One liner ffmpeg (or other) to get only resolution?

I'm not really well versed in the command line so hopefully this isn't too stupid of a question.

If I run:

ffmpeg -i videofile.avi

I get an output such as this:

ffmpeg version git-2013-11-21-6a7980e Copyright (c) 2000-2013 the FFmpeg develop    ers
  built on Nov 21 2013 12:06:32 with gcc 4.6 (Ubuntu/Linaro 4.6.3-1ubuntu5)
  configuration: --prefix=/home/user/ffmpeg_build --extra-cflags=-I/home/user/ffmpeg_build/include --extra-ldflags=-L/home/user/ffmpeg_build/lib --b        indir=/home/user/bin --extra-libs=-ldl --enable-gpl --enable-libass --enable    -libfdk-    aac --enable-libmp3lame --enable-libtheora --enable-libvorbis --enable-l    ibx264 --enable-    nonfree
  libavutil      52. 53.100 / 52. 53.100
  libavcodec     55. 44.100 / 55. 44.100
  libavformat    55. 21.100 / 55. 21.100
  libavdevice    55.  5.100 / 55.  5.100
  libavfilter     3. 91.100 /  3. 91.100
  libswscale      2.  5.101 /  2.  5.101
  libswresample   0. 17.104 /  0. 17.104
  libpostproc    52.  3.100 / 52.  3.100
Input #0, avi, from 'videofile.avi':
  Metadata:
    encoder         : Lavf52.68.0
  Duration: 00:23:07.68, start: 0.000000, bitrate: 2390 kb/s
    Stream #0:0: Video: mpeg4 (Simple Profile) (XVID / 0x44495658), yuv420p, 640x480     [SAR 1:1 DAR 4:3], 23.98 fps, 23.98 tbr, 23.98 tbn, 1199 tbc
    Stream #0:1: Audio: mp3 (U[0][0][0] / 0x0055), 44100 Hz, stereo, s16p, 128 k    b/s

If I were only interested in the command outputting "640x480", how might I do that?

I imagine I have to pipe the output to, and perform, a regular expression? No idea how to do that. Thanks!

Using ffprobe from the ffmpeg package:

ffprobe -v error -select_streams v:0 -show_entries stream=width,height -of csv=s=x:p=0 input.mp4

Example result:

1280x720

What these options do:

• -v error makes the output less verbose.
• -select_streams v:0 selects only the first video stream.
• -show_entries stream=width,height chooses only width and height from the big list of parameters that ffprobe can provide.
• -of csv=s=x:p=0 formats the text output. The csv formatting type is used because it makes a simple output. s=x makes it use an x to separate the width and height values. p=0 makes it omit the stream prefix in the output.

See the ffprobe documentation and FFmpeg Wiki: ffprobe tips for more info.

Get video resolution with ffmpeg:

ffmpeg -i filename 2>&1 | grep -oP 'Stream .*, \K[0-9]+x[0-9]+'

Output (e.g.):

640x480

exiftool -b metavideo.mp4 -ImageWidth
exiftool -b metavideo.mp4 -ImageHeight

do the job without any greps

One-liner looks like:

exiftool -b metavideo.mp4 -ImageSize

That returns WxH string that you was looking for.

I only see here samples of using exiftool and parsing it's output, which seems to be a weird choice. exiftool can be used in a direct manner to get the resolution (and almost any other metadata) :

exiftool -ImageSize -s3 filename

Will results in output in a form WIDTHxHEIGHT (eq: 1280×720)

NOTE: -s3 instructs exiftool to produce the shortest output format (without tag names, quotes, etc)

You can do this using regular expressions and ffmpeg, but I'd prefer to use exiftool, part of the libimage-exiftool-perl package:

$ exiftool some/video.ogv | awk -F' *: *' '$1 == "Image Size"{print $2}'
1242x480

Why exiftool?

• The ffmpeg/avprobe utilities print lines of the form Stream #0:0: Video: mpeg4 (Simple Profile) (XVID / 0x44495658), yuv420p, 640x480 where one or more values of the form AAAxBBB are possible (as you can see in your example). Determining which one is the resolution might be difficult.
• exiftool, on the other hand, outputs data of the form Type : Value. So it's easier to get the exact field, namely Image Size. Plus, given the way the fields are specified (separated by colon and whitespace), you can easily extract the value.

What the awk command does is:

• Set the field separator to ' *: *' - a colon : surrounded by any number of spaces.
• Check if the first field ($1) is Image Size
• And print the second field (print $2).

An alternate version would be:

awk '/^Image Size/{print $4}'

• By default, awk uses whitespace to split fields, so Image, Size, : and 1242x480 are all different fields.
• By matching the line's beginning (^) to Image Size, we get the correct line.
• And we print the fourth field (since Image, Size and : are the first three fields).