Version: .NET Framework 2.0
In my Windows Forms application, how can I open a file in its native
program?. Let’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
2  1400 
Nam wrote:
Version: .NET Framework 2.0
In my Windows Forms application, how can I open a file in its native
program?. Let’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.Diagnost ics.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.Diagnost ics.Process.Sta rt(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
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?. Let’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.Diagnost ics.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.Diagnost ics.Process.Sta rt(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
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