<html>
<head>
<style>
.hmmessage P
{
margin:0px;
padding:0px
}
body.hmmessage
{
FONT-SIZE: 9pt;
FONT-FAMILY:Tahoma
}
</style>
</head>
<body class='hmmessage'>Hi,<BR>
<BR>
Anybod can help me?<BR>
<BR>
Thanks a lot!<BR>
<BR>
<BR>
Message: 4<BR>> Date: Thu, 27 Sep 2007 08:58:57 +0000<BR>> From: ChengYQ <crashboat@hotmail.com><BR>> Subject: [vlc-devel] sout transcode error when using libvlc.<BR>> To: <vlc-devel@videolan.org><BR>> Message-ID: <BAY144-W39534FAB43606527E1195FBCB10@phx.gbl><BR>> Content-Type: text/plain; charset="gb2312"<BR><BR>
Hi,<BR>
I got a problem streaming video by udp using libvlc.When I run vlc(0.8.6c) from command:<BR>"vlc v4l:/dev/video0:size=320x240 --sout '#transcode{vcodec=mp4v,vb=1024,scale=1}:duplicate{dst=display,dst=std{access=udp,mux=ts,dst=192.168.0.231:1234}}',the sout works fine and stream the video, I think the drivers and my Gentoo Linux system environment is fine.<BR><BR>But when I follow the example <A href="http://wiki.videolan.org/LibVLC_Tutorial_086c" target=_blank>http://wiki.videolan.org/LibVLC_Tutorial_086c</A>, and changed the code a little bit to stream video out by UDP.it says that: <BR>"main private error: no sout stream module mathched 'transcode' <BR>main stream output error:stream chain failed for 'transcode{vcodec=mp4v,vb=1024,scale=1}:duplicate{dst=display,dst=std{access=udp,mux=ts,dst=192.168.0.231:1234}}'" <BR><BR>My code is as following:<BR>#include <stdio.h><BR>#include <stdlib.h><BR>#include <vlc/libvlc.h><BR>static void quit_on_exception (libvlc_exception_t *excp) { <BR>if (libvlc_exception_raised (excp)) { <BR> fprintf(stderr, "error: %s\n", libvlc_exception_get_message(excp)); <BR> exit(-1); <BR>}}<BR><BR>int main(int argc, char **argv){<BR>libvlc_exception_t excp; <BR>libvlc_instance_t *inst; <BR>char *myarg0 = "test"; <BR>char *myarg1 = "v4l://"; <BR>char *myarg2 = ":v4l-vdev=/dev/video0"; <BR>char *myarg3 = ":v4l-adev="; <BR>char *myarg4 = ":sout=#transcode{vcodec=mp4v,vb=1024,scale=1}:duplicate{dst=display,dst=std{access=udp,mux=ts,dst=192.168.0.231:1234}}"; char *myargs[5] = {myarg0,myarg1,myarg2,myarg3,myarg4}; <BR>libvlc_exception_init (&excp); <BR>inst = libvlc_new (5, myargs, &excp); <BR>quit_on_exception (&excp); <BR>libvlc_playlist_play (inst, -1, 0, NULL, &excp); <BR>usleep (20000000); <BR>libvlc_destroy (inst); <BR>return 0;<BR>} <BR><BR>If I feed only the first 4 args, the video display fine without streaming. Is that because I missed some arguments or I need to link some more modules, or ...?<BR><BR>Please help!!!<BR><BR>Thanks a lot!<BR>Yours<BR><BR><BR><br /><hr />使用新一代 Windows Live Messenger 轻松交流和共享！ <a href='http://messenger.live.cn/' target='_new'>立即体验！</a></body>
</html>