I wish to know how to calculate the difference among two video timecodes tha are in frames (in this case, a second equals 30 frames).
Let say the point A is 600 (00:00:02) and the point B is 120 (00:00:04).
How can I calculate the difference among pont 开发者_如何学编程A and B and echo the result in the 00:00:00.00 format (h:m:s) using bash?
UPDATE:
This is perfection: http://www.1728.com/angle.htm
First off, the export
statements are redundant since the current shell is expanding $A
and $B
.
To get those numbers from the timestamps, GNU date
will help:
jinx:752 Z$ date -d 00:01:18.44 +%T.%N # to show that it works
00:01:18.440000000
jinx:753 Z$ date -d 00:01:18.44 +%s.%N
1299214878.440000000
Now you can do your calculations, then translate back at the end for display (as Edwin Buck said, don't try to store intermediate results in this format):
jinx:754 Z$ date -d @1299214878.44 +%T.%N
00:01:18.440000000
Keep all of the times in "frames", then display the frame as a time through a single formatting routine. If you attempt to convert to times, you'll make the mathematical conversions much harder (as time is a base 60 system).
frames2date () {
local secs=$((2 * $1));
date --utc --date "1970-01-01 ${secs} sec" "+%T"
}
should convert the number of frames into a time, note that you need to add years, months, and days if your times exceed 24 hours.
(Not Perfect) SOLUTION:
With the help of my friends, here's my complete bash script:
#!/bin/bash
#ffmpeg [inputfile] [starttimecode] [videoduration] [video details] [size] [audio details] [alltherest] [outputfile]
video=-'vcodec libx264 -vpre slower -crf 22'
size='-s 320x240'
audio='-acodec libmp3lame -aq 0 -ac 1'
in=$(date --utc --date "$3" +%s)
out=$(date --utc --date "$4" +%s)
dur=$(date --utc --date "1970-01-01 $[$out-$in] sec" "+%T")
ss=$(date --utc --date "1970-01-01 $[$in] sec" "+%T")
ffmpeg -i $1.$2 $video $size $audio -threads 0 -y $1.mp4
#ffmpeg -i t1.flv -ss 00:00:32 -t 00:00:04 -vcodec libx264 -vpre slower -crf 22 -s 320x240 -acodec libmp3lame -aq 0 -ac 1 -threads 0 -y t1.mp4
I saved the script as "v" and here's how I call it from cli:
v t1 flv 00:00:32.658 00:00:36.060
This way I can cut a part of a video file directly from ffmpeg cli. But the calculations among timecodes are not quite precise, so I need to work it on a little bit more to achieve this: http://www.1728.com/angle.htm
Here is another approach not using the date tool:
function timecode_to_int() {
if [[ "$1" =~ ^((0*([0-9]+):)?0*([0-9]+):)?0*([0-9]+)(\.([0-9]{1,3}))? ]]; then
VALUE=$((${BASH_REMATCH[5]}))
[[ -n "${BASH_REMATCH[4]}" ]] && VALUE=$((${BASH_REMATCH[4]}*60+$VALUE))
[[ -n "${BASH_REMATCH[3]}" ]] && VALUE=$((${BASH_REMATCH[3]}*60*60+$VALUE))
VALUE=$(($VALUE*1000))
if [[ -n "${BASH_REMATCH[7]}" ]]; then
VALUE=$(($VALUE+10#$(printf %-3s ${BASH_REMATCH[7]} | sed s_\ _0_g)))
fi
echo "${VALUE}"
else
echo "Invalid timecode '$1', aborting." >&2
exit 1
fi
}
function int_to_timecode() {
MILLISECONDS="$(($1%1000))"
SECS="$(($1/1000%60))"
MINUTES="$(($1/1000/60%60))"
HOURS="$(($1/1000/60/60))"
if [[ $HOURS -ne 0 ]]; then
printf "%d:%02d:%02d" $HOURS $MINUTES $SECS
elif [[ $MINUTES -ne 0 ]]; then
printf "%d:%02d" $MINUTES $SECS
else
printf "%d" $SECS
fi
if [[ ${MILLISECONDS} -ne 0 ]]; then
printf ".%03d" $MILLISECONDS | sed s_0*\$__
fi
}
# To answer your question completely:
echo $(int_to_timecode $(($(timecode_to_int "00:00:04")-$(timecode_to_int "00:00:02"))))
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