Question

有一个我的网络视频流示例。

当我打开第二个VlC并再次流式传输视频时，从头开始，但是我想进行同步，例如，第一个视频在20号。

秒，当我打开它时，秒应该从20秒开始。我正在使用Windows Visual Studio 2017。我该如何解决这个问题？

#include "pch.h"
#include <iostream>
#include <gst/gst.h>
#include <gst/rtsp-server/rtsp-media.h>
#include <gst/rtsp-server/rtsp-server.h>
#include <gst/rtsp-server/rtsp-media-factory-uri.h>

 #define PORT "8554"
static char *port = (char *)PORT;

static GOptionEntry entries[] = {
{"port",'p', 0, G_OPTION_ARG_STRING, &port,"Port " PORT "", "PORT"}, 
{NULL} };

int main(int argc, gchar * argv[])
{
GMainLoop *loop;
GstRTSPServer *server;
GstRTSPMountPoints *mounts;
GstRTSPMediaFactoryURI *factory;
GOptionContext *sample;
GError *error = NULL;
gchar *uri;

sample = g_option_context_new("<uri> - Test RTSP Server, URI");
g_option_context_add_main_entries(sample, entries, NULL);
g_option_context_add_group(sample, gst_init_get_option_group());
if (!g_option_context_parse(sample, &argc, &argv, &error)) {
    return -1;
}
loop = g_main_loop_new(NULL, FALSE);

server = gst_rtsp_server_new();
g_object_set(server, "service", port, NULL);

mounts = gst_rtsp_server_get_mount_points(server);

factory = gst_rtsp_media_factory_uri_new();

const char* streamUri = "https://www.freedesktop.org/software/gstreamer-sdk/data/media/sintel_trailer-480p.webm";

if (gst_uri_is_valid(streamUri)) {
    uri = g_strdup(streamUri);
}
else if (g_file_test(streamUri, G_FILE_TEST_EXISTS)) {
    uri = gst_filename_to_uri(streamUri, NULL);
}
else {
    printf("There is no uri");
    return -1;
}

gst_rtsp_media_factory_uri_set_uri(factory, uri);
g_free(uri);

gst_rtsp_mount_points_add_factory(mounts, "/deneme", GST_RTSP_MEDIA_FACTORY(factory));

g_object_unref(mounts);

if (gst_rtsp_server_attach(server, NULL) == 0)
    printf("FAILED!");

g_print("stream ready at rtsp://127.0.0.1:%s/deneme\n", port);
g_main_loop_run(loop);

return 0;
}

Answer 1

您需要将工厂设置为共享，您可以通过调用以下函数来做到这一点：

/* make the factory shared as we will stream same data */
gst_rtsp_media_factory_set_shared(factory, TRUE);

我如何创建缓冲区并使流同步

1 个答案: