPEG.

I have used the following command and it generates the required file.


ffmpeg -f concat -i mylist.txt -c copy output.wav

File :

file '1.wav'
file '2.wav'
file '3.wav'
file '4.wav'

But as you can see the problem is that I have to create a text file to specify the list of WAV files to concatenate.

I can do all these tasks, but I would prefer a single command something that looks like

ffmpeg -i 1.wav -i 2.wav -i 3.wav -i 4.wav output.wav 


ffmpeg -i "concat:1.wav|2.wav|3.wav|4.wav" -c copy output.wav

I have tried these two simple commands but they return just the voice of 1.wav Please help me write a single command( or correct the above 2 commands ) that achieves the desired result.

Please don't suggest other Media Encoders/Editors, I want to use FFMPEG only, as it is already installed and used at other places.

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migrated from Apr 25 '13 at 10:34

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3 Answers 3

up vote 7 down vote accepted

You could try using the concat filter; it requires re-encoding, and so will take more system resources (a pretty tiny amount on any vaguely modern computer in this particular case), but PCM -> PCM audio should be mathematically lossless. In your case, you would use something like:

ffmpeg -i input1.wav -i input2.wav -i input3.wav -i input4.wav \
-filter_complex '[0:0][1:0][2:0][3:0]concat=n=4:v=0:a=1[out]' \
-map '[out]' output.wav

If you have five input files, use n=5.

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Note: Don't customize [0:0][1:0][2:0][3:0] to match the number of files. Use that exactly as it says. The only thing you should change is the input files and n=4. – Keavon Mar 7 at 22:40

The FFmpeg wiki mentions using the concat protocol is not possible with all file types. It works fine with most MPEG containers and bitstreams, but obviously not WAV files with PCM audio.

You don't necessarily have to create a temporary file and use that. With Bash (or other shells that support process substitution), you can do everything in a single command:

ffmpeg -f concat -i <( for f in *.wav; do echo "file '$(pwd)/$f'"; done ) output.wav

The process substitution <( ) creates a file—or, to be precise, a file descriptor—on the fly, which ffmpeg can read. This file will contain the path to every .wav file in the current directory. We have to prefix it with $(pwd) to get the correct path relative to your current directory, and not relative to the file descriptor.

Of course, the command within the substitution can be changed to something else.

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You could use shntool for wav-files.

shnjoin -r none 01.wav 02.wav ...
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