Hello,Being French, if there are mistakes, I apologize on behalf of Google Trad.I'm currently working on a daily calendar project to manage a team that should have appointments that are either for a single owner or multiple owners.My client being demanding, he does not want there to be two separate appointments but only one appointment that appears.So I have to find a way to override the appointment on the column of the first owner on the column of the second owner so that, visually, we see only one appointment for two owner.I enclose a screen of what I must have visually.
I'm using UltraDayView with GroupingStyle = OwnerWithinDate, I've attached UltraCalendarInfo and UltraCalendarLook with ViewStyle = Office2007.
Through the CreateFilter, I manage to enlarge the width of the appointment:
Public Sub AfterCreateChildElements(parent As UIElement) Implements IUIElementCreationFilter.AfterCreateChildElements
If (Not (TypeOf (parent) Is DayView.AppointmentUIElement)) Then Return
parent.Rect = New Rectangle(New Point(parent.Rect.X, parent.Rect.Y), New Drawing.Size(1000, parent.Rect.Height))
But the part of the rendezvous that overtakes on the second owner is not displayed.
Appointments do not have the capability to span several owners in our tool. If you would like to see this in a future version, I recommend submitting a product idea. You can suggest new product ideas for future versions on our ideas site.
Remember when submitting your idea to explain the context in which a feature would be used and why it is needed as well as anything that would prevent you from accomplishing this today. You can even add screenshots to build a stronger case.
Thank you in advance for submitting your product idea.
I would also like to use caution while attempting to implement this in application code. The control is not designed to have appointments spanning across owners, so many issues will need to be dealt with:
I would ensure that you work with your team to resolve these issues before beginning to implement this functionality. Please let me know if you have any questions.