setAuthorName() публичный Метод

public setAuthorName ( String newName ) : void
newName String
Результат void
Пример #1
0
        public static void Main(string[] args)
        {
            Book abook = new Book();
            //Here's an instance of our type to serialize

            abook.setAuthorName("Michael Feathers");
            abook.setBookID(1337);
            abook.setTitle("Working Effectively with Legacy Code");
            //
            //

            SimplTypesScope book_example_sts = new SimplTypesScope("book_example", typeof(Book));

                //Serialize to JSON
            String jsonResult = SimplTypesScope.Serialize(abook, StringFormat.Json);

                //Serialize to XML
            // (Just change the StringFormat parameter!)
            String xmlResult  = SimplTypesScope.Serialize(abook, StringFormat.Xml);

            Object result1 = book_example_sts.Deserialize(jsonResult, StringFormat.Json);
            Book r1 = (Book) result1;

            Object result2 = book_example_sts.Deserialize(xmlResult, StringFormat.Xml);
            Book r2 = (Book) result2;

            Console.WriteLine(jsonResult);

            Console.WriteLine(xmlResult);
            Console.ReadLine();
        }
Пример #2
0
        public static void Main(string[] args)
        {
            Book abook = new Book();

            //Here's an instance of our type to serialize

            abook.setAuthorName("Michael Feathers");
            abook.setBookID(1337);
            abook.setTitle("Working Effectively with Legacy Code");
            //
            //

            SimplTypesScope book_example_sts = new SimplTypesScope("book_example", typeof(Book));

            //Serialize to JSON
            String jsonResult = SimplTypesScope.Serialize(abook, StringFormat.Json);

            //Serialize to XML
            // (Just change the StringFormat parameter!)
            String xmlResult = SimplTypesScope.Serialize(abook, StringFormat.Xml);

            Object result1 = book_example_sts.Deserialize(jsonResult, StringFormat.Json);
            Book   r1      = (Book)result1;

            Object result2 = book_example_sts.Deserialize(xmlResult, StringFormat.Xml);
            Book   r2      = (Book)result2;

            Console.WriteLine(jsonResult);

            Console.WriteLine(xmlResult);
            Console.ReadLine();
        }