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Programmatically opening a file in its native program

P: n/a
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
Jul 28 '08 #1
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P: n/a
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.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
Jul 28 '08 #2

P: n/a
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.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
Jul 28 '08 #3

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