How to bulk extract bitrate data from videos and output them in a text file?











up vote
1
down vote

favorite
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I have several videos in a folder;



~/Downloads/movie1.mkv
~/Downloads/movie2.mkv
~/Downloads/movie3.mkv


I would like to extract the bitrate for each file, and output the data into a single text file, or output list which I can copy and paste into a text file.



I have installed ffmpeg.



So, for example, the output of fffmpeg -i movie1.mkv is;



Metadata:
encoder : libebml v1.2.0 + libmatroska v1.1.0
creation_time : 2011-04-09T18:18:05.000000Z
Duration: 00:04:27.71, start: 0.000000, bitrate: 10698 kb/s
Stream #0:0(eng): Video: h264 (High), yuv420p(progressive), 1920x1038, SAR 1:1 DAR 320:173, 23.98 fps, 23.98 tbr, 1k tbn, 47.95 tbc (default)
Metadata:
title : movie1
Stream #0:1(eng): Audio: dts (DTS), 48000 Hz, 5.1(side), fltp, 1536 kb/s (default)
Metadata:
title : DTS-ES 5.1 @ 1509 Kbps
Stream #0:2(eng): Audio: ac3, 48000 Hz, stereo, fltp, 192 kb/s
Metadata:
title : Commentary
Stream #0:3(eng): Subtitle: subrip
Stream #0:4(eng): Subtitle: subrip


The "bitrate: 10698 kb/s" is the crucial part I am after here.



Lets pretend all three movies have the same bitrate.



I would like to extract the bitrate information from all three videos, and have them output as;



movie1.mkv 10698
movie2.mkv 10698
movie3.mkv 10698


How would I go about extracting and outputing this information in bulk?



I've been trying a combination of finding by file name, ffmpeg, and then | to grep. e.g ; find . -name "*.mkv" -exec ffmpeg -i "${1%.mkv}" | grep "bitrate:"



Current attempts;



1. The command; ffmpeg -i movie1.mkv 2>&1 | grep bitrate | sed 's/bitrate: (.*), kb/1/g'



Returns;



Duration: 00:04:27.71, start: 0.000000, bitrate: 10698 kb/s


2. The command; find . -name "*.mkv" -exec ffprobe -v error -show_entries format=bit_rate -of default=noprint_wrappers=1:nokey=1 "${~/Downloads/1%.mkv}" ';'



Returns



~/Downloads/1%.mkv: No such file or directory
~/Downloads/1%.mkv: No such file or directory
~/Downloads/1%.mkv: No such file or directory


I feel that I'm close here, and that there's just something wrong with the find and recalling the output of find into ffprobe.










share|improve this question




























    up vote
    1
    down vote

    favorite
    1












    I have several videos in a folder;



    ~/Downloads/movie1.mkv
    ~/Downloads/movie2.mkv
    ~/Downloads/movie3.mkv


    I would like to extract the bitrate for each file, and output the data into a single text file, or output list which I can copy and paste into a text file.



    I have installed ffmpeg.



    So, for example, the output of fffmpeg -i movie1.mkv is;



    Metadata:
    encoder : libebml v1.2.0 + libmatroska v1.1.0
    creation_time : 2011-04-09T18:18:05.000000Z
    Duration: 00:04:27.71, start: 0.000000, bitrate: 10698 kb/s
    Stream #0:0(eng): Video: h264 (High), yuv420p(progressive), 1920x1038, SAR 1:1 DAR 320:173, 23.98 fps, 23.98 tbr, 1k tbn, 47.95 tbc (default)
    Metadata:
    title : movie1
    Stream #0:1(eng): Audio: dts (DTS), 48000 Hz, 5.1(side), fltp, 1536 kb/s (default)
    Metadata:
    title : DTS-ES 5.1 @ 1509 Kbps
    Stream #0:2(eng): Audio: ac3, 48000 Hz, stereo, fltp, 192 kb/s
    Metadata:
    title : Commentary
    Stream #0:3(eng): Subtitle: subrip
    Stream #0:4(eng): Subtitle: subrip


    The "bitrate: 10698 kb/s" is the crucial part I am after here.



    Lets pretend all three movies have the same bitrate.



    I would like to extract the bitrate information from all three videos, and have them output as;



    movie1.mkv 10698
    movie2.mkv 10698
    movie3.mkv 10698


    How would I go about extracting and outputing this information in bulk?



