我正在尝试创建可加重多个RTSP的应用程序,以用于移动应用程序。我的想法是,我在连接到IP摄像机的Windows PC上运行一个简单的WPF应用,然后将xamarin应用连接到它。我希望WPF程序可以根据需要将所需的RTSP中继到应用程序,但是这样做我很茫然。我学习了如何在Windows和Android应用程序上接收流,但是我正在寻找一种将Windows应用程序接收到的流转发到Android设备中的方法。
根据我发现的潜伏在互联网上的情况,有一种使用VLC库进行流式传输的方法:How to streaming video via VLC api in C#
但是我不知道如何从另一个流而不是一个文件流。
这就是我所得到的-在应用程序上简单显示RTSP:
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
xmlns:d="http://schemas.microsoft.com/expression/blend/2008"
xmlns:mc="http://schemas.openxmlformats.org/markup-compatibility/2006"
xmlns:local="clr-namespace:Serwer"
xmlns:wpf="clr-namespace:LibVLCSharp.WPF;assembly=LibVLCSharp.WPF"
mc:Ignorable="d"
Title="MainWindow" Height="650" Width="800">
<Grid>
<Grid.RowDefinitions>
<RowDefinition Height="*"/>
<RowDefinition Height="*"/>
</Grid.RowDefinitions>
<Grid.ColumnDefinitions>
<ColumnDefinition Width="*"/>
<ColumnDefinition Width="*"/>
</Grid.ColumnDefinitions>
<Label Grid.Row="0" Grid.Column="0">Placeholder for settings</Label>
<Label Grid.Row="0" Grid.Column="1">Placeholder for more streams</Label>
<Label Grid.Row="1" Grid.Column="1"></Label>
<wpf:VideoView x:Name="VideoView" Grid.Row="1" Grid.Column="0"/>
</Grid>
</Window>
这是XAML背后的代码:
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Windows;
using System.Windows.Controls;
using System.Windows.Data;
using System.Windows.Documents;
using System.Windows.Input;
using System.Windows.Media;
using System.Windows.Media.Imaging;
using System.Windows.Navigation;
using System.Windows.Shapes;
namespace Serwer
{
/// <summary>
/// Interaction logic for MainWindow.xaml
/// </summary>
public partial class MainWindow : Window
{
const string VIDEO_URL = "rtsp://wowzaec2demo.streamlock.net/vod/mp4:BigBuckBunny_115k.mov";
readonly LibVLC _libvlc;
public MainWindow()
{
InitializeComponent();
// this will load the native libvlc library (if needed, depending on the platform).
Core.Initialize();
// instantiate the main libvlc object
_libvlc = new LibVLC();
VideoView.MediaPlayer = new LibVLCSharp.Shared.MediaPlayer(_libvlc);
VideoView.MediaPlayer.Play(new Media(_libvlc, VIDEO_URL, FromType.FromLocation));
}
}
}
好吧,我终于找到了命令列表并正确使用了它们。原来我爱上了LibVLC。 :D这是描述各种命令的链接:https://wiki.videolan.org/Documentation:Streaming_HowTo/Advanced_Streaming_Using_the_Command_Line/#rtp
这是我完成代码的方式:
using LibVLCSharp.Shared;
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Windows;
using System.Windows.Controls;
using System.Windows.Data;
using System.Windows.Documents;
using System.Windows.Input;
using System.Windows.Media;
using System.Windows.Media.Imaging;
using System.Windows.Navigation;
using System.Windows.Shapes;
namespace Serwer
{
/// <summary>
/// Interaction logic for MainWindow.xaml
/// </summary>
public partial class MainWindow : Window
{
const string VIDEO_URL = "rtsp://wowzaec2demo.streamlock.net/vod/mp4:BigBuckBunny_115k.mov";
readonly LibVLC _libvlc;
public MainWindow()
{
InitializeComponent(); // this will load the native libvlc library (if needed, depending on the platform).
Core.Initialize(); // instantiate the main libvlc object
_libvlc = new LibVLC();
VideoView.MediaPlayer = new LibVLCSharp.Shared.MediaPlayer(_libvlc);
var rtsp1 = new Media(_libvlc, VIDEO_URL, FromType.FromLocation); //Create a media object and then set its options to duplicate streams - 1 on display 2 as RTSP
rtsp1.AddOption(":sout=#duplicate" +
"{dst=display{noaudio}," +
"dst=rtp{mux=ts,dst=192.168.0.110,port=8080,sdp=rtsp://192.168.0.110:8080/go.sdp}"); //The address has to be your local network adapters addres not localhost
VideoView.MediaPlayer.Play(rtsp1);
VideoView1.MediaPlayer = new LibVLCSharp.Shared.MediaPlayer(_libvlc);
VideoView1.MediaPlayer.Mute=true;
VideoView1.MediaPlayer.Play(new Media(_libvlc, "rtsp://192.168.0.110:8080/go.sdp", FromType.FromLocation));
}
}
}
感谢立方体指向正确的方向!