我想使用vlc库来流式传输屏幕的一部分。我写了一个小例子:
#include <iostream>
#include <cstdlib>
#include <vlc/vlc.h>
int main(int argc, char**argv)
{
libvlc_instance_t * inst = libvlc_new(argc, argv);
libvlc_vlm_add_broadcast(inst, "mybroad",
"screen://", "#transcode{vcodec=h264, venc=x264,vb=0,scale=0, acodec=mpga,ab=128,channels=2, samplerate=44100}:http{mux=ffmpeg{mux=flv}, dst=:7777/}",
0, NULL, 1, 0);
libvlc_vlm_play_media(inst, "mybroad");
std::cout << "ready" << std::endl;
// next two lines - it just for waitint
int i;
std::cin >> i;
// omit the code that frees libvlc
return 0;
}这个代码流遍了我的屏幕。如果我在控制台中,我可以流传输屏幕的一部分:
vlc -I "dummy" screen:// --screen-left=0 --screen-top=0 \
--screen-width=640 --screen-height=480 \
--screen-fps=1 \
--sout '#transcode{vcodec=h264,vb=800,scale=1,\
acodec=mpga,ab=128,channels=2,\
samplerate=44100}:http{mux=ts,dst=:7777/}'我试着通过修改一行代码来实现:
libvlc_vlm_add_broadcast(inst, "mybroad",
"screen:// :screen-fps=24 :screen-top=0 :screen-left=0 :screen-width=320 :screen-height=240",
"#transcode{vcodec=h264,venc=x264, vb=0,scale=0,acodec=mpga,ab=128,channels=2, samplerate=44100}:http{mux=ffmpeg{mux=flv},dst=:7777/}",
0, NULL, 1, 0);但这一修改并没有改变任何东西。
老实说,我想从一个监视器流(我有两个监视器),但我可以计算监视器的边界。
发布于 2013-04-25 15:44:44
我找到了解决方案。
#include <iostream>
#include <cstdlib>
#include <vlc/vlc.h>
int main(int argc, char**argv)
{
// the array with parameters
const char* params[] = {"screen-top=0",
"screen-left=0",
"screen-width=640",
"screen-height=480",
"screen-fps=10"};
libvlc_instance_t * inst = libvlc_new(argc, argv);
libvlc_vlm_add_broadcast(inst, "mybroad",
"screen://",
"#transcode{vcodec=h264,vb=800,scale=1,acodec=mpga,ab=128,channels=2,samplerate=44100}:http{mux=ts,dst=:7777/}",
5, params, // <= 5 == sizeof(params) == count of parameters
1, 0);
libvlc_vlm_play_media(inst, "mybroad");
std::cout << "ready" << std::endl;
int i;
std::cin >> i;
return 0;
}https://stackoverflow.com/questions/16186401
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