    I've been trying a combination of finding by file name, ffmpeg, and then | to grep. e.g ; find . -name "*.mkv" -exec ffmpeg -i "${1%.mkv}" | grep "bitrate:"



    Current attempts;



    1. The command; ffmpeg -i movie1.mkv 2>&1 | grep bitrate | sed 's/bitrate: (.*), kb/1/g'



    Returns;



    Duration: 00:04:27.71, start: 0.000000, bitrate: 10698 kb/s


    2. The command; find . -name "*.mkv" -exec ffprobe -v error -show_entries format=bit_rate -of default=noprint_wrappers=1:nokey=1 "${~/Downloads/1%.mkv}" ';'



    Returns



    ~/Downloads/1%.mkv: No such file or directory
    ~/Downloads/1%.mkv: No such file or directory
    ~/Downloads/1%.mkv: No such file or directory


    I feel that I'm close here, and that there's just something wrong with the find and recalling the output of find into ffprobe.










    share|improve this question


























      up vote
      1
      down vote

      favorite
      1









      up vote
      1
      down vote

      favorite
      1






      1





      I have several videos in a folder;



      ~/Downloads/movie1.mkv
      ~/Downloads/movie2.mkv
      ~/Downloads/movie3.mkv


      I would like to extract the bitrate for each file, and output the data into a single text file, or output list which I can copy and paste into a text file.



      I have installed ffmpeg.



      So, for example, the output of fffmpeg -i movie1.mkv is;



      Metadata:
      encoder : libebml v1.2.0 + libmatroska v1.1.0
      creation_time : 2011-04-09T18:18:05.000000Z
      Duration: 00:04:27.71, start: 0.000000, bitrate: 10698 kb/s
      Stream #0:0(eng): Video: h264 (High), yuv420p(progressive), 1920x1038, SAR 1:1 DAR 320:173, 23.98 fps, 23.98 tbr, 1k tbn, 47.95 tbc (default)
      Metadata:
      title : movie1
      Stream #0:1(eng): Audio: dts (DTS), 48000 Hz, 5.1(side), fltp, 1536 kb/s (default)
      Metadata:
      title : DTS-ES 5.1 @ 1509 Kbps
      Stream #0:2(eng): Audio: ac3, 48000 Hz, stereo, fltp, 192 kb/s
      Metadata:
      title : Commentary
      Stream #0:3(eng): Subtitle: subrip
      Stream #0:4(eng): Subtitle: subrip


      The "bitrate: 10698 kb/s" is the crucial part I am after here.



      Lets pretend all three movies have the same bitrate.



      I would like to extract the bitrate information from all three videos, and have them output as;



      movie1.mkv 10698
      movie2.mkv 10698
      movie3.mkv 10698


      How would I go about extracting and outputing this information in bulk?



      I've been trying a combination of finding by file name, ffmpeg, and then | to grep. e.g ; find . -name "*.mkv" -exec ffmpeg -i "${1%.mkv}" | grep "bitrate:"



      Current attempts;



      1. The command; ffmpeg -i movie1.mkv 2>&1 | grep bitrate | sed 's/bitrate: (.*), kb/1/g'



      Returns;



      Duration: 00:04:27.71, start: 0.000000, bitrate: 10698 kb/s


      2. The command; find . -name "*.mkv" -exec ffprobe -v error -show_entries format=bit_rate -of default=noprint_wrappers=1:nokey=1 "${~/Downloads/1%.mkv}" ';'



      Returns



      ~/Downloads/1%.mkv: No such file or directory
      ~/Downloads/1%.mkv: No such file or directory
      ~/Downloads/1%.mkv: No such file or directory


      I feel that I'm close here, and that there's just something wrong with the find and recalling the output of find into ffprobe.










      share|improve this question















      I have several videos in a folder;



      ~/Downloads/movie1.mkv
      ~/Downloads/movie2.mkv
      ~/Downloads/movie3.mkv


      I would like to extract the bitrate for each file, and output the data into a single text file, or output list which I can copy and paste into a text file.



      I have installed ffmpeg.



      So, for example, the output of fffmpeg -i movie1.mkv is;



      Metadata:
      encoder : libebml v1.2.0 + libmatroska v1.1.0
      creation_time : 2011-04-09T18:18:05.000000Z
      Duration: 00:04:27.71, start: 0.000000, bitrate: 10698 kb/s
      Stream #0:0(eng): Video: h264 (High), yuv420p(progressive), 1920x1038, SAR 1:1 DAR 320:173, 23.98 fps, 23.98 tbr, 1k tbn, 47.95 tbc (default)
      Metadata:
      title : movie1
      Stream #0:1(eng): Audio: dts (DTS), 48000 Hz, 5.1(side), fltp, 1536 kb/s (default)
      Metadata:
      title : DTS-ES 5.1 @ 1509 Kbps
      Stream #0:2(eng): Audio: ac3, 48000 Hz, stereo, fltp, 192 kb/s
      Metadata:
      title : Commentary
      Stream #0:3(eng): Subtitle: subrip
      Stream #0:4(eng): Subtitle: subrip


      The "bitrate: 10698 kb/s" is the crucial part I am after here.



      Lets pretend all three movies have the same bitrate.



      I would like to extract the bitrate information from all three videos, and have them output as;



      movie1.mkv 10698
      movie2.mkv 10698
      movie3.mkv 10698


      How would I go about extracting and outputing this information in bulk?



      I've been trying a combination of finding by file name, ffmpeg, and then | to grep. e.g ; find . -name "*.mkv" -exec ffmpeg -i "${1%.mkv}" | grep "bitrate:"



      Current attempts;



      1. The command; ffmpeg -i movie1.mkv 2>&1 | grep bitrate | sed 's/bitrate: (.*), kb/1/g'



      Returns;



      Duration: 00:04:27.71, start: 0.000000, bitrate: 10698 kb/s


      2. The command; find . -name "*.mkv" -exec ffprobe -v error -show_entries format=bit_rate -of default=noprint_wrappers=1:nokey=1 "${~/Downloads/1%.mkv}" ';'



      Returns



      ~/Downloads/1%.mkv: No such file or directory
      ~/Downloads/1%.mkv: No such file or directory
      ~/Downloads/1%.mkv: No such file or directory


      I feel that I'm close here, and that there's just something wrong with the find and recalling the output of find into ffprobe.







      ffmpeg grep cat mkv






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 13 at 12:28

























      asked Nov 13 at 10:05









      Martin JS

      153




      153






















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          1
          down vote



          accepted










          To screen:



          find . -name "*.mkv" -print0 | xargs -0 -i{} sh -c " echo -n '{} ' && ffmpeg -i '{}' 2>&1 | sed -n -e 's/^.*bitrate: //p' "


          To file result.txt:



          find . -name "*.mkv" -print0 | xargs -0 -i{} sh -c " echo -n '{} ' && ffmpeg -i '{}' 2>&1 | sed -n -e 's/^.*bitrate: //p' " > result.txt


          You will see in file result.txt:



          somefile1.mkv 1788 kb/s
          somefile2.mkv 1681 kb/s
          ...


          Formatted output example:



          find . -name "*.mkv" -print0 | xargs -0 -i{} sh -c " echo -n 'Filename: {}, Bitrate is: ' && ffmpeg -i '{}' 2>&1 | sed -n -e 's/^.*bitrate: //p' " > result.txt


          You will see in file result.txt:



          Filename: somefile1.mkv, Bitrate is: 1788 kb/s
          Filename: somefile2.mkv, Bitrate is: 1681 kb/s
          ...





          share|improve this answer























          • I keep getting the "missing argument to `-exec " error
            – Martin JS
            Nov 13 at 11:56












          • Try: find . -name "*.mkv" -exec ffmpeg -i "${1%.mkv}" | sed -n -e 's/^.*bitrate: //p' What output you get?
            – S_Flash
            Nov 13 at 12:01










          • Still getting the "missing argument to `-exec " error. I have found that : ffprobe -v error -show_entries format=bit_rate -of default=noprint_wrappers=1:nokey=1 "movie1.mkv" returns just the bit rate, so if somehow the output of find, and the use echo can be used with that, it might do the trick
            – Martin JS
            Nov 13 at 12:03












          • I added the ';', and it returned with a "No such file or directory error three times, so I think there's an issue in the finding of the file name, and passing it to ffmpeg.
            – Martin JS
            Nov 13 at 12:39












          • I found errors and edited my answer with some new examples
            – S_Flash
            Nov 13 at 12:43











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          1
          down vote



          accepted










          To screen:



          find . -name "*.mkv" -print0 | xargs -0 -i{} sh -c " echo -n '{} ' && ffmpeg -i '{}' 2>&1 | sed -n -e 's/^.*bitrate: //p' "


          To file result.txt:



          find . -name "*.mkv" -print0 | xargs -0 -i{} sh -c " echo -n '{} ' && ffmpeg -i '{}' 2>&1 | sed -n -e 's/^.*bitrate: //p' " > result.txt


          You will see in file result.txt:



          somefile1.mkv 1788 kb/s
          somefile2.mkv 1681 kb/s
          ...


