How to get video duration in seconds? [duplicate]

Answer 1

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.

Answer 2

I used this in php:

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

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

In case you have ffmpeg installed, most likely you would have ffprobe installed too.

ffprobe -i file.mp4 -show_format -v quiet | sed -n 's/duration=//p' | xargs printf %.0f

The command will return an integer value.

Output example:

190

Answer 5

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

Answer 6

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

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