https://www.infopathdev.com:443/forums/41.aspxIf you're not sure where to ask your question, ask it here. If necessary, our moderators will make sure it gets classified in the correct board.enCommunityServer 2007 SP3 (Build: 31118.962)https://www.infopathdev.com:443/forums/thread/8641.aspxWed, 28 Dec 2005 11:51:14 GMT033a2e2d-04e2-4a9d-be01-a4634161eefd:8641sjc7240https://www.infopathdev.com:443/forums/thread/8641.aspxhttps://www.infopathdev.com:443/forums/commentrss.aspx?SectionID=41&PostID=8641That worked! Thank you very much.https://www.infopathdev.com:443/forums/thread/8640.aspxWed, 28 Dec 2005 07:59:12 GMT033a2e2d-04e2-4a9d-be01-a4634161eefd:8640Nagesh0https://www.infopathdev.com:443/forums/thread/8640.aspxhttps://www.infopathdev.com:443/forums/commentrss.aspx?SectionID=41&PostID=8640Hi sjc724,<br /><br /> After getting the full file name you use the following code,<br /><br /> System.Diagnostics.ProcessStartInfo psi = new ProcessStartInfo(@"Give your full file path here");<br /> System.Diagnostics.Process process = new Process();<br /> process.StartInfo = psi;<br /> process.Start();<br /><br /> I think this would work for you,<br /><br /> Thank you<br /><br /><br />Nagesh<br />InfoPath Dev Indiahttps://www.infopathdev.com:443/forums/thread/5228.aspxTue, 27 Dec 2005 15:40:12 GMT033a2e2d-04e2-4a9d-be01-a4634161eefd:5228sjc7240https://www.infopathdev.com:443/forums/thread/5228.aspxhttps://www.infopathdev.com:443/forums/commentrss.aspx?SectionID=41&PostID=5228I 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?<br /><br />Thank you!