XmlSerializer Convert C# object to xml string

Question

I have created a C# class:

public class books {
    public int bookNum { get; set; }
    public class book {
        public string name { get; set; }
        public class record {
            public string borrowDate { get; set; }
            public string returnDate { get; set; }
        }
        public record[] records { get; set; }
    }
    public book[] books { get; set; }
}

But is when I use XmlSerializer convert to XML string. The result is not the same as below xml.

What is the problem of my C# class? I want to use XmlSerializer to ouput the result instead of using XmlDocument.

Any ideas? Thanks in advance!

<books>
    <bookNum>2</bookNum>
    <book>
        <name>Book 1</name>
        <record>
            <borrowDate>2013-7-1</borrowDate>
            <returnDate>2013-7-12</returnDate>
        </record>
        <record>            
            <borrowDate>2013-8-1</borrowDate>
            <returnDate>2013-8-5</returnDate>
        </record>
    </book>
    <book>
        <name>Book 2</name>
        <record>
            <borrowDate>2013-6-1</borrowDate>
            <returnDate>2013-6-12</returnDate>
        </record>
        <record>            
            <borrowDate>2013-7-1</borrowDate>
            <returnDate>2013-7-5</returnDate>
        </record>
    </book>
</books>

EDIT

Below is my C# code and the ouput result:

books books = new books {
        bookNum = 2,
        Books = new books.book[] { 
            new books.book {  
                name = "Book1", 
                records = new books.book.record[] {
                    new books.book.record {
                        borrowDate = "2013-1-3",
                        returnDate = "2013-1-5"
                    },
                     new books.book.record {
                        borrowDate = "2013-2-3",
                        returnDate = "2013-4-5"
                    }
                }
            },
             new books.book {  
                name = "Book1", 
                records = new books.book.record[] {
                    new books.book.record {
                        borrowDate = "2013-1-3",
                        returnDate = "2013-1-5"
                    },
                     new books.book.record {
                        borrowDate = "2013-2-3",
                        returnDate = "2013-4-5"
                    }
                }
            }
        }
    };


    XmlSerializer xsSubmit = new XmlSerializer(typeof(books));

    XmlDocument doc = new XmlDocument();

    System.IO.StringWriter sww = new System.IO.StringWriter();
    XmlWriter writer = XmlWriter.Create(sww);
    xsSubmit.Serialize(writer, books);
    var xml = sww.ToString(); // Your xml
    context.Response.Write(xml);

XML:

<books>
    <bookNum>2</bookNum>
    <Books>
        <book>
            <name>Book1</name>
            <records>
                <record>
                    <borrowDate>2013-1-3</borrowDate>
                    <returnDate>2013-1-5</returnDate>
                </record>
                <record>
                    <borrowDate>2013-2-3</borrowDate>
                    <returnDate>2013-4-5</returnDate>
                </record>
            </records>
        </book>
        <book>
            <name>Book1</name>
            <records>
                <record>
                    <borrowDate>2013-1-3</borrowDate>
                    <returnDate>2013-1-5</returnDate>
                </record>
                <record>
                    <borrowDate>2013-2-3</borrowDate>
                    <returnDate>2013-4-5</returnDate>
                </record>
            </records>
         </book>
    </Books>
</books>

"The result is not the same as below xml" - how it is "not the same"? What do you get instead? (Side note: you use unusual format for dates in XML - ISO8601 YYYY-MM-DD) — Alexei Levenkov, Jul 19 '13 at 06:24
Actually Mac, you can. You need to attribute the properties to change their Element names, but you can. — David Colwell, Jul 19 '13 at 06:27
You can get such output using implementation of [IXmlSerializable](http://msdn.microsoft.com/ru-ru/library/system.xml.serialization.ixmlserializable.aspx) attribute - see 3rd abstract in my answer. — Andrii Kalytiiuk, Jul 19 '13 at 07:06

score 10 · Accepted Answer · edited May 23 '17 at 12:01

You cannot serialize class from your question using standard serialization tools so that it will have <book> entries on the same level as <bookNum> node.

When class saved with standard serialization tools list of your <book> nodes will always be nested into separate array node that will be on the same level as <bookNum> node. Same concerns records array field on book class.

To generate XML output that you want to - with <book> nodes on same level as <bookNum> node - you will have to implement IXmlSerializable interface in your books class for custom serialization. To see examples of IXmlSerializable implementation visit these links: StackOverflow answer, CodeProject article.

Another solution will be - as stated user Alexandr in comment to my answer - to inherit your books class from List<book> type and to have on your book class field records of class type that is inherited from List<record> type.

