How to get video duration in seconds?

Question

How can I get video duration in seconds?

What I've tried:

ffmpeg -i file.flv 2>&1 | grep "Duration"
  Duration: 00:39:43.08, start: 0.040000, bitrate: 386 kb/s


mediainfo file.flv | grep Duration
Duration : 39mn 43s

This what close, but it's not so accurate, 2383 is 39.71 minutes

ffmpeg -i file.flv 2>&1 | grep "Duration"| cut -d ' ' -f 4 | sed s/,// | sed 's@\..*@@g' | awk '{ split($1, A, ":"); split(A[3], B, "."); print 3600*A[1] + 60*A[2] + B[1] }'
2383

llogan · Answer 1 · 2020-06-05T18:04:55.323

238

Just use ffprobe directly. No need for sed, grep, etc. There are several "durations" you can acquire (depending on your input).

Format (container) duration

ffprobe -v error -show_entries format=duration -of default=noprint_wrappers=1:nokey=1 input.mp4

Result:

30.024000

Adding the -sexagesimal option will use the HOURS:MM:SS.MICROSECONDS time unit format:

0:00:30.024000

Video stream duration

If you want the duration of a particular video or audio stream:

ffprobe -v error -select_streams v:0 -show_entries stream=duration -of default=noprint_wrappers=1:nokey=1 input.mp4

Result:

30.000000

The above commands are from FFmpeg Wiki: FFprobe Tips.

With `ffmpeg`

You can use ffmpeg to get duration by decoding the input:

ffmpeg -i input.mp4 -f null -
…
frame= 1587 fps=0.0 q=0.0 Lsize=N/A time=00:01:03.48 bitrate=N/A

In this example time=00:01:03.48 is the duration.

This may take a long time depending on your input file.

edited Jun 05 '20 at 18:04

answered Jul 26 '15 at 22:07

llogan

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These do not work for .m2v files do you have a solution for them aswell – utdev Jan 05 '17 at 09:47
@utdev See the "With `ffmpeg`" section. – llogan Jan 05 '17 at 18:39
Your third solution gives me a wrong time, my video has a length of 01:19:00 but the command returns me time=01:09:15.32, do you have a guess why this is happening – utdev Feb 15 '17 at 10:59
1

For anyone using avprobe, the parameters are slightly different: avprobe -v error -show_format_entry duration .\Sample.mp4 – Brad Jun 18 '19 at 22:25
Certain container types (e.g. Matroska [.mkv]) can be queried for duration at the container level ("format"), but you won't get a result at the video stream level. – MrPotatoHead Sep 22 '20 at 13:18
Bizarrely enough, this method prints the duration to stderr. – Ivan C May 09 '21 at 19:15
@IvanC By default ffmpeg logs to stderr as stdout is reserved for multimedia outputs. ffprobe will output to stdout by default. – llogan May 10 '21 at 16:28

score 71 · Accepted Answer · answered Sep 25 '13 at 19:41

71

To get minutes, you have to divide 2383 seconds by 60.

39.7167

and then multiply the fractional part .7167 by 60 to get the remaining seconds.

43.002

So it's 39 minutes, 43 seconds. The application appears to be giving you an accurate value.

answered Sep 25 '13 at 19:41

Robert Harvey

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Best answer because you identified the problem in the given solution. Most people would suggest a new solution. – user1735921 Apr 25 '20 at 14:51

score 41 · Answer 3 · edited Nov 08 '14 at 00:00

41

If you have ffmpeg, you should also have ffprobe:

ffprobe -i input.file -show_format | grep duration
ffprobe -i input.file -show_format -v quiet | sed -n 's/duration=//p'

This will also give fractions of seconds, if that's a problem you can further process that away with sed.

