How to get the content of an XML element using XmlSerializer?

Question

I have an XML reader in this XML line:

<?xml version="1.0" encoding="UTF-8" ?>
<story id="1224488641nL21535800" date="20 Oct 2008" time="07:44">
<title>PRESS DIGEST - PORTUGAL - Oct 20</title>
<text>
<p>    LISBON, Oct 20 (Reuters) - Following are some of the main
 stories in Portuguese newspapers on Monday. Reuters has not
verified these stories and does not vouch for their accuracy. </p>
<p>More HTML stuff here</p>
</text>
</story>

I created an XSD and the corresponding class for deserialization.

[System.Xml.Serialization.XmlRootAttribute(Namespace="", IsNullable=false)]
public class story {
    [System.Xml.Serialization.XmlAttributeAttribute()]
    public string id;
    [System.Xml.Serialization.XmlAttributeAttribute()]
    public string date;
    [System.Xml.Serialization.XmlAttributeAttribute()]
    public string time;
    public string title;
    public string text;
}

Then I instantiate the class using the DeserializeXmlSerializer method .

XmlSerializer ser = new XmlSerializer(typeof(story));
return (story)ser.Deserialize(xr);

Now the member is text storyalways null. How to change the class storyso that XML is parsed as expected?

EDIT:

Using XmlText does not work, and I do not control the XML analysis.

Answer 1

I found a very unsatisfactory solution.

Change the class as follows (ugh!)

// ...
[XmlElement("HACK - this should never match anything")]
public string text;
// ...

And change the calling code as follows (yuck!)

XmlSerializer ser = new XmlSerializer(typeof(story));
string text = string.Empty;
ser.UnknownElement += delegate(object sender, XmlElementEventArgs e) {
    if (e.Element.Name != "text")
        throw new XmlException(
              string.Format(CultureInfo.InvariantCulture, 
             "Unknown element '{0}' cannot be deserialized.",
             e.Element.Name));
    text += e.Element.InnerXml;
};

story result = (story)ser.Deserialize(xr);
result.text = text;
return result;

This is a very bad way to do this because it destroys encapsulation. Is there a better way to do this?

Answer 2

, , - p-, : .

. XmlArray ( , XmlArrayItemAttribute), , :

<text>
   <p>blah</p>
   <p>blib</p>
</text>

, , .

- :

public class Text //Obviously a bad name for a class...
{
   public string[] p;
   public string[] pre;
}

XmlArray, , , , , .

Edit:

:

[System.Xml.Serialization.XmlRootAttribute(Namespace = "", IsNullable = false)]
    public class story
    {
        [System.Xml.Serialization.XmlAttributeAttribute()]
        public string id;
        [System.Xml.Serialization.XmlAttributeAttribute()]
        public string date;
        [System.Xml.Serialization.XmlAttributeAttribute()]
        public string time;
        public string title;

        [XmlArrayItem("p")]
        public string[] text;

    }

XML, . :

    <text>
       <p>
          <p>qwertyuiop</p>
          <p>asdfghjkl</p>
       </p>
       <pre>
          <pre>stuff</pre>
          <pre>nonsense</pre>
       </pre>
   </text>

, , .

Answer 3

IXmlSerializable , , ( , ). , .

Answer 4

, XML . HTML- , HTML XML. CDATA < > .

Answer 5

XML, StreamReader. XmlReader HTML XML, , .

XmlSerializer HTML .

Answer 6

, XmlAnyElement UnknownElement .

Answer 7

xsd.exe? xsd xml xsd, xml.

Answer 8

, , ...

Answer 9

I ran into the same problem after using XSD.exe to generate XSD from XML and then XSD for classes. I added the [XmlText] tag in front of the object class in the generated class file (in my case, it is called P by virtue of the tag <p>that it displayed as an XML node), and it worked instantly. pulling out the full HTML content that was inside the parent node and putting P into that object, which then renamed it something more useful.