I have this command in a bash script:
mv audio.mp3 $datadir/$newname
where $datadir is a directory created at the beginning of the script:
datadir=$(date +%Y-%m-%d_%H-%M)
mkdir -p $datadir
while the rest if the script is
wget -q "$url" -O audio.mp3
poddate=$(stat -c "%y" "audio.mp3"|awk '{print $1"_"$2}'|sed 's/\..*$//')
extmp3=.mp3
seprdr=_
newname=$podname$seprdr$poddate$extmp3
echo -e "${GREEN}\t$newname"
mv audio.mp3 $datadir/$newname
the script runs fine when it is located on the hard drive.
Now, if it is run from an SD card, I get the following error:
cannot move audio.mp3 to a subdirectory of itself
what is wrong?
thanks
Related
How do I change the sample rate for every file in the folder?
The following code converts the files, but it erases the file. After this command, every wav file is empty.
for i in wav/*.wav; do
sox -r 8000 -e unsigned -b 16 -c 1 "$i" "$i"
done
How do I run the code to every file in the directory?
Something like:
for i in wav/*.wav; do
sox -r 8000 -e unsigned -b 16 -c 1 "$i" "$i_modified"
done
I wrote a skript to quickly create short preview clips from vides I recorded on timestamps that I found worth checking out later for cutting.
My file with the timestamps is written like this
FILE_NAME1#MM:SS MM:SS
FILE_NAME2#MM:SS MM:SS MM:SS MM:SS
example:
MAH01728#02:47 03:34 03:44 05:00 06:08 06:55
The script looks like this:
#!/bin/bash
while read f
do
file=$(echo $f | cut -d"#" -f1)
filename=${file}".MP4"
timestamps=$(echo $f | cut -d"#" -f2)
for time in $timestamps
do
ffmpeg -ss 00:${time}.0 -i "orig/${filename}" -c copy -t 10 "preview/${file}_${time}.MP4"
done
done < $1
The script gets half of the previews that I want and on the other the filename is messed up and ffmpeg complains that the file is not found:
orig/714.MP4: No such file or directory
orig/00:58 01:25.MP4: No such file or directory
So I modified the script for trouble shooting and just put an echo in front of the ffmpeg command - now all file names are correct. What am I missing?
ffmpeg -ss 00:01:47.0 -i orig/MAH01714.MP4 -c copy -t 10 preview/MAH01714_01:47.MP4
ffmpeg -ss 00:02:00.0 -i orig/MAH01713.MP4 -c copy -t 10 preview/MAH01713_02:00.MP4
ffmpeg -ss 00:00:58.0 -i orig/MAH01712.MP4 -c copy -t 10 preview/MAH01712_00:58.MP4
ffmpeg -ss 00:01:25.0 -i orig/MAH01712.MP4 -c copy -t 10 preview/MAH01712_01:25.MP4
ffmpeg reads from standard input, consuming data from $1 that was intended for the read command at the top of the loop. Redirect its standard input from /dev/null:
while IFS="#" read file timestamps; do
filename="$file.MP4"
for time in $timestamps; do
ffmpeg -ss 00:${time}.0 -i "orig/${filename}" \
-c copy -t 10 "preview/${file}_${time}.MP4" < /dev/null
done
done < "$1"
echo does not read from standard input, which is why your modification made it appear to be working correctly.
Using this code is it possible for it run this code then create a folder with file name if folder already exits move same name to same folder?
for i in *.mkv;
do name=`echo "${i%.*}.mkv"`;
ass="ass='$name'.ass";
echo "$ass"; ffmpeg -i "$i" -vf "$ass" anime2/"${i%.mkv}.mp4";
done
You can just use mkdir -p and avoid having to use if/then:
for i in *.mkv; do mkdir -p "${i%.*}" && ffmpeg -i "$i" -vf ass="${i%.*}.ass" "${i%.*}/${i%.*}.mp4"; done
There is something weird going on when running following script with Docker.
The dockerfile for this is:
FROM debian:9
WORKDIR /app
RUN apt-get -y update && \
apt-get -y install ffmpeg curl
COPY . /app
The script run.sh:
#!/bin/bash
find /pfs/in -maxdepth 1 -name "*.flac" -print0 | while IFS= read -r -d $'\0' inFile; do
echo "\n##### Process '${inFile}' #####"
ffmpeg -y -i ${inFile} -ar 16000 tmp.wav # use 16kHz - default for EML
done
Starting this, when mounting 3 files into the container:
$ ls pfs/in/
Testaudio16k_2.flac Testaudio16k.flac Testaudio16k.wav TestSprache_Saetze.flac
$ docker run --rm -t -v $(pwd)/pfs/in:/pfs/in test-img:latest /bin/bash run.sh
I get an error on processing the second file: pfs/in/Testaudio16k_2.flac: No such file or directory. The leading / is missing. It is also missing in the preceeding echo. Indeed this happens every second file (if I put more than 3 files in that folder).
Now coming to the counter example:
If I comment out the ffmpeg line in the script, rebuild and run, The echo prints for every file the correct path.
Does anybody have an idea about this?
Is it about the find or is the ffmpegdoing something weird? Something completely different?
Use -nostdin:
ffmpeg -nostdin -i input output
From the ffmpeg documentation:
-stdin
Enable interaction on standard input. On by default unless
standard input is used as an input. To explicitly disable interaction
you need to specify -nostdin.
Disabling interaction on standard input is useful, for example, if
ffmpeg is in the background process group. Roughly the same result
can be achieved with ffmpeg ... < /dev/null but it requires a shell.
I was able to reproduce on a regular Bash console. It's indeed a weird interaction between ffmpeg and the IFS or null delimiter.
This method work as intended
find ~/tmp -maxdepth 1 -name "*.flac" -print0 | xargs -r0 -i bash -c 'echo {}; ffmpeg -hide_banner -y -i {} -ar 16000 tmp.wav'
For your code, add < /dev/null at the end of ffmpeg command as explained here.
ffmpeg -hide_banner -y -i {} -ar 16000 tmp.wav < /dev/null
Can't reproduce that but I strongly recommend to quote the filename when passing it to ffmpeg like this:
ffmpeg -y -i "${inFile}" -ar 16000 tmp.wav.
A space at the beginning or end of the filename would explain your error when you don't use quotes
Btw, you don't need a bash loop. You can just use find:
docker run --rm -t -v $(pwd)/pfs/in:/pfs/in foo \
find /pfs/in -maxdepth 1 -name "*.flac" \
-printf "## Process file %f ##\n" \
-exec ffmpeg -y -i {} -ar 16000 tmp.wav \;
I'm working on some Haskell project using FFmpeg. I need to batch create from a media folder with MP4 files and create screenshots from all of them. I got the code and am using it on a terminal in Unix. It works, but how do I make it in one line to be executed in system "xxxx" in Haskell?
If not using several system"xx"...
#/bin/sh
for i in $(ls *.mp4)
do
ffmpeg -i $i -vframes 7 -y -ss 10 -s 150x150 -an -sameq -f image2 -r 1/5 $i%1d.jpg
done
I tried:
import System.Cmd
function = do{system "#/bin/sh";
system "for i in $(ls *.mp4)";
system "do";
system "ffmpeg -i $i -vframes 7 -y -ss 10 -s 150x150 -an -sameq -f image2 -r 1/5 $i%1d.jpg";
system "done";}
but it gives a error:
-vframes: No such file or directory
/bin/sh: Syntax error: "done" unexpected
The problem is that you're trying to execute each line of your script as a separate, independent invocation of the shell. You just need to do it all with one system call, and separate each line of the script with \n:
system "for i in $(ls *.mp4)\ndo\n..."
but you can write the shell command on one logical line, instead:
system "for i in $(ls *.mp4); do ...; done"
The first line (which should be #!/bin/sh, by the way) is not necessary when using system.
I'm not sure why you want to use Haskell for this purpose, though, if you're just going to execute a single shell script. You should write the loop over the directory contents in Haskell, and only call out to the system to do an individual conversion. At the very least, you should probably put this script into its own file and invoke it with system "sh convert.sh" or similar.
(If you want a more convenient syntax for multi-line strings like these scripts in Haskell, try the interpolatedstring-perl6 or string-qq packages.)
First, It's #!/bin/sh. Notice the exclamation mark.
Second, you're trying to execute a series of commands one after another, so no state is kept between them. Try to execute it as a single command:
function = system "for i in $(ls *.mp4); do ffmpeg -i $i -vframes 7 -y -ss 10 -s 150x150 -an -sameq -f image2 -r 1/5 $i%1d.jpg; done"
Another option is to save your whole script, with the #! corrected, as a .sh file, make it executable and:
function = system "./myscript.sh"
Bash 4.X Solution
system "/bin/bash -c 'shopt -s globstar; for i in **.mp4; do ffmpeg -i \"$i\" -vframes 7 -y -ss 10 -s 150x150 -an -sameq -f image2 -r 1/5 \"$i\"%1d.jpg; done'"
You don't need #!/bin/bash with system (don't forget the bang !)
Quote your variables otherwise files with spaces in their names wont work
Don't use ls like that, it will break when it comes across a file with spaces in its name
Posix Solution
system "find /some/path -type f -name \"*.mp4\" -exec sh -c 'for f; do ffmpeg -i \"$f\" -vframes 7 -y -ss 10 -s 150x150 -an -sameq -f image2 -r 1/5 \"$f%1d.jpg\"; done' _ {} +"
You should not echo the shell script like this but create a shell command like this:
system "for i in $(ls *.mp4); do ffmpeg -i $i -vframes 7 -y -ss 10 -s 150x150 -an -sameq -f image2 -r 1/5 $i%1d.jpg; done"