我尝试在WPF上制作媒体播放器。
我做到了这一点:
public partial class MyMediaPlayer : Window
{
public MyMediaPlayer()
{
InitializeComponent();
//
OpenFileDialog dlg = new OpenFileDialog();
dlg.InitialDirectory = "c:\\"; // init
dlg.Filter = "All Files (*.*)|*.*"; // filter
dlg.RestoreDirectory = true;
// dialog window
if (dlg.ShowDialog() == true) // checked ?
{
string selectedFileName = dlg.FileName; // path of the media
MediaPlayer player = new MediaPlayer();
player.Open(new Uri(selectedFileName, UriKind.Relative));
VideoDrawing aVideoDrawing = new VideoDrawing();
aVideoDrawing.Rect = new Rect(0, 0, 100, 100);
aVideoDrawing.Player = player; // play
// never play
player.Play();
}
}
}
和XAML文件:
<Window ... >
<Grid>
<MediaElement Margin="10,10,10,0 " Source="D:\test.avi"
Name="McMediaElement"
Width="450" Height="250" LoadedBehavior="Manual" UnloadedBehavior="Stop" Stretch="Fill"
/>
</Grid>
</Window>
但是,视频永远不会启动,窗口会保持白色。
请帮助:)
ps:抱歉我的英文不好
答案 0 :(得分:0)
MediaPlayer类(msdn):
MediaPlayer与MediaElement的不同之处在于它不是 可以直接添加到用户界面(UI)的控件 应用。要显示使用MediaPlayer加载的媒体,请使用VideoDrawing 或者必须使用DrawingContext。
因此,如果您想使用MediaPlayer,则应使用DrawingBrush
class:
...
string selectedFileName = dlg.FileName;
MediaPlayer player = new MediaPlayer();
player.Open(new Uri(selectedFileName, UriKind.Relative));
VideoDrawing aVideoDrawing = new VideoDrawing();
aVideoDrawing.Rect = new Rect(0, 0, 100, 100);
aVideoDrawing.Player = player;
player.Play();
DrawingBrush DBrush = new DrawingBrush(aVideoDrawing);
this.Background = DBrush;
...
在此解决方案中,您不必在XAML中添加MediaElement
。
要仅在XAML中播放媒体,请使用MediaElement
(msdn)。
XAML:
<MediaElement Name="McMediaElement" Source="D:\test.avi"
LoadedBehavior="Play" UnloadedBehavior="Stop" Stretch="Fill"
Margin="10,10,10,0" Width="450" Height="250"
/>
代码隐藏:
public MainWindow()
{
InitializeComponent();
OpenFileDialog dlg = new OpenFileDialog();
dlg.InitialDirectory = "c:\\";
dlg.Filter = "All Files (*.*)|*.*"; // filter
dlg.RestoreDirectory = true;
if (dlg.ShowDialog() == true)
{
string selectedFileName = dlg.FileName;
McMediaElement.Source = new Uri(selectedFileName, UriKind.Absolute);
}
}