XML serialization and "<" , ">" when generating an HTML document

Question

I'm trying to dynamically generate an HTML table by using XmlSerializer and a set of classes like this:

[XmlInclude(typeof(Th))]
public class Td
{
    [XmlElement("span")]
    public string Designation { get; set; }

    [XmlAttribute("colspan")]
    public int ColSpan { get; set; }

    [XmlAttribute("rowspan")]
    public int RowSpan { get; set; }

    public Td(string designation, int colspan)
    {
        Designation = designation;
        ColSpan = colspan;
        RowSpan = 1;
    }

    public Td()
    {            
    }

}

The problem here is that the Designation property can have a tag as value like <option...>, so when I serialize my model I get <option...> instead of <option...>

I can solve the problem by using the string.Replace method like this: Replace("<", "<").Replace(">", ">");

Is there a clean way to get the expected result without the use of string.Replace?

XML != HTML. Most XHTML work has been discontinued, afaik.

Marc Gravell
– Marc Gravell

2012-06-06 14:38:40 +00:00
Commented Jun 6, 2012 at 14:38 — Marc Gravell
– Marc Gravell, Commented Jun 6, 2012 at 14:38

Michael Liu · Accepted Answer · 2012-06-06 14:43:06Z

2

You can create another property that exposes Designation as an XmlNode:

[XmlIgnore]
public string Designation { get; set; }

[XmlElement("span")]
public XmlNode DesignationAsXml
{
    get
    {
        XmlDocument doc = new XmlDocument();
        doc.InnerXml = "<root>" + this.Designation + "</root>";
        return doc.DocumentElement.FirstChild;
    }
    set
    {
        throw new NotSupportedException();
    }
}

answered Jun 6, 2012 at 14:43

Michael Liu

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2 Comments

Joel Peltonen Over a year ago

How would Deserialization work here? A serialized object will not deserialize with the same information here.

Michael Liu Over a year ago

@Nenotlep: The OP's scenario didn't require deserialization so I didn't bother implementing the property setter. To do so, simply assign value.OuterXml to this.Designation instead of throwing an exception.

Johnny_D · Accepted Answer · 2012-06-06 14:38:39Z

0

Replacing it manually is needless. I think the best way is using System.XML.XmlWriter.WriteString class. Look for class definition here

answered Jun 6, 2012 at 14:38

Johnny_D

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