This is the bash code
read -t 10 -p "Video path ==>" VIDEO_SOURCE
[ "$VIDEO_SOURCE" = q ]
if [ "$VIDEO_SOURCE" = "$NONE" ]; then
./run.sh
Basically what i wanted it to do is to run ./run.sh after 10 seconds if i didn't give it any input, the problem is that is runs the ./run.sh in less then 10 seconds.
Edit:
This is the full script
#! /bin/bash
VBR="1500k"
FPS="24"
QUAL="superfast"
RTMP_URL="rtmp://live.live"
KEY="xxx-xxx-xxx-xxx"
VIDEO_SOURCE="video.mp4"
while :; do
ffmpeg \
-re -f lavfi -i "movie=filename=$VIDEO_SOURCE:loop=0, setpts=N/(FRAME_RATE*TB)" \
-vcodec libx264 -pix_fmt yuv420p -preset $QUAL -r $FPS -g $(($FPS * 2)) -b:v $VBR \
-f flv "$RTMP_URL/$KEY"
read -t 10 -p "Video path ==>" VIDEO_SOURCE
[ "$VIDEO_SOURCE" = q ]
if [ -z "$VIDEO_SOURCE"]; then
./run.sh
fi
done
What i wanted it to do is to wait 10 seconds after a crash or after i press q , after 10 seconds either is going to run the new input if specified it, if the input is invalid or empty (in case i was too slow typing the input or in case of a crash) i wanted it to run ./run.sh, i'm running this script in tmux just to keep it running and if i want to update the input i just go back to the tmux session and press q and give it the new input.
I couldn't reproduce your problem. Nevertheless, if there really was such a strange problem, you could try to work around it using the following hack:
read var < <(timeout --foreground 10 head -n1)
This should behave somewhat like read -t 10 but relies on the external program timeout instead of read's built-in timeout functionality.
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 $RANDOM | xargs -i sh -c" at the begining and the "{}" curly brackets after the "-passlogfile" command.
There are probably even simpler or more elegant ways of doing it, but this works exactly as I wanted, so I'm happy.
I wrote a skript to quickly create short preview clips from vides I recorded on timestamps that I found worth checking out later for cutting.
My file with the timestamps is written like this
FILE_NAME1#MM:SS MM:SS
FILE_NAME2#MM:SS MM:SS MM:SS MM:SS
example:
MAH01728#02:47 03:34 03:44 05:00 06:08 06:55
The script looks like this:
#!/bin/bash
while read f
do
file=$(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.
I am trying to execute ffmpeg in a loop over multiple files. I only want one instance to run at a time, and to only use 50% of the cpu. I've been trying cpulimit but it isn't playing nice with the loop.
for i in {1..9}; do cpulimit -l 50 -- ffmpeg <all the options>; done
This spawns all nine jobs at once, and they are all owned by init so I have to open htop to kill them.
for i in {1..9}; do ffmpeg <all the options> & cpulimit -p $! -l 50; done
This hangs, ctrl+c continues to the next loop iteration. These instances can only be killed by SIGKILL.
Using a queue is the way to go. A simple solution that I use is Task Spooler. You can limit the number of cores ffmpeg uses with -threads also. Here's some code for you:
ts sh -c "ffmpeg -i INPUT.mp4 -threads 4 OUTPUT.mp4"
You can set the max number of simultaneous tasks to 1 with: ts -S 1
To see the current queue just run ts
You should run it in foreground. In this way the loop will work as expected.
$ cpulimit --help
...
-f --foreground launch target process in foreground and wait for it to exit
This works for me.
for file in *.mp4; do
cpulimit -f -l 100 -- ffmpeg -i "$file" <your options>
done
If you want the -threads option to have an effect on the encoder, you should put it after the -i argument, before the output filename - your current option only tells the decoding part to use a single thread. So to keep it all using a single thread, you want -threads 1 both before and after the -i option. so you can do it like:
ffmpeg -threads 1 -i INPUT.mp4 -threads 1 OUTPUT.mp4
I would like to do what the title says
This is a ffmpeg command to download from a specific time in a video, offline or online.
ffmpeg -ss (stop time) -i (direct video link) -t (start time) -c:v copy -c:a copy (title.mp4)
I am going to be downloading this on OSX.
I dont care what the title is.
I think* there is a bash command that allows me to change the timings in this command up by a specific amount (+300 seconds per, the counter for start and stop time is in raw seconds)
So, bash script that runs that command but increases the start and stop times incrementally by 300 (the stop timing being 60+ seconds ahead), downloads, then repeats.
here it is:
contents of youtube-dl:
#!/usr/bin/env bash
# set start to 0, 300, 600... up to 72000 (20 hours)
for start in `seq 0 300 72000`; do
# set the outfile name
file="$2.$start.60.mp4"
ffmpeg -ss $start -i "$1" -t 60 -c:v copy -c:a copy "$file"
# get the duration of the last outfile
last_duration=`ffprobe -i "$file" -show_entries format=duration -v quiet -of csv="p=0"`
# if last outfile's duration isn't greater than a second, delete it and stop
[[ ! "$last_duration" -gt 1 ]] && rm -f $file && exit
done
then do:
chmod +x youtube-dl
usage:
./youtube-dl "http://your/movie.flv" title
ps: i discovered that your ffmpeg command was a little broken: it's -t (duration), not -t (start time).
refs:
ffmpeg usage (slhck, 2012)
ffprobe usage (ivan-neeson, 2014)
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"