InfoPath Dev

I am developing a .NET Windows application that will interface with InfoPath and I would like the user to be able to double click an .xsn filename from within a ListView control and automatically fire up InfoPath and display the InfoPath form based on the filename selected. I capture the full path to the file in the double click event but not sure how to open the file with InfoPath programmatically. Can someone point me in the right direction?

Thank you!

Hi sjc724,

After getting the full file name you use the following code,

System.Diagnostics.ProcessStartInfo psi = new ProcessStartInfo(@"Give your full file path here");
System.Diagnostics.Process process = new Process();
process.StartInfo = psi;
process.Start();

I think this would work for you,

Thank you


Nagesh
InfoPath Dev India

That worked! Thank you very much.