Hi Walter,
Hopefully the attachment will work this time.
I've also added the bit of code you recommended into my InitializeComponent method and now when I try to load the form in design mode I get the following error:
You must have a license to use this ActiveX control.
at System.Windows.Forms.AxHost.CreateInstance()
at System.Windows.Forms.AxHost.GetOcxCreate()
at System.Windows.Forms.AxHost.set_Site(ISite value)
at System.ComponentModel.Container.Add(IComponent component, String name)
at System.ComponentModel.Design.DesignerHost.Add(IComponent component, String name)
at System.ComponentModel.Design.DesignerHost.System.ComponentModel.Design.IDesignerHost.CreateComponent(Type componentType, String name)
at System.ComponentModel.Design.Serialization.DesignerSerializationManager.CreateInstance(Type type, ICollection arguments, String name, Boolean addToContainer)
at System.ComponentModel.Design.Serialization.DesignerSerializationManager.System.ComponentModel.Design.Serialization.IDesignerSerializationManager.CreateInstance(Type type, ICollection arguments, String name, Boolean addToContainer)
at System.ComponentModel.Design.Serialization.CodeDomSerializerBase.DeserializeInstance(IDesignerSerializationManager manager, Type type, Object[] parameters, String name, Boolean addToContainer)
at System.ComponentModel.Design.Serialization.ComponentCodeDomSerializer.DeserializeInstance(IDesignerSerializationManager manager, Type type, Object[] parameters, String name, Boolean addToContainer)
at System.ComponentModel.Design.Serialization.CodeDomSerializerBase.DeserializeExpression(IDesignerSerializationManager manager, String name, CodeExpression expression)
at System.ComponentModel.Design.Serialization.CodeDomSerializer.DeserializeStatementToInstance(IDesignerSerializationManager manager, CodeStatement statement)
at System.ComponentModel.Design.Serialization.CodeDomSerializer.Deserialize(IDesignerSerializationManager manager, Object codeObject)
at System.Windows.Forms.Design.ControlCodeDomSerializer.Deserialize(IDesignerSerializationManager manager, Object codeObject)
at System.ComponentModel.Design.Serialization.TypeCodeDomSerializer.DeserializeName(IDesignerSerializationManager manager, String name, CodeStatementCollection statements)
Regards,
Liviu