On 24-01-2022 15:07, Børge Wedel Müller wrote:
Hi Again
When I run the command below, I got a "larger than sign" eg this sign ">"
What Iis I doing wrong?
Best
/Børge Wedel Müller
Den 23. januar 2022 kl. 19.22.19 +01.00, skrev LuKaRo <[email protected]>:
Hi,
you can use standard Unix methods for batch processing, e.g. in bash:
for i in *.mp3; do ffmpeg -ss 00:00:30.000 -i "$i" -c copy "$(basename $i
.mp3)-output.mp3"
This will cut the first 30 seconds from each file ending in .mp3, creating a
file with the same name and -output at the end. Instead of -c copy, you can add
any additional ffmpeg arguments regarding the output.
If you want to leverage parallel processing, you can use xargs instead (note
that this will disable interactive input in ffmpeg, e.g. asking for overwrites):
ls *.mp3 | xargs -P 4 -I '{}' bash -c 'ffmpeg -ss 00:00:30.000 -i "{}" -c copy
"$(basename {} .mp3)-output.mp3"
Again, replace -c copy by any output parameters you'd like to use, and replace
-P 4 by the number of processes you want to start in parallel.
Let us know if you have any further questions.
Kind regards,
LuKaRo
On 1/23/22 17:28, Børge Wedel Müller wrote:
Hi
I have searched over and over again, so this is last resort :-)
I have approx 3000 audio files MP3, and I need to remove the first 30 seconds
of each and everyone of the files.
I am using a LINUX system. (Debian)
I hope you can learn me how to remove these first annoying 30 seconds i one
batch.
Thanks.
Best
/Børge Wedel Müller
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The bash command is not complete, the do statement needs to be closed
with done;
like this:
for i in *.mp3; do ffmpeg -ss 00:00:30.000 -i "$i" -c copy "$(basename $i
.mp3)-output.mp3" done
otherwise bash will prompt for more commands to follow do.
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