When serializing class from your question, assuming that your assigned proper XmlRoot, XmlElement, XmlArray and XmlArrayItem attributes as follows:

[XmlRoot("books")]
public class books
{
    [XmlElement("bookNum")]
    public int bookNum { get; set; }

    [XmlRoot("book")]
    public class book
    {
        [XmlElement("name")]
        public string name { get; set; }

        [XmlRoot("record")]
        public class record
        {
            [XmlElement("borrowDate")]
            public string borrowDate { get; set; }

            [XmlElement("returnDate")]
            public string returnDate { get; set; }
        }

        [XmlArray("borrowRecords")]
        [XmlArrayItem("record")]
        public record[] records { get; set; }
    }

    [XmlArray("booksList")]
    [XmlArrayItem("book")]
    public book[] books { get; set; }
}

you will get XML output as follows:

<books>
    <bookNum>2</bookNum>
    <booksList>
        <book>
            <name>Book 1</name>
            <borrowRecords>
                <record>
                    <borrowDate>2013-1-3</borrowDate>
                    <returnDate>2013-1-5</returnDate>
                </record>
                <record>            
                    <borrowDate>2013-2-3</borrowDate>
                    <returnDate>2013-4-5</returnDate>
                </record>
            </borrowRecords>
        </book>
        <book>
            <name>Book 2</name>
            <borrowRecords>
                <record>
                    <borrowDate>2013-1-3</borrowDate>
                    <returnDate>2013-1-5</returnDate>
                </record>
                <record>            
                    <borrowDate>2013-2-3</borrowDate>
                    <returnDate>2013-4-5</returnDate>
                </record>
            </borrowRecords>
        </book>
    </booksList>
</books>

You sure about that? Cause I think I did it a while ago by having the root-class (`books` in this case) extend `List` and then applying properties to it with `XmlElement`-attribute... — Alxandr, Jul 19 '13 at 07:06
@Alxandr Inheriting list is another option - it changes a bit conditions of the problem. +1 for good idea. — Andrii Kalytiiuk, Jul 19 '13 at 07:11
I'm also guessing this is doable without inheriting from list, simply by providing an `Add` method, but I don't know how the internals of the XmlSerializer works... — Alxandr, Jul 19 '13 at 07:13
@Alxandr After quick search with google I have not found solutions with `Add` method - what I came up is only different variations with `IXmlSerializable`. Will appreciate links to alternative solutions. — Andrii Kalytiiuk, Jul 19 '13 at 07:25
It was only a guess based on the fact that the `Add` method enables stuff like `var a = new List() { "a", "b", "c" };` Also, `List` doesn't seem to implement `IXmlSerializable`, though it might just be a serializing constructor for all I know. I just figured that since .NET gives special meaning to an `Add` method, maybe the XmlSerializer does too... — Alxandr, Jul 19 '13 at 07:35
I did not find a serializing-constructor using JustDecompile. Maybe `List` or `IList` or `ICollection` are handled explicitly by the XmlSerializer? — Alxandr, Jul 19 '13 at 07:39

score 8 · Answer 2 · answered Jul 19 '13 at 06:46

I made the following change to your class code. I am unable to duplicate the XML serialization using the default serializer, because it will not duplicate the 'Record' element without giving it a container element.

[System.Xml.Serialization.XmlRoot("books")]
public class books 
{
    public int bookNum { get; set; }
    public class book {
        public string name { get; set; }
        public class record {
            public string borrowDate { get; set; }
            public string returnDate { get; set; }
        }
        public record[] records { get; set; }
    }
    public book[] books { get; set; }
}

Serializing this gives me the following output

<books xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <bookNum>2</bookNum>
  <books>
    <book>
      <name>first</name>
      <records>
        <record>
          <borrowDate>19/07/2013 4:41:29 PM</borrowDate>
          <returnDate>19/07/2013 4:41:29 PM</returnDate>
        </record>
      </records>
    </book>
  </books>
</books>

using this test code

books bks = new books();
bks.bookNum = 2;
bks.books = new books.book[]{ new books.book{name="first", records = new books.book.record[] {new books.book.record{borrowDate = DateTime.Now.ToString(), returnDate = DateTime.Now.ToString()}}}};

System.Xml.Serialization.XmlSerializer serializer = new System.Xml.Serialization.XmlSerializer(typeof(books));

XmlWriterSettings settings = new XmlWriterSettings();
settings.Encoding = new UnicodeEncoding(false, false); // no BOM in a .NET string
settings.Indent = true;
settings.OmitXmlDeclaration = true;

using(StringWriter textWriter = new StringWriter()) {
    using(XmlWriter xmlWriter = XmlWriter.Create(textWriter, settings)) {
        serializer.Serialize(xmlWriter, bks);
    }
    return textWriter.ToString(); //This is the output as a string
}

books is the container element for book. That is how it knows that they belong in the same property when re-serializing. — David Colwell, Jul 19 '13 at 07:14
Bad coding standards for C#. You can tell the writer is a Java developer. — csharpforevermore, Jun 15 '22 at 17:03

score 5 · Answer 3 · answered Feb 03 '16 at 00:59

I realize this is a couple of years late, but I have been able to achieve the structure that you desired just by using XmlElementAttribute.

