How to get video duration in seconds?
18,107
Solution 1
2383 is correct. There are 60 seconds in a minute, not 100. 43/60 = .71
https://www.google.com/#q=39+minutes+43.08+seconds+in+seconds
Solution 2
There is a better, faster and low CPU/HD footprint solution, just with mediainfo without relying into awk:
mediainfo --Inform="General;%Duration%" input.m4v
Solution 3
Using awk
mediainfo file.flv | awk '/Duration/ {print $3*60+$4}'
2383
ffmpeg -i file.flv 2>&1 | awk '/Duration/ {split($2,a,":");print a[1]*3600+a[2]*60+a[3]}'
2383.08
To handle different formats, use this:
cat file
Duration : 39mn 43s
Duration : 39s 43ms
awk '/Duration/ {for (i=3;i<=NF;i++) if ($i~/[0-9]+mn$/) s+=$i*60; else if ($i~/[0-9]+s$/) s+=$i; else if ($i~/[0-9]+ms$/) s+=$i/10; print s;s=0}' file
2383
43.3
Solution 4
Its Working for me try it :)
$duration = shell_exec("ffmpeg -i \"". $input . "\" 2>&1");
preg_match("/Duration: (\d{2}:\d{2}:\d{2}\.\d{2})/",$duration,$matches);
$time = explode(':',$matches[1]);
$hour = $time[0];
$minutes = $time[1];
$seconds = round($time[2]);
$total_seconds = 0;
$total_seconds += 60 * 60 * $hour;
$total_seconds += 60 * $minutes;
echo $total_seconds += $seconds;
Solution 5
You can use
ffprobe -v quiet -print_format xml -show_format -show_streams inputfile.flv
it will return an xml like this
<?xml version="1.0" encoding="UTF-8"?>
<ffprobe>
<streams>
<stream index="0" codec_name="h264" codec_long_name="H.264 / AVC / MPEG-4 AVC /
MPEG-4 part 10" profile="High" codec_type="video" codec_time_base="1/50" codec_tag_string="avc1" codec_tag="0x31637661" width="1920" height="1080" has_b_frames="2" sample_aspect_ratio="0:1" display_aspect_ratio="0:1" pix_fmt="yuvj420p" level="40" r_frame_rate="25/1" avg_frame_rate="25/1" time_base="1/12800" start_pts="0" start_time="0.000000" duration_ts="50176" duration="3.920000" bit_rate="4347640" nb_frames="98">
<disposition default="0" dub="0" original="0" comment="0" lyrics="0" karaoke="0" forced="0" hearing_impaired="0" visual_impaired="0" clean_effects="0" attached_pic="0"/>
<tag key="language" value="eng"/>
<tag key="handler_name" value="VideoHandler"/>
</stream>
<stream index="1" codec_name="aac" codec_long_name="AAC (Advanced Audio Coding)" codec_type="audi
o" codec_time_base="1/48000" codec_tag_string="mp4a" codec_tag="0x6134706d" sample_fmt="fltp" sample_rate="48000" channels="2" bits_per_sample="0" r_frame_rate="0/0" avg_frame_rate="0/0" time_base="1/48000" start_pts="-1024" start_time="-0.021333" duration_ts="189184" duration="3.941333" bit_rate="122277" nb_frames="185">
<disposition default="0" dub="0" original="0" comment="0" lyrics="0" karaoke="0" forced="0" hearing_impaired="0" visual_impaired="0" clean_effects="0" attached_pic="0"/>
<tag key="language" value="eng"/>
<tag key="handler_name" value="SoundHandler"/>
</stream>
</streams>
<format filename="inputfile.flv" nb_streams="2" format_name="mov,mp4,m4a,3gp,3g2,mj2" format_long_name="QuickTime / MOV" start_time="-0.021333" duration="3.942000" size="2194901" bit_rate="4454390">
<tag key="major_brand" value="isom"/>
<tag key="minor_version" value="512"/>
<tag key="compatible_brands" value="isomiso2avc1mp41"/>
<tag key="encoder" value="Lavf54.63.100"/>
</format>
</ffprobe>
then filter by the video stream i.e. usually that with attribute index ="0" and get the value of the attribute "duration" which is already in seconds.
Author by
user2783132
Updated on June 12, 2022Comments
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user2783132 almost 2 years
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
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Ilia about 10 yearsThanks Jotne, just what I was looking for!
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Ilia about 10 yearsFound a bug. If the video's length is smaller than 1min, then your mediainfo example doesn't do it correctly, as the mediainfo outputs time in different format, e.g.:
Duration : 36s 360ms
and it results using your code in 2520. Please fix. -
Jotne about 10 years@IliaRostovtsev are you the same person as the poster of the question
user2783132
? -
Ilia about 10 yearsNo! :) He had the same issue, I can tell now..
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Jotne about 10 years@IliaRostovtsev see my updated post
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ubiquibacon almost 10 yearsI also wanted an output in the format of hh:mm:ss.ms, but I had to slightly modify the command you suggest to get what I needed. I replaced
Video
withGeneral
to give the command--Inform="General;%Duration/String3%"
. MediaInfo version 0.7.69 -
trs over 5 yearsNote this gives you milliseconds and is therefore more accurate than any of the
grep
/awk
variations. For example:mediainfo --Inform="General;%Duration%" video.mp4
-->204466
whilemediainfo video.mp4 | grep '^Duration'
-->Duration : 3 min 24 s
(= 204000ms) -
trs over 5 yearsYou're not converting the value to seconds, you are cutting the ms portion off instead of rounding. So your result is not accurate. You could just as easily cut the last 3 digits of the value given by Rodrigo's answer and leave out the detour through php and string parsing.
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Ilia over 5 years@trs Agree. I didn't like my own, old answer, so I've edited it.