Question

我有以下C＃类：

public class Books
{

public List<Book> BookList;

}

public class Book
{

public string Title;
public string Description;
public string Author;
public string Publisher;

}

如何将此类序列化为以下XML？

<Books>
  <Book Title="t1" Description="d1"/>
  <Book Description="d2" Author="a2"/>
  <Book Title="t3" Author="a3" Publisher="p3"/>
</Books>

我希望XML只包含那些值为null / empty的属性。例如：在第一个Book元素中，author是空白的，因此它不应出现在序列化的XML中。

Answer 1

您应该可以使用ShouldSerialize*模式：

public class Book
{
    [XmlAttribute] 
    public string Title {get;set;}

    public bool ShouldSerializeTitle() {
        return !string.IsNullOrEmpty(Title);
    }

    [XmlAttribute]
    public string Description {get;set;}

    public bool ShouldSerializeDescription() {
        return !string.IsNullOrEmpty(Description );
    }

    [XmlAttribute]
    public string Author {get;set;}

    public bool ShouldSerializeAuthor() {
        return !string.IsNullOrEmpty(Author);
    }

    [XmlAttribute]
    public string Publisher {get;set;}

    public bool ShouldSerializePublisher() {
        return !string.IsNullOrEmpty(Publisher);
    }
}

Answer 2

替代方案：

将您的公共字段切换为属性
使用DefaultValueAttribute属性
使用ContentPropertyAttribute属性
使用XamlWriter / XamlReader

你最终得到这样的东西：

 [ContentProperty("Books")]
 public class Library {

   private readonly List<Book> m_books = new List<Book>();

   public List<Book> Books { get { return m_books; } }

 }

 public class Book
 {

    [DefaultValue(string.Empty)]
    public string Title { get; set; }

    [DefaultValue(string.Empty)]
    public string Description { get; set; }

    [DefaultValue(string.Empty)]
    public string Author { get; set; }

 }

Answer 3

public class Books
{
    [XmlElement("Book")]
    public List<Book> BookList;
}

public class Book
{
    [XmlAttribute]
    public string Title;
    [XmlAttribute]
    public string Description;
    [XmlAttribute]
    public string Author;
    [XmlAttribute]
    public string Publisher;
}

class Program
{
    static void Main()
    {
        var books = new Books
        {
            BookList = new List<Book>(new[] 
            {
                new Book 
                {
                    Title = "t1",
                    Description = "d1"
                },
                new Book 
                {
                    Author = "a2",
                    Description = "d2"
                },
                new Book 
                {
                    Author = "a3",
                    Title = "t3",
                    Publisher = "p3"
                },
            })
        };

        var serializer = new XmlSerializer(books.GetType());
        serializer.Serialize(Console.Out, books);
    }
}

如果要从根节点中删除命名空间：

var namespaces = new XmlSerializerNamespaces();
namespaces.Add(string.Empty, string.Empty);
serializer.Serialize(Console.Out, books, namespaces);

另外，我建议您在模型类中使用properties而不是fields来更好地封装：

public class Books
{
    [XmlElement("Book")]
    public List<Book> BookList { get; set; }
}

Answer 4

另一种选择是使用“指定”属性，例如（类似于“ ShouldSerialize（）”的作品）：

public bool <<Your XML-property name here>>Specified
{
    get { return <<your condition here (i.e. not null check)>>; }
}

条件xml序列化

4 个答案: