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| | #1 (permalink) |
| | 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?. 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 |
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| | #2 (permalink) |
| | Re: Programmatically opening a file in its native program Nam wrote: Quote: > 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 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 |
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| | #3 (permalink) |
| | Re: Programmatically opening a file in its native program Arto, Thank you very much. You explained it very well and your example was very easy to understand. Nam "Arto Viitanen" wrote: Quote: > Nam wrote: Quote: > > 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 > |
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