我用了windowsmediaplayer控件,想利用它打开并播放媒体文件,即有一个commanddialog控件,用一个按钮启动它
Private Sub Command1_Click()
CommonDialog1.ShowOpenCommonDialog1.Filter = "(*.avi;*.mpeg;*.dat;*.mp3;*.mov)|*.avi;*.mpeg;*.mp3;*.dat;*.mov"
Dim a$
a = CommonDialog1.FileName
WindowsMediaPlayer1.FileName = a
WindowsMediaPlayer1.autoStart = True
End Sub却提示说对象不支持该属性或方法!如何解决?
Private Sub Command1_Click()
CommonDialog1.ShowOpenCommonDialog1.Filter = "(*.avi;*.mpeg;*.dat;*.mp3;*.mov)|*.avi;*.mpeg;*.mp3;*.dat;*.mov"
Dim a$
a = CommonDialog1.FileName
WindowsMediaPlayer1.FileName = a
WindowsMediaPlayer1.autoStart = True
End Sub却提示说对象不支持该属性或方法!如何解决?
MediaPlayer1.FileName = a
MediaPlayer1.autoStart = True
MediaPlayer1.FileName = a
MediaPlayer1.autoStart = True呵呵,这样的错误建议输代码的时候,前面先输入me.这样应该就不会类似的错误了
' 初始化窗口
Private Sub Form_Load()
CommonDialog1.Filter = "影像文件(*.Avi;*.Mpg;*.Dat;*.Asf)|*.Avi;*.MPG;*.DAT;*.Asf| All files (*.*)| *.*"
End Sub
' 打开文件并自动播放
Private Sub open_Click(Index As Integer)
Dim MFile As String
CommonDialog1.ShowOpen
MFile = CommonDialog1.FileName
MediaPlayer1.FileName = MFile
MediaPlayer1.AutoStart = True
MediaPlayer1.AutoSize = True
End Sub
这个是原码,大家帮忙看看
[email protected]
谢谢
WMPOCX1.URL = MFile