Question

I’m using an XmlSerializer. My class:

[Serializable]
[XmlRoot(ElementName="MyClass")]
public class MyClass
{
    public string Value;
}

I would like to serialize it so that Value ends up as an attribute of a subelement named (for instance) “Text”.

Desired outcome:

<MyClass>
    <Text Value="3"/>
</MyClass>

But NOT (which would be the effect of marking Value as an XmlAttribute)

<MyClass Value="3">
</MyClass>

And NOT (which would be the effect of marking Value as an XmlElement):

<MyClass>
    <Value>3</Value>
</MyClass>

How do I achieve this?

I am aware that I could change the type of Value from string to another serializable custom class.

Unfortunately, I have plenty of such properties, so I’d need to create dozens of tiny classes.

Is there any quicker solution?

---

EDIT:

In response to your comments:

• No, not every property has to be serialized to a subelement named “Text”. Subelement’s name is unique and unambiguous.

• Sample output XML:

  <visibility>
      <site visible="yes"/>
      <comparator visible="no"/>
      <expiration days="7"/>
      <comment>blahblahblah</comment>
  <visibility>
  

• Sample class:

!

[XmlRoot(ElementName="Visibility")]
public class Visibility
{
    [XPath("/site@visible")] // if only this was possible!
    public string OnSite
    {
        get { return SiteVisible ? "yes" : "no"; }
    }

    [XPath("/comparator@visible")] // as above...
    public string InComparator
    {
        get { return ComparatorVisible ? "yes" : "no"; }
    }

    [XmlIgnore]
    public bool SiteVisible;
    [XmlIgnore]
    public bool ComparatorVisible;

    [XPath("/expiration@days")] // as above...
    public int ExpiresAfterDays; 

    [XmlElement("comment")] // this is easy
    public string Comment;
}

Answer 1

Without changing the type of Value I think it’s not possible. You can add the attribute XmlElement(ElementName="Text") on Value but you will obtain a result similar to this:

<MyClass> 
    <Text>3</Text> 
</MyClass> 

Edited:
Another solution can involve XSLT transformation: you can generate the xml using .Net serialization and after apply a xml transformation.

XslTransform myXslTransform = new XslTransform();
myXslTransform.Load(xsltDoc);
myXslTransform.Transform(sourceDoc, resultDoc);

The trasformation of my example should be something like this:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:template match="/">
    <root>
        <xsl:apply-templates/>
    </root>
    </xsl:template>
    <xsl:template match="MyClass">
        <MyClass>
            <Text>
               <xsl:attribute name="Value">
                    <xsl:value-of select="Text"/>
               </xsl:attribute>
            </Text>
        </MyClass>
    </xsl:template>
</xsl:stylesheet>