C# Changing the element names of items in a list when serializing/deserializing XML

Question

I have a class defined as below:

[XmlRoot("ClassName")]
public class ClassName_0
{
    //stuff...
}

I then create a list of ClassName_0 like such:

var myListInstance= new List<ClassName_0>();

This is the code I use to serialize:

var ser = new XmlSerializer(typeof(List<ClassName_0>));
ser.Serialize(aWriterStream, myListInstance);

This is the code I use to deserialize:

var ser = new XmlSerializer(typeof(List<ClassName_0>));
var wrapper = ser.Deserialize(new StringReader(xml));

If I serialize it to xml, the resulting xml looks like below:

<?xml version="1.0" encoding="utf-8"?>
<ArrayOfClassName_0 xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
    <ClassName_0>
        <stuff></stuff>
    </ClassName_0>
    <ClassName_0>
        <stuff></stuff>
    </ClassName_0>
</ArrayOfClassName_0>

Is there a way to serialize and be able to deserialize the below from/to a list of ClassName_0?

<?xml version="1.0" encoding="utf-8"?>
<ArrayOfClassName xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
    <ClassName>
        <stuff></stuff>
    </ClassName>
    <ClassName>
        <stuff></stuff>
    </ClassName>
</ArrayOfClassName>

Thanks!

I think this link might be useful for you, http://stackoverflow.com/questions/364253/how-to-deserialize-xml-document — sagar, Jul 19 '13 at 08:11
Thanks for the link. Unfortunately, this is generated code, so I cannot create a custom list of the desired type and add all the element attributes to that. I can only add attributes to the classes themselves. — VARAK, Jul 19 '13 at 08:12

score 2 · Answer 1 · answered Jul 19 '13 at 08:13

2

In your example ClassName isn't the real root. The real root is your list. So you have to mark the list as the root element. Your class is just an XmlElement.

answered Jul 19 '13 at 08:13

Jan Peter

912
7
19

So XMLRoot has no effect in this