          Formatted output example:



          find . -name "*.mkv" -print0 | xargs -0 -i{} sh -c " echo -n 'Filename: {}, Bitrate is: ' && ffmpeg -i '{}' 2>&1 | sed -n -e 's/^.*bitrate: //p' " > result.txt


          You will see in file result.txt:



          Filename: somefile1.mkv, Bitrate is: 1788 kb/s
          Filename: somefile2.mkv, Bitrate is: 1681 kb/s
          ...





          share|improve this answer























          • I keep getting the "missing argument to `-exec " error
            – Martin JS
            Nov 13 at 11:56












          • Try: find . -name "*.mkv" -exec ffmpeg -i "${1%.mkv}" | sed -n -e 's/^.*bitrate: //p' What output you get?
            – S_Flash
            Nov 13 at 12:01










          • Still getting the "missing argument to `-exec " error. I have found that : ffprobe -v error -show_entries format=bit_rate -of default=noprint_wrappers=1:nokey=1 "movie1.mkv" returns just the bit rate, so if somehow the output of find, and the use echo can be used with that, it might do the trick
            – Martin JS
            Nov 13 at 12:03












          • I added the ';', and it returned with a "No such file or directory error three times, so I think there's an issue in the finding of the file name, and passing it to ffmpeg.
            – Martin JS
            Nov 13 at 12:39












          • I found errors and edited my answer with some new examples
            – S_Flash
            Nov 13 at 12:43















          up vote
          1
          down vote



          accepted










          To screen:



          find . -name "*.mkv" -print0 | xargs -0 -i{} sh -c " echo -n '{} ' && ffmpeg -i '{}' 2>&1 | sed -n -e 's/^.*bitrate: //p' "


          To file result.txt:



          find . -name "*.mkv" -print0 | xargs -0 -i{} sh -c " echo -n '{} ' && ffmpeg -i '{}' 2>&1 | sed -n -e 's/^.*bitrate: //p' " > result.txt


          You will see in file result.txt:



          somefile1.mkv 1788 kb/s
          somefile2.mkv 1681 kb/s
          ...


          Formatted output example:



          find . -name "*.mkv" -print0 | xargs -0 -i{} sh -c " echo -n 'Filename: {}, Bitrate is: ' && ffmpeg -i '{}' 2>&1 | sed -n -e 's/^.*bitrate: //p' " > result.txt


          You will see in file result.txt:



          Filename: somefile1.mkv, Bitrate is: 1788 kb/s
          Filename: somefile2.mkv, Bitrate is: 1681 kb/s
          ...





          share|improve this answer























          • I keep getting the "missing argument to `-exec " error
            – Martin JS
            Nov 13 at 11:56












          • Try: find . -name "*.mkv" -exec ffmpeg -i "${1%.mkv}" | sed -n -e 's/^.*bitrate: //p' What output you get?
            – S_Flash
            Nov 13 at 12:01










          • Still getting the "missing argument to `-exec " error. I have found that : ffprobe -v error -show_entries format=bit_rate -of default=noprint_wrappers=1:nokey=1 "movie1.mkv" returns just the bit rate, so if somehow the output of find, and the use echo can be used with that, it might do the trick
            – Martin JS
            Nov 13 at 12:03












          • I added the ';', and it returned with a "No such file or directory error three times, so I think there's an issue in the finding of the file name, and passing it to ffmpeg.
            – Martin JS
            Nov 13 at 12:39












          • I found errors and edited my answer with some new examples
            – S_Flash
            Nov 13 at 12:43













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          To screen:



          find . -name "*.mkv" -print0 | xargs -0 -i{} sh -c " echo -n '{} ' && ffmpeg -i '{}' 2>&1 | sed -n -e 's/^.*bitrate: //p' "


          To file result.txt:



          find . -name "*.mkv" -print0 | xargs -0 -i{} sh -c " echo -n '{} ' && ffmpeg -i '{}' 2>&1 | sed -n -e 's/^.*bitrate: //p' " > result.txt


          You will see in file result.txt:



          somefile1.mkv 1788 kb/s
          somefile2.mkv 1681 kb/s
          ...


          Formatted output example:



          find . -name "*.mkv" -print0 | xargs -0 -i{} sh -c " echo -n 'Filename: {}, Bitrate is: ' && ffmpeg -i '{}' 2>&1 | sed -n -e 's/^.*bitrate: //p' " > result.txt


          You will see in file result.txt:



          Filename: somefile1.mkv, Bitrate is: 1788 kb/s
          Filename: somefile2.mkv, Bitrate is: 1681 kb/s
          ...





          share|improve this answer














          To screen:



          find . -name "*.mkv" -print0 | xargs -0 -i{} sh -c " echo -n '{} ' && ffmpeg -i '{}' 2>&1 | sed -n -e 's/^.*bitrate: //p' "


          To file result.txt:



          find . -name "*.mkv" -print0 | xargs -0 -i{} sh -c " echo -n '{} ' && ffmpeg -i '{}' 2>&1 | sed -n -e 's/^.*bitrate: //p' " > result.txt


          You will see in file result.txt:



          somefile1.mkv 1788 kb/s
          somefile2.mkv 1681 kb/s
          ...