I discovered this by using XSD.exe to generate schema definitions from xml and generate .Net code from xsd files. As far as I know, this works in .Net 3.5 through 4.6.

Here is the class definition I used:

public class books
{
    public int bookNum { get; set; }
    public class book {
        public string name { get; set; }
        public class record {
            public string borrowDate { get; set; }
            public string returnDate { get; set; }
        }
        [XmlElement("record")]
        public record[] records { get; set; }
    }
    [XmlElement("book")]
    public book[] allBooks { get; set; }
}

And here is a LinqPad snippet that illustrates the serialization/deserialization (based on David Colwell's code snippet, thanks BTW for the tip on how to exclude BOM, it was just what I was looking for):

books bks = new books();
books bks2 = null;
bks.bookNum = 2;
bks.allBooks = new books.book[] 
        { 
            new books.book {
                name="book 1", 
                records = new books.book.record[] {
                        new books.book.record{borrowDate = DateTime.Now.ToString(), returnDate = DateTime.Now.ToString()}
                    }
                },
            new books.book { 
                name="book 2", 
                records = new books.book.record[] { 
                        new books.book.record{borrowDate = DateTime.Now.ToString(), returnDate = DateTime.Now.ToString()}, 
                        new books.book.record{borrowDate = DateTime.Now.ToString(), returnDate = DateTime.Now.ToString()}}
                    },
        };
string xmlString;

System.Xml.Serialization.XmlSerializer serializer = new System.Xml.Serialization.XmlSerializer(typeof(books));

XmlWriterSettings settings = new XmlWriterSettings();
settings.Encoding = new UnicodeEncoding(false, false); // no BOM in a .NET string
settings.Indent = true;
settings.OmitXmlDeclaration = true;

XmlSerializerNamespaces ns = new XmlSerializerNamespaces();
// exclude xsi and xsd namespaces by adding the following:
ns.Add(string.Empty, string.Empty);

using(StringWriter textWriter = new StringWriter()) {
    using(XmlWriter xmlWriter = XmlWriter.Create(textWriter, settings)) {
        serializer.Serialize(xmlWriter, bks, ns);
    }
    xmlString = textWriter.ToString(); //This is the output as a string
}

xmlString.Dump();

// Deserialize the xml string now       
using ( TextReader reader = new StringReader(xmlString) ) {
    bks2 = ( books )serializer.Deserialize(reader);
}

bks2.Dump();

This produced XML that is can be serialized and deserialized without implementing IXmlSerializable, such as:

<books>
  <bookNum>2</bookNum>
  <book>
    <name>book 1</name>
    <record>
      <borrowDate>2/2/2016 5:57:25 PM</borrowDate>
      <returnDate>2/2/2016 5:57:25 PM</returnDate>
    </record>
  </book>
  <book>
    <name>book 2</name>
    <record>
      <borrowDate>2/2/2016 5:57:25 PM</borrowDate>
      <returnDate>2/2/2016 5:57:25 PM</returnDate>
    </record>
    <record>
      <borrowDate>2/2/2016 5:57:25 PM</borrowDate>
      <returnDate>2/2/2016 5:57:25 PM</returnDate>
    </record>
  </book>
</books>

Up vote for the idea of using xsd.exe to create a class from the desired XML! Definitely going to add that to my tool bag. — Andrew Steitz, Jul 03 '18 at 15:55

score 0 · Answer 4 · answered Mar 01 '18 at 18:11

if you need other classes, such as book2 inside the books class, you have some special instructions to implement it. example

public class books
{     
   public int bookNum {get; set; }
   public class book {
         public string name {get; set; }
         public class record {
             public string borrowDate {get; set; }
             public string returnDate {get; set; }
         }
         [XmlElement ("record")]
         public record [] records {get; set; }
     }
     [XmlElement ("book")]
     public book [] allBooks {get; set; }

     public int book2Num {get; set; }
     public class book2 {
         public string name {get; set; }
         public class record {
             public string borrowDate {get; set; }
             public string returnDate {get; set; }
         }
         [XmlElement ("record")]
         public record [] records {get; set; }
     }
     [XmlElement ("book2")]
     public book2 [] allBook2 {get; set; }
}`

When I try to run the program I have the following error:

"Additional information: Error when reflecting the type "

Please edit answer to elaborate on the topic of special instructions. — sorak, Mar 01 '18 at 18:32

XmlSerializer Convert C# object to xml string

4 Answers4