Getting Mediainfo Data into a CSV

Mediainfo is a pretty handy tool to examine media files like MP4 containers and the streams in it such as the video and audio streams. It will also spit out a couple of formats in order parse the data such as XML. But I just want to get a certain set of data and want to move it into a CSV file so I can bring it into something like Google Sheets or Excel. Seems the “Inform” argument can get me some of the way there. You can tell it to give you multiple data points about a particular stream or “General” aspect about the file. You can’t mix say “Audio” and “Video”. That’s ok. I just want a handful of things about the Video stream and the filename. Woops. The file name is in the “General” bucket. So I am going to cheat a little with the “echo” command and tell it to print the file name and a comma and not to print out a new line in order to have it create the first column for the CSV row.

So this little script will find my MP4 movie files in the “/foo/*” directories and subdirectories, assign the name to the “movie” variable, print out the name and comma without a new line and then spit out a bunch of stuff about the video stream to get me a nice CSV output…

echo "Filename,PixelWidth,PixelHeight,DisplayRatio,Bitrate,FrameRate"
find /foo/ -type f | grep -i \.MP4 | while read movie;do
echo -n "$movie,"
mediainfo --Inform="Video;%Width%,%Height%,%DisplayAspectRatio/String%,%BitRate%,%FrameRate%" $movie

And I get something like:


You can get a list of Mediainfo parameters to use with “–Info-Parameters” argument.

[Update: I created a mediainfo template and script that will create a nice CSV with lots of info from an MP4 container. Assuming one video and audio track for the container.  You can see it on github at:]