Пример #1
0
        private void FromXml(XmlLuaSolutionDocument objectToDeserialize)
        {
            base.FromXml(objectToDeserialize);

            if (objectToDeserialize.ActiveProject != null)
            {
                _activeProject = FindDocumentGroup(objectToDeserialize.ActiveProject.FileName) as ILuaEditDocumentProject;
            }
        }
Пример #2
0
        private XmlLuaSolutionDocument ToXml(XmlLuaSolutionDocument objectToSerialize)
        {
            if (objectToSerialize == null)
            {
                objectToSerialize = new XmlLuaSolutionDocument();
            }

            base.ToXml(objectToSerialize);

            objectToSerialize.ActiveProject = _activeProject == null ? null : new DocumentRef(_activeProject.FileName);

            return(objectToSerialize);
        }