Hey sachin,
you can achieve what you are trying to do by inspecting the errors collection of the XmlForm object. For example, if your control is called XmlFormView1, you can do the following in your code-behind:
this.XmlFormView1.DataBind(); // or else XmlForm will be null
XmlForm xmlForm = this.XmlFormView1.XmlForm;
And then you can party on xmlForm all you want.
However there is one problem: you can only do this when the form posts back to the server. You can either achieve it by having a button doing a postback (saying "display errors") or by setting all the controls to always send data to server when modified. The later is not advised because of bad performance and user experience.
Hope that helps,
David.
David Airapetyan / Software Services ArchitectQdabra® Software / Streamline data gathering to turn process into knowledge