Convert an Object to an Xml String

Convert an object to an XML string

Here are conversion method for both ways.
this = instance of your class

public string ToXML()
    {
        using(var stringwriter = new System.IO.StringWriter())
        { 
            var serializer = new XmlSerializer(this.GetType());
            serializer.Serialize(stringwriter, this);
            return stringwriter.ToString();
        }
    }

 public static YourClass LoadFromXMLString(string xmlText)
    {
        using(var stringReader = new System.IO.StringReader(xmlText))
        {
            var serializer = new XmlSerializer(typeof(YourClass ));
            return serializer.Deserialize(stringReader) as YourClass ;
        }
    }

Convert Java object to XML string

You can use the Marshaler's method for marshaling which takes a Writer as parameter:

marshal(Object,Writer)

and pass it an Implementation which can build a String object

Direct Known Subclasses:
BufferedWriter, CharArrayWriter, FilterWriter, OutputStreamWriter, PipedWriter, PrintWriter, StringWriter

Call its toString method to get the actual String value.

So doing:

StringWriter sw = new StringWriter();
jaxbMarshaller.marshal(customer, sw);
String xmlString = sw.toString();

Convert a list of objects to an xml string

I made a fiddle here that illustrates a way to serialize your object all at once, instead of piecing strings together.

I suspect your extra <Tweet></Tweet> is because of a null or empty value in the list, because I am not experiencing it in my test above.

XmlSerializer Convert C# object to xml string

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>

how to convert xml document object to string?

You will need to serialize your xmlDoc back to XML once you have made the changes:

var s = new XMLSerializer();
var newXmlStr = s.serializeToString(xmlDoc);

Now you can do what you need to do with the string of updated XML, overwrite your xml variable, or send it to the server, or whatever...

See the MDN docs for further info

c# How to convert Object to XML

var writer = new StringWriter();
var serializer = new XmlSerializer(typeof(YourData));
serializer.Serialize(writer, obj);
string xml = writer.ToString();

See also MSDN

Convert XML String to Object

You need to use the xsd.exe tool which gets installed with the Windows SDK into a directory something similar to:

C:\Program Files\Microsoft SDKs\Windows\v6.0A\bin

And on 64-bit computers:

C:\Program Files (x86)\Microsoft SDKs\Windows\v6.0A\bin

And on Windows 10 computers:

C:\Program Files (x86)\Microsoft SDKs\Windows\v7.0A\bin

On the first run, you use xsd.exe and you convert your sample XML into a XSD file (XML schema file):

xsd yourfile.xml

This gives you yourfile.xsd, which in a second step, you can convert again using xsd.exe into a C# class:

xsd yourfile.xsd /c

This should give you a file yourfile.cs which will contain a C# class that you can use to deserialize the XML file you're getting - something like:

XmlSerializer serializer = new XmlSerializer(typeof(msg));
msg resultingMessage = (msg)serializer.Deserialize(new XmlTextReader("yourfile.xml"));

Should work pretty well for most cases.

Update: the XML serializer will take any stream as its input - either a file or a memory stream will be fine:

XmlSerializer serializer = new XmlSerializer(typeof(msg));
MemoryStream memStream = new MemoryStream(Encoding.UTF8.GetBytes(inputString));
msg resultingMessage = (msg)serializer.Deserialize(memStream);

or use a StringReader:

XmlSerializer serializer = new XmlSerializer(typeof(msg));
StringReader rdr = new StringReader(inputString);
msg resultingMessage = (msg)serializer.Deserialize(rdr);

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