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  I'm trying to open a *.avi file with OpenCV using Python(x, y) and every time I run the following code, the file fails to open.
  import cv2
  from cv2 import cv
  # Set up
  fn = "./video.avi"
  v = cv2.VideoCapture()
  # Configure codec:
  # From ffprobe, codec is M-JPEG ==> MJPG;
  codec = cv.CV_FOURCC("M", "J", "P", "G")
  v.set(cv.CV_CAP_PROP_FOURCC, codec)
  # Test
  v.open(fn)
  print fn
  if not v.isOpened():
  print "Video failed to open"
  else:
  print "Video opened!"
  The following is the output of running
  ffprobe video.avi
  libpostproc 52. 3.100 / 52. 3.100
  Input #0, avi, from 'L4R01CA_T05032300.avi':
  Duration: 00:00:06.01, start: 0.000000, bitrate: 16136 kb/s
  Stream #0:0: Video: mjpeg (MJPG / 0x47504A4D), yuvj420p(pc), 720x480, 29.97 tbr, 29.97 tbn, 29.97 tbc
  Stream #0:1: Audio: ac3 ([0] [0][0] / 0x2000), 48000 Hz, stereo, fltp, 256 kb/s
  Anyone have any ideas as to why OpenCV cannot open this file? I have also tried to open the file without setting cv.CV_CAP_PROP_FOURCC to MJPG with the same results.
  Thank you.