xml element list to array of common base type

Question

I have an xml file with a list like this:

<TestFile>
  <string>Foo</string>
  <bool>false</bool>
  <bool>true</bool>
  <string>Bar</string>
</TestFile>

I want to deserialize it to an Array of type "Value". That type has two subtypes "ValueString" and "ValueBool":

[XmlRoot("TestFile")]
public class TestFile
{
    public List<Test> Tests;
}

public class Value
{
}

public class ValueString : Value
{
    [XmlText]
    public string Value;
}

public class ValueBool : Value
{
    [XmlText]
    public bool Value;
}

I can't figure out how to do this. I've tried XmlIncludeAttributes but it's not enough because the element names do not match the class names. I've tried XmlChoiceIdentifier but the examples I found don't compile when I adapt them...

I need to preserve the order of the elements so separating the elements into two lists won't work. I also can't change the xml structure because it comes from an external source. Obviously it's just an example - my real types are more complex...

Personally I'd recommend writing your own xml parsing code for this as xml serialization really isn't intended as a way to read xml. — juharr
– juharr, Commented Apr 11, 2011 at 13:29

Kempeth · Accepted Answer · 2011-04-11 13:35:25Z

2

I finally stumbled over the solution:

[XmlRoot("TestFile")]
public class TestFile
{
    [XmlElement(ElementName = "string", Type = typeof(ValueString))]
    [XmlElement(ElementName = "bool", Type = typeof(ValueBool))]
    public List<Test> Tests;
}

I thought I tried that before... Well, at least it works now. There was another issue with this example: You can't have a boolean field mapped to an element text, but that specific issue isn't present in my true scenario anyway...

answered Apr 11, 2011 at 13:35

Kempeth

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Comments

Bryan Allred · Accepted Answer · 2011-04-11 13:53:30Z

Often we find ourselves having to transform the XML file given to us. Yes, it would be great if everyone subscribed to the same structure, but in a mid- to large-sized company this can be more difficult to achieve.

I started with your XML file:

<?xml version="1.0" encoding="utf-8" ?>
<TestFile>
  <string>Foo</string>
  <bool>false</bool>
  <bool>true</bool>
  <string>Bar</string>
</TestFile>

Then created the transform file (this is just an example and is all up to your own preference):

<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:msxsl="urn:schemas-microsoft-com:xslt" exclude-result-prefixes="msxsl"
>
  <xsl:output method="xml" indent="yes"/>

  <xsl:template match="@* | node()">
      <xsl:copy>
        <ParameterCollection>
          <xsl:apply-templates select="@* | node()"/>
        </ParameterCollection>
      </xsl:copy>
  </xsl:template>
  <xsl:template match="bool">
    <Parameter type="bool">
      <xsl:apply-templates select="node()"/>
    </Parameter>
  </xsl:template>
  <xsl:template match="string">
    <Parameter type="string">
      <xsl:apply-templates select="node()"/>
    </Parameter>
  </xsl:template>
</xsl:stylesheet>

Then I started piecing all the necessary classes together:

[XmlRootAttribute("TestFile", IsNullable = false)]
public class TestFile
{
    [XmlArrayAttribute("ParameterCollection")]
    public Parameter[] Parameters;
}

public class Parameter
{
    [XmlAttribute("type")]
    public string ObjectType;

    [XmlText]
    public string ObjectValue;
}

Then apply everything (hopefully in a more thoughtful manner than I have done):

class Program
{
    static void Main(string[] args)
    {
        FileInfo xmlFile = new FileInfo(@"Resources\TestFile.xml");
        FileInfo transformFile = new FileInfo(@"Resources\TestFileTransform.xslt");
        FileInfo prettyFile = new FileInfo(@"Resources\PrettyFile.xml");

        if (xmlFile.Exists && transformFile.Exists)
        {
            // Perform transform operations.
            XslCompiledTransform trans = new XslCompiledTransform();
            trans.Load(transformFile.FullName);
            trans.Transform(xmlFile.FullName, prettyFile.FullName);
        }

        if (prettyFile.Exists)
        {
            // Deserialize the new information.
            XmlSerializer serializer = new XmlSerializer(typeof(TestFile));
            XDocument doc = XDocument.Load(prettyFile.FullName);
            TestFile o = (TestFile)serializer.Deserialize(doc.CreateReader());

            // Show the results.
            foreach (Parameter p in o.Parameters)
            {
                Console.WriteLine("{0}: {1}", p.ObjectType, p.ObjectValue);
            }
        }

        // Pause for effect.
        Console.ReadKey();
    }
}

Hopefully this helps someone, or at least gives them another option. Typically, IMHO I would prefer to parse the file or stream, but that is just me.

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