edited Nov 08 '14 at 00:00

Community

1

answered Sep 25 '13 at 21:10

evilsoup

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how can I remove the fraction part of seconds? – Tina J Aug 18 '17 at 13:42
@TinaJ Just pipe it with `printf`, like `| xargs printf %.0f` - this will return an integer value. – Ilia Ross Aug 14 '18 at 09:21
@TinaJ Also if you don't want to round value mathematically but want just to remove decimal part you can assign the duration to a variable `duration=$(ffprobe -i input.file -show_format -v quiet | sed -n 's/duration=//p')` And remove decimal part `echo ${duration%.*}` – dmitry1100 May 18 '20 at 11:18

score 9 · Answer 4 · answered May 05 '18 at 08:00

9

mediainfo --Output='General;%Duration%' file.flv

This outputs the duration in milliseconds as a single integer value. No need for grep/cut/sed/...

answered May 05 '18 at 08:00

Josef Kufner

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Thanks - this was just what I was looking for. Can be installed on MacOS with `brew install mediainfo` – Alex K Nov 22 '18 at 17:32

score 3 · Answer 5 · edited Aug 03 '18 at 03:29

3

Solution with mplayer that gives seconds directly:

mplayer -identify -frames 0 -vo null -nosound file.flv 2>&1 | awk -F= '/LENGTH/{print $2}'

edited Aug 03 '18 at 03:29

Giacomo1968

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answered Sep 25 '13 at 19:56

micke

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score 2 · Answer 6 · answered Jun 16 '14 at 22:06

2

For my ffmpeg-0.6.5-1.el6.rf.x86_64, to get just the second, ffprobe command format is:

ffprobe <file> -show_format 2>&1 | sed -n 's/duration=//p'

answered Jun 16 '14 at 22:06

Sunry

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how can I remove the fraction part of seconds? – Tina J Aug 18 '17 at 13:42
To remove fraction part, use: `sed -n -E 's/duration=([0-9]+).*/\1/p` which extracts the first matching numbers (discarding anything after a non-numeric symbol) using a backreference (`\1`, which only works with the `-E` option). – Yeti Jul 16 '20 at 12:46

score 1 · Answer 7 · answered Aug 23 '21 at 11:53

Python

You can extract streams using following code

import json

infile = "<path to your file>"
command = "ffprobe -v quiet -print_format json -show_format  -show_streams {}".format(infile)
res = json.loads(subprocess.check_output(command, shell=True))
streams = res.get('streams')

Each stream would have a field duration.

Special case of Webm

If your input file is webm, Your method of extracting stream duration is going to be different, since the standard store it in a different location. In each stream you would see a field 'tags'

"tags": {
                "language": "eng",
                "HANDLER_NAME": "ISO Media file produced by Google Inc. Created on: 12/09/2019.",
                "VENDOR_ID": "[0][0][0][0]",
                "ENCODER": "Lavc59.4.101 libvorbis",
                "DURATION": "00:03:14.703000000"
            }

You would have to convert this human readable time format into seconds. Which you can easily do with following utility function.

def convert_secs(text):
    if isinstance(text, float):
        num = str(text)
        nums = num.split('.')
    else:
        nums = text.split(':')
    if len(nums) == 2:
        st_sn = int(nums[0]) * 60 + float(nums[1])
        return st_sn
    elif len(nums) == 3:
        st_sn = int(nums[0]) * 3600 + int(nums[1]) * 60 + float(nums[2])
        return st_sn
    else:
        raise ValueError("Not correct time")

score 1 · Answer 8 · answered Aug 11 '23 at 11:07

I was a bit surprised to find that the simplest way of doing this isn't here, and that the closest thing doesn't give integers, but adds incorrect decimal points that need to be rounded off.

It's quite easy: just use your languages modulus operator.

In php:

$vidLength = 2383; //seconds
$secondsPerMinute = 60;

$seconds = $vidLength % $secondsPerMinute;
//$seconds = 43

$minutes = ($vidLength - $seconds) / $secondsPerMinute;
//$minutes = 39

No rounding errors or weird fractional parts. Just simple, plain math using one of the 6 basic arithmetic operators just like + or - that can be found in any programming language.

Also, the actual perceived error is that you've done whole minutes and then have divided the seconds by 60. 43/60 = 0,7167, effectively turning it from actual seconds to hundredths of a minute.

score 0 · Answer 9 · edited Aug 03 '18 at 03:30

I came across the issue of getting some strange and incorrect metadata from some video files I was working with and I couldn't succeed on finding a pattern or any type of handling using code and tools like ffmpeg, mp4box, ffprobe, mediainfo, mplayer, to get the real duration of the video.

