Programmatically opening a file in its native program

Version: .NET Framework 2.0

In my Windows Forms application, how can I open a file in its native
program?. Letâ??s say a listbox displays a list of file names with full paths;
when a user double clicks on a file name, the file opens in its native
program, a Word document opens in a Word Application, an HTML file opens in
IE etc. etc. Much like search result pane of Windows Explorer Search utility.

Thank you

Arto
Mon Jul 28 00:39:02 CDT 2008

Nam wrote:
> Version: .NET Framework 2.0
>
> In my Windows Forms application, how can I open a file in its native
> program?. Letâ??s say a listbox displays a list of file names with full paths;
> when a user double clicks on a file name, the file opens in its native
> program, a Word document opens in a Word Application, an HTML file opens in
> IE etc. etc. Much like search result pane of Windows Explorer Search utility.
>
> Thank you

System.Diagnostics.Process has method Start, that "Starts a process
resource by specifying the name of a document or application file and
associates the resource with a new Process component.".

Following code works fine:


OpenFileDialog ofd = new OpenFileDialog();
if (ofd.ShowDialog() == DialogResult.OK)
System.Diagnostics.Process.Start(ofd.FileName);

The file open dialog shows, and when the user selects a file, and
clicks open, the file is opened on its associated program.

--
Arto Viitanen

Nam
Mon Jul 28 11:16:01 CDT 2008

Arto,
Thank you very much. You explained it very well and your example was very
easy to understand.

Nam

"Arto Viitanen" wrote:

> Nam wrote:
> > Version: .NET Framework 2.0
> >
> > In my Windows Forms application, how can I open a file in its native
> > program?. Letâ??s say a listbox displays a list of file names with full paths;
> > when a user double clicks on a file name, the file opens in its native
> > program, a Word document opens in a Word Application, an HTML file opens in
> > IE etc. etc. Much like search result pane of Windows Explorer Search utility.
> >
> > Thank you
>
> System.Diagnostics.Process has method Start, that "Starts a process
> resource by specifying the name of a document or application file and
> associates the resource with a new Process component.".
>
> Following code works fine:
>
>
> OpenFileDialog ofd = new OpenFileDialog();
> if (ofd.ShowDialog() == DialogResult.OK)
> System.Diagnostics.Process.Start(ofd.FileName);
>
> The file open dialog shows, and when the user selects a file, and
> clicks open, the file is opened on its associated program.
>
> --
> Arto Viitanen
>