          Formatted output example:



          find . -name "*.mkv" -print0 | xargs -0 -i{} sh -c " echo -n 'Filename: {}, Bitrate is: ' && ffmpeg -i '{}' 2>&1 | sed -n -e 's/^.*bitrate: //p' " > result.txt


          You will see in file result.txt:



          Filename: somefile1.mkv, Bitrate is: 1788 kb/s
          Filename: somefile2.mkv, Bitrate is: 1681 kb/s
          ...






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Nov 13 at 13:38

























          answered Nov 13 at 10:49









          S_Flash

          886117




          886117












          • I keep getting the "missing argument to `-exec " error
            – Martin JS
            Nov 13 at 11:56












          • Try: find . -name "*.mkv" -exec ffmpeg -i "${1%.mkv}" | sed -n -e 's/^.*bitrate: //p' What output you get?
            – S_Flash
            Nov 13 at 12:01










          • Still getting the "missing argument to `-exec " error. I have found that : ffprobe -v error -show_entries format=bit_rate -of default=noprint_wrappers=1:nokey=1 "movie1.mkv" returns just the bit rate, so if somehow the output of find, and the use echo can be used with that, it might do the trick
            – Martin JS
            Nov 13 at 12:03












          • I added the ';', and it returned with a "No such file or directory error three times, so I think there's an issue in the finding of the file name, and passing it to ffmpeg.
            – Martin JS
            Nov 13 at 12:39












          • I found errors and edited my answer with some new examples
            – S_Flash
            Nov 13 at 12:43


















          • I keep getting the "missing argument to `-exec " error
            – Martin JS
            Nov 13 at 11:56












          • Try: find . -name "*.mkv" -exec ffmpeg -i "${1%.mkv}" | sed -n -e 's/^.*bitrate: //p' What output you get?
            – S_Flash
            Nov 13 at 12:01










          • Still getting the "missing argument to `-exec " error. I have found that : ffprobe -v error -show_entries format=bit_rate -of default=noprint_wrappers=1:nokey=1 "movie1.mkv" returns just the bit rate, so if somehow the output of find, and the use echo can be used with that, it might do the trick
            – Martin JS
            Nov 13 at 12:03












          • I added the ';', and it returned with a "No such file or directory error three times, so I think there's an issue in the finding of the file name, and passing it to ffmpeg.
            – Martin JS
            Nov 13 at 12:39












          • I found errors and edited my answer with some new examples
            – S_Flash
            Nov 13 at 12:43
















          I keep getting the "missing argument to `-exec " error
          – Martin JS
          Nov 13 at 11:56






          I keep getting the "missing argument to `-exec " error
          – Martin JS
          Nov 13 at 11:56














          Try: find . -name "*.mkv" -exec ffmpeg -i "${1%.mkv}" | sed -n -e 's/^.*bitrate: //p' What output you get?
          – S_Flash
          Nov 13 at 12:01




          Try: find . -name "*.mkv" -exec ffmpeg -i "${1%.mkv}" | sed -n -e 's/^.*bitrate: //p' What output you get?
          – S_Flash
          Nov 13 at 12:01












          Still getting the "missing argument to `-exec " error. I have found that : ffprobe -v error -show_entries format=bit_rate -of default=noprint_wrappers=1:nokey=1 "movie1.mkv" returns just the bit rate, so if somehow the output of find, and the use echo can be used with that, it might do the trick
          – Martin JS
          Nov 13 at 12:03






          Still getting the "missing argument to `-exec " error. I have found that : ffprobe -v error -show_entries format=bit_rate -of default=noprint_wrappers=1:nokey=1 "movie1.mkv" returns just the bit rate, so if somehow the output of find, and the use echo can be used with that, it might do the trick
          – Martin JS
          Nov 13 at 12:03














          I added the ';', and it returned with a "No such file or directory error three times, so I think there's an issue in the finding of the file name, and passing it to ffmpeg.
          – Martin JS
          Nov 13 at 12:39






          I added the ';', and it returned with a "No such file or directory error three times, so I think there's an issue in the finding of the file name, and passing it to ffmpeg.
          – Martin JS
          Nov 13 at 12:39














          I found errors and edited my answer with some new examples
          – S_Flash
          Nov 13 at 12:43




          I found errors and edited my answer with some new examples
          – S_Flash
          Nov 13 at 12:43


















           

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