How to get video duration in seconds?

Question

How can I get the duration of a video in seconds?

what i 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 

that's close, but it's not so accurate, 2383 - 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 

Answer 1

2383 is true. There are 60 seconds per minute, not 100. 43/60 = .71

https://www.google.com/#q=39+minutes+43.08+seconds+in+seconds

Answer 2

There is a better, faster and lower solution to solve problems with CPU / HD, only with mediainfo, not relying on awk:

 mediainfo --Inform="General;%Duration%" input.m4v 

Answer 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 

Answer 4

you can use

 ffprobe -v quiet -print_format xml -show_format -show_streams inputfile.flv 

it will return 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 the video stream, that is, usually with the attribute index = "0" and get the value of the attribute "duration", which is already in seconds.

Answer 5

I’ll try his work for me :)

 $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; 

Answer 6

As Rodrigo Polo suggested using mediainfo --Inform="General;%Duration%" input.m4v , perhaps, but returns strange results.

For example, for a video with a duration of 41 s 564 ms, the output of this command will be:

 41564 

This is not suitable for me, and I came up with the following solution:

 mediainfo --Inform="Video;%Duration/String3%" input.m4v 

Return value:

 00:00:41.564 

In the end, you can simply format the return value with PHP to convert it to seconds, for example:

 $parsed = date_parse( '00:00:41.564' ); echo $parsed['hour'] * 3600 + $parsed['minute'] * 60 + $parsed['second']; 

Final conclusion:

41

Example

Learn more about MediaInfo CLI

Answer 7

I used this in php:

 exec('mediainfo --Inform="General;%Duration%" ' . $filename, $output); // type 86 sec = 85962.695312 $durationInSec = ceil(ceil($output[0])/1000); 

Answer 8

If you have ffmpeg, you should also have ffprobe. It displays information in STDOUT in a series of key = value pairs and gives the default duration in seconds (you should use the -sexagesimal option if you want hh: mm: ss):

 ffprobe -show_format file.mp4 | grep -F duration | cut -d= -f2