Process.Start(filename);
Regards
Stuart
On Thu, Jun 12, 2008 at 4:47 PM, Holly Styles
<[EMAIL PROTECTED]> wrote:
> Hi There,
>
> I'm wondering if anyone has an opinion on the best way to appraoch this.
> I am using SpreadsheetGear to create an xls file from a windows forms
> application.
> I hav
Process.Start("filename");
--
Steve Johnson
On Thu, Jun 12, 2008 at 11:47 AM, Holly Styles <[EMAIL PROTECTED]>
wrote:
> Hi There,
>
> I'm wondering if anyone has an opinion on the best way to appraoch this.
> I am using SpreadsheetGear to create an xls file from a windows forms
> application.
>
Hi Holly, this will do what you've described:
Process process = new Process();
process.StartInfo.FileName = @"C:\spreadsheet.xls";
process.StartInfo.UseShellExecute = true;
process.Start();
On Thu, 12 Jun 2008 16:47:51 +0100, Holly Styles
<[EMAIL PROTECTED]> wrote:
>Hi There,
>
>I'm wondering if
Check System.Diagnostics.Process.Start(string filename)
Sébastien
On 6/12/08, Holly Styles <[EMAIL PROTECTED]> wrote:
> Hi There,
>
> I'm wondering if anyone has an opinion on the best way to appraoch this.
> I am using SpreadsheetGear to create an xls file from a windows forms
> application.
You can just use System.Diagnostics.Process to do this directly. Here's an
example done in IronPython that should trivially convert to any other
language:
import System
from System.Diagnostics import Process
process = Process()
process.StartInfo.FileName = 'f:\\book1.xlsx'
process.StartInfo.UseSh
Hi There,
I'm wondering if anyone has an opinion on the best way to appraoch this.
I am using SpreadsheetGear to create an xls file from a windows forms
application.
I have looked at office interop for opening this file in Excel once it
has been written to disk. But I'm not comfortable with that.
