Conditional xml serialization

Question

I have the following C# classes :

public class Books
{

public List<Book> BookList;

}

public class Book
{

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

}

How can I serialize this class into the following XML?

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

I want the XML to have only those attributes whose values are not null/empty. For example: In the first Book element, author is blank, so it should not be present in the serialized XML.

T.S. · Accepted Answer · 2018-09-24 14:57:39Z

47

You should be able to use the ShouldSerialize* pattern:

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);
    }
}

edited Sep 24, 2018 at 14:57

T.S.

19.7k11 gold badges70 silver badges96 bronze badges

answered Nov 4, 2010 at 9:19

Marc Gravell

1.1m273 gold badges2.6k silver badges3k bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Dean Over a year ago

I was unclear on how the above solution worked until I read the following post: kjellsj.blogspot.com/2006/02/…

iCollect.it Ltd Over a year ago

+1 Coolest thing I never knew about :) Just solved a particularly sticky backward-compatibility problem with this gem.

Stelios Adamantidis Over a year ago

Note that ShouldSerialize method should be public here, unlike other cases like PropertyGrid serialization control where the method can be private and yet not be ignored.

Nicolas Repiquet · Accepted Answer · 2010-11-04 08:57:24Z

8

Alternative :

Switch your public fields to properties
Define default values with the DefaultValueAttribute attribute
Define content property with the ContentPropertyAttribute attribute
Use XamlWriter/XamlReader

You end up with something like this:

 [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; }

 }

answered Nov 4, 2010 at 8:57

Nicolas Repiquet

9,3373 gold badges36 silver badges54 bronze badges

1 Comment

Ivan Krivyakov Over a year ago

This should be the answer. ShouldSerialize* works, but [DefaultValue("")] is by far simpler method

Darin Dimitrov · Accepted Answer · 2010-11-04 08:52:09Z

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);
    }
}

And if you want to remove the namespace from the root node:

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

Also I would recommend you using properties instead of fields in your model classes for better encapsulation:

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

edited Nov 4, 2010 at 8:52

answered Nov 4, 2010 at 8:46

Darin Dimitrov

1.0m277 gold badges3.3k silver badges3k bronze badges

4 Comments

Will Dean Over a year ago

Is this actually answering the question about skipping the null/blank items?

Darin Dimitrov Over a year ago

Indeed my code works for null elements, but it doesn't work for blank elements (they will still be present in the generated XML).

user40907 Over a year ago

I believe, when u use properties, by default, the value is null and hence the attributes wont be present in the generated xml unless they r set to empty or any other value.

Darin Dimitrov Over a year ago

Fields are also null by default for reference types and the default value for value types.

Martin.Martinsson · Accepted Answer · 2020-09-30 05:27:16Z

2

Another option is to use "Specified"-property like (works like "ShouldSerialize()"):

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

answered Sep 30, 2020 at 5:27

Martin.Martinsson

2,20125 silver badges28 bronze badges

Collectives™ on Stack Overflow

Conditional xml serialization

4 Answers 4

3 Comments

1 Comment

4 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

3 Comments

1 Comment

4 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related