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.
Related
Normally, when using ffmpeg, you define a name for the "pass" information to be stored in a file when enconding videos with 2 (or more) passes, e.g.:
ffmpeg -i "INPUT_FILE" -pass 1 -passlogfile videopass.log /dev/null -y && ffmpeg -i "INPUT_FILE" -pass 2 -passlogfile videopass.log -y "OUTPUT_FILE"
I'd love to be able to create a random pass name automatically in bash in the simplest way possible, preferably a "one-liner" using the system's default tools (Linux)... something like:
$(tr -dc A-Za-z0-9 </dev/urandom | head -c 8).log | ffmpeg -i "INPUT_FILE" -pass 1 -passlogfile > /dev/null -y && ffmpeg -i "INPUT_FILE" -pass 2 -passlogfile > -y "OUTPUT_FILE"
I'm too stupid to figure out a way of doing something similar that actually makes sense.
Thank you very much in advance!
Assuming the requirement is to create a random name for the password file:
$ pwdlog=$(mktemp XXXXXXXX.log) # have mktemp create a file with 8 random characters + ".log"
$ typeset -p pwdlog
declare -- pwdlog="Em6GeMdc.log"
$ ls -l "${pwdlog}"
-rw-------+ 1 username None 0 Apr 26 15:13 Em6GeMdc.log
This file could then be referenced in the ffmpeg call like such:
ffmpeg -i "INPUT_FILE" -pass 1 -passlogfile "${pwdlog}" ...
After countless tries and googling, I found a way to do exactly what I was looking for... A one-liner, nothing too complicated or fancy and using my system's default tools:
printf $RANDOM | xargs -i sh -c 'ffmpeg -i "INPUT_VIDEO" -pass 1 -passlogfile {} -an -f mp4 /dev/null -y && ffmpeg -i "INPUT_VIDEO" -pass 2 -passlogfile {} -y "OUTPUT_VIDEO"'
This will create a very simple 5-digit random number, which will be used as the initial name for the pass files in ffmpeg, ffmpeg automatically adds the file extension(s). Of course, you can (and probably should) add video/audio parameters to your ffmpeg commands. The important elements are the "printf $RANDOM | xargs -i sh -c" at the begining and the "{}" curly brackets after the "-passlogfile" command.
There are probably even simpler or more elegant ways of doing it, but this works exactly as I wanted, so I'm happy.
How do I change the sample rate for every file in the folder?
The following code converts the files, but it erases the file. After this command, every wav file is empty.
for i in wav/*.wav; do
sox -r 8000 -e unsigned -b 16 -c 1 "$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" "$i_modified"
done
Why is my code below not working as expected?
shopt -s extglob || {
echo "Unable to enable exglob."
exit 1
}
TARGETEXT='.jpg'
TARGETPREFIX='./' ## Make sure it ends with /.
while IFS= read -r FILE; do
BASE=${FILE##*/}
NOEXT=${BASE%.*}
TARGETFILEPATH=${TARGETPREFIX}${NOEXT}${TARGETEXT}
duration=$(ffprobe -i "$FILE" -show_format -v quiet | sed -n 's/duration=//p'| sed 's/\.[^\.]*$//')
space=$(awk '{print $1/$2}' <<<"$duration 5")
echo ffmpeg -ss "$space" -i "$FILE" -vf "select='isnan(prev_selected_t)+gte(t-prev_selected_t,"$space")',sca
le='if(gt(a,4/3),206,-1):if(gt(a,4/3),-1,154)',pad=w=206:h=155:x=(ow-iw)/2:y=(oh-ih)/2:color=black,tile=4x1" -aspect
4:3 -frames:v 1 -vsync vfr "$TARGETFILEPATH" -y
done < <(exec find -type f -name '*.mp4')
I have 2 mp4 files in the directory. If I run this code it work fine with 'echo' in front of the ffmpeg command. It'll output 2 lines of ffmpeg, one line for each file. Filenames are correct and all looks perfect.
When I run the same, removing 'echo', the first ffmpeg command will run successfully and my script will give an error for the 2nd file name. I can see in the ffmpeg output that there is one character missing at the beginning of the filename(should be uscenes.mp4).
scenes.mp4: No such file or directory
I've started over with a for loop and without find. All working now!
IEXT='.jpg'
for fname in ./*.mp4; do
[ -e "$fname" ] || continue
NOEXT=${fname::-4}
OUPUT=${NOEXT}${IEXT}
duration=$(ffprobe -i "$fname" -show_format -v quiet | sed -n 's/duration=//p'| sed 's/\.[^\.]*$//')
space=$(awk '{print $1/$2}' <<<"$duration 5")
ffmpeg -ss "$space" -i "$fname" -vf "select='isnan(prev_selected_t)+gte(t-prev_selected_t,"$space")',scale='
if(gt(a,4/3),206,-1):if(gt(a,4/3),-1,154)',pad=w=206:h=155:x=(ow-iw)/2:y=(oh-ih)/2:color=black,tile=4x1" -aspect 4:3
-frames:v 1 -vsync vfr "$OUPUT" -y
done
I'm new to shell script, trying to concat listed movies in a folder like:
filename="list.txt"
cat ${filename} | while read line;
do
ffmpeg -y -i sum.mov -i $line -filter_complex concat tmp.mov
cp tmp.mov sum.mov
done
However, this loop runs only one time.
What is the best practice?
Thanks
SOLVED:
#list.txt
file 'movie1.mov'
file 'movie2.mov'
file 'movie3.mov'
Don't need to write a shell script. Just
ffmpeg -f concat -i list.txt -c copy output.mov
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"