在C＃中打开视频文件

Question

I am new to C# and this might sound stupid, I did some research and I think I am confused. I want my c# program to open a video file (c:\\abc.mov), I have set the .mov files to open automatically with quick time player and I want the program to open the file with quick time player just like double clicking on that file. When I use this code it does not do anything!

File.Open(@"c:\abc.mov", FileMode.Open);

Please help me?

Answer 1

You should use Process.Start instead. Here's the MSDN page on that.

You can specify which program you want to start with whetever arguments you need, like in this example.

Edit: Added another example. Thanks @DJBurb

Process.Start(@"c:\\abc.mov");

This code should open the .mov file with the default movie player associated with the .mov extension.

Answer 2

我相信open（）将打开您的文件以供当前程序编辑，而不是使用系统的默认播放器实际打开文件

Answer 3

File.Open返回FileStream因此您可以读取该文件，相反，您绝对想使用Process.Start(@"c:\\abc.mov");

Answer 4

这将使用dafault视频播放器打开您的视频文件

System.Diagnostics.Process.Start(filepath);

Answer 5

 private void buttonOpen_Click(object sender, EventArgs e)
    { 
        if (ofd.ShowDialog()==DialogResult.OK)
        {
            Process.Start(ofd.FileName);
        }
    }