InfoPath Dev

Hi

I have a button with form code (vb)

Application.XmlForms.Open(http://server/site/filename.xml)

and get this error code

System.Runtime.InteropServices.COMException
Die folgende Datei kann von InfoPath nicht geöffnet werden: http://server/site/filename.xml) 
   at Microsoft.Office.Interop.InfoPath.SemiTrust.XDocumentsCollectionWrapper.Open(Object varURI, Int32 dwBehavior)
   at Microsoft.Office.InfoPath.Internal.XmlFormsHost.Open(String fileLocation, XmlFormOpenMode behavior)
   at Microsoft.Office.InfoPath.Internal.XmlFormsHost.Open(String fileLocation)
   at ErfassungSchuldaten_V03_C_Bereiche.FormCode.btn_b1_oeffnen_Clicked(Object sender, ClickedEventArgs e)
   at Microsoft.Office.InfoPath.Internal.ButtonEventHost.OnButtonClick(DocActionEvent pEvent)
   at Microsoft.Office.Interop.InfoPath.SemiTrust._ButtonEventSink_SinkHelper.OnClick(DocActionEvent pEvent)

The form has full trust and certificate.

Can somebody help me?

thanks

Susanne  

Hi Susanne.

My german is not so good to know the exact error message but I think you have to specify the template (.xsn file) and then specify the XML to open.

I have not tried opening the xml directly in this way but is it supposed to use the local installed form as the template?

I am assuming if you open it directly the form opens correctly.