Identifying the real duration of the video was a requirement for a project I was working and the only way I found to get it always right was to reencode the video file using ffmpeg and forcing encoded files to ignore original file's metadata, like:

ffmpeg -i INPUT_FILENAME -acodec copy -vcodec copy -map_metadata -1 OUTPUT_FILENAME"

(This might run faster than you expect. I got surprised, for the type of content and on the environment I was using, average time was 2 seconds)

... and then get duration using a tool at your choice. I like mediainfo, btw: - "mediainfo FILE --Inform="Video;%Duration%" gives you duration in miliseconds.

You did not provide more information or better solution than other answers — Sam, Jun 01 '17 at 07:36

score 0 · Answer 10 · answered Aug 30 '19 at 19:39

If you only need to query metadata:

ffprobe -hide_banner -v quiet -show_streams -print_format flat video.mp4

[...]

streams.stream.0.duration="5221.146009"

[...]

So you can parse it:

while read -r; do
  if [[ "$REPLY" =~ ^streams\.stream\.([0-9])+\.duration=\"([^"]+)\"$ ]]; then
    echo -E Duration of stream "${BASH_REMATCH[1]}": "${BASH_REMATCH[2]}"
  fi
done < <(ffprobe -hide_banner -v quiet -show_streams -print_format flat video.mp4)

But if you want to get the effective container's duration, you need to decode it:

AV_LOG_FORCE_NOCOLOR=y ffmpeg -nostdin -hide_banner -nostats -loglevel info -i video.mp4 -f null -c copy - 2>&1 | tail -n 2

It will take some CPU time to decode it, until:

[...]

frame=130527 fps=53271 q=-1.0 Lsize=N/A time=01:27:01.12 bitrate=N/A speed=2.13e+03x
video:152014kB audio:40790kB subtitle:0kB other streams:0kB global headers:0kB muxing overhead: unknown

Then, you can parse it:

if [[ "$(AV_LOG_FORCE_NOCOLOR=y ffmpeg -nostdin -hide_banner -nostats -loglevel info -i video.mp4 -f null -c copy - 2>&1 | tail -n 2 | head -n 1)" =~ \ time=([0-9]+):([0-9]{2}):([0-9]{2})\.([0-9]+) ]]; then
  declare duration=0 us="${BASH_REMATCH[4]}" t
  for t in "${BASH_REMATCH[@]:1:3}"; do
    ((duration *= 60))
    ((duration += ${t#0} ))
  done
  while [ ${#us} -lt 6 ]; do us+=0; done
  ((us >= 500000)) && ((duration++))
  ((duration)) || ((duration++))
fi
echo -E Duration: "$duration"

score 0 · Answer 11 · answered Oct 26 '21 at 05:45

# Returns duration (in seconds) of a video $1 (uses ffmpeg).
get_video_duration() {
  OUTPUT=$(ffmpeg -i "$1" -vframes 1 -f rawvideo -y /dev/null 2>&1) ||
    { debug -e "get_video_duration: error running ffmpeg:\n$OUTPUT"; return 1; }
  DURATION=$(echo "$OUTPUT" | grep -m1 "^[[:space:]]*Duration:" |
    cut -d":" -f2- | cut -d"," -f1 | sed "s/[:\.]/ /g") || 
    { debug -e "get_video_duration: error parsing duration:\n$OUTPUT"; return 1; }
  read HOURS MINUTES SECONDS DECISECONDS <<< "$DURATION"
  echo $((10#$HOURS * 3600 + 10#$MINUTES * 60 + 10#$SECONDS))      
}

Usage:DURATION=$(get_video_duration "$VIDEO")

How to get video duration in seconds?

11 Answers11

Format (container) duration

Video stream duration

With `ffmpeg`

Python

Special case of Webm

Linked

Related

How to get video duration in seconds?

11 Answers11

Format (container) duration

Video stream duration

With ffmpeg

Python

Special case of Webm

Linked

Related

With `ffmpeg`