Deserialize array elements to list

Question

I have the following XML...

<configuration>
    <img name="name1" />
    <img name="name2" />
    <warn>
        <img name="warn1" />
    </warn>
</configuration>

...which I try to deserialize into...

[XmlType("img")]
public class ImageNameExceptionItemXml
{
    [XmlAttribute("name")]
    public string Filename;
}

[XmlRoot("configuration")]
public class ImageNameExceptionListXml: List<ImageNameExceptionItemXml>
{
    [XmlArray("warn")]
    [XmlArrayItem("img")]
    public ImageNameExceptionListXml WarnList { get; set; }
}

...but I end up with the WarnList property null.

I already tried...

[XmlElement("warn"}]
public ImageNameExceptionListXml WarnList { get; set; }

...or...

[XmlElement("warn"}]
public List<ImageNameExceptionItemXml> WarnList { get; set; }

...but I still end up with WarnList property null. Why is that?

Not sure, but I don't think the Serializer likes deserializing the extra properties on what it think is an array. So this may be tricky without explicitly implementing IXmlSerializable. — Balthy
– Balthy, Commented Feb 16, 2012 at 11:51
@findcaiyzh I can only change the <warn> node. Maybe I can try <configuration> <img name="name1" /> <img name="name2" /> <warnImg name="warn1" /> </configuration> — sjlewis
– sjlewis, Commented Feb 16, 2012 at 11:55
@sjlewis the problem is XmlSerializer just allow one type in array/list.<configuration><imgs><img... /><img ... /></imgs><warn ... /> is fine. — findcaiyzh
– findcaiyzh, Commented Feb 16, 2012 at 12:00

Vinicius Ottoni · Accepted Answer · 2012-02-16 12:03:25Z

4

Let's go to the problem. The warn is a "root" element, so, you have to transform it in a class too:

The xml:

<?xml version="1.0" encoding="utf-8" ?>
<configuration>
  <img name="name1" />
  <img name="name2" />
  <warn>
      <img name="warn1" />
      <img name="warn2" />
  </warn>
</configuration>

The class:

[XmlType("img")]
public class ImageNameExceptionItemXml 
{ 
    [XmlAttribute("name")]     
    public string Filename; 
}

[XmlType("warn")]
public class WarnExceptionItemXml
{
    [XmlElement("img")]
    public List<ImageNameExceptionItemXml> ImgList { get; set; }
}

[XmlRoot("configuration")]
public class ImageNameExceptionListXml
{
    [XmlElement("img")]
    public List<ImageNameExceptionItemXml> ImgList { get; set; }

    [XmlElement("warn")]
    public WarnExceptionItemXml WarnList { get; set; } 
}

And the deserialize test:

XmlSerializer xml = new XmlSerializer(typeof(ImageNameExceptionListXml));

ImageNameExceptionListXml teste = (ImageNameExceptionListXml)xml.Deserialize(new FileStream("XMLFile1.xml", FileMode.Open));

answered Feb 16, 2012 at 12:03

Vinicius Ottoni

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3 Comments

svick Over a year ago

Do you have to repeat the same name both in [XmlElement] and [XmlType]?

Vinicius Ottoni Over a year ago

@svick nope, you can use only in [XmlElement], but is better to give the idea easier. And you using in a different class it's better to understand who class is connected to xml element.

Vinicius Ottoni Over a year ago

To better understand the difference: msdn.microsoft.com/en-US/library/… and msdn.microsoft.com/en-US/library/…

findcaiyzh · Accepted Answer · 2012-02-16 12:04:10Z

0

XmlSerializer just allow one type in array/list. is fine. Post My test code:

[XmlType("img")]
[Serializable]
public class ImageNameExceptionItemXml
{
    [XmlAttribute("name")]
    public string Filename;
}

[XmlType("warn")]
[Serializable]
public class Warnning
{
    [XmlArrayItem(typeof(ImageNameExceptionItemXml))]
    public List<ImageNameExceptionItemXml> imgs { get; set; }
}

[XmlRoot("configuration")]
[Serializable]
public class ImageNameExceptionListXml
{
    [XmlArrayItem(typeof(Warnning))]
    public List<Warnning> warns{ get; set; }

    [XmlArrayItem(typeof(ImageNameExceptionItemXml))]
    public List<ImageNameExceptionItemXml> imgs { get; set; }

}

xml:

<configuration>
<imgs>
    <img name="name1" />
    <img name="name2" />
</imgs>
<warns>
    <warn>
      <imgs>
        <img name="warn2" />
        <img name="warn1" />
      </imgs>
    </warn>
    <warn>
      <imgs>
        <img name="warn3" />
        <img name="warn4" />
      </imgs>
    </warn>
 </warns>
</configuration>

answered Feb 16, 2012 at 12:04

findcaiyzh

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1 Comment

findcaiyzh Over a year ago

Vinicius Ottoni's answer is better.

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