Hello,
I've got a problem with a video file.
Here is the informations when I do ffmpeg -i myFile :
FFmpeg version SVN-r13206, Copyright (c) 2000-2008 Fabrice Bellard, et al.
configuration: --disable-debug
libavutil version: 49.6.0
libavcodec version: 51.57.0
libavformat version: 52.13.0
libavdevice version: 52.0.0
built on Jun 26 2008 10:48:47, gcc: 3.4.2 20041017 (Red Hat 3.4.2-6.fc3)
Input #0, avi, from 'Enregistrement_2008.07.10_11.49.09.avi':
Duration: 00:01:31.29, start: 0.000000, bitrate: 224 kb/s
Stream #0.0: Audio: mp2, 44100 Hz, mono, 64 kb/s
Stream #0.1: Video: mpeg4, yuv420p, 1600x1200 [PAR 1:1 DAR 4:3], 10.00
tb(r)
Stream #0.2: Video: mpeg4, yuv420p, 1600x1200 [PAR 1:1 DAR 4:3], 10.00
tb(r)
This file is a screen capture made with my programm. There is one video
stream per screen (dual screen) and I captured 1 image each 10 seconds. I
can read this file with ffplay.
But when I do :
int err = av_open_input_file(&_formatContext, filename, NULL, 0,NULL);
if(err!=0)
{
ERROR(.........);
return;
}
err = av_find_stream_info(_formatContext);
There is no problem for opening the file, but
av_find_stream_info(_formatContext) returns a value different of zero.
What is wrong ?
Nicolas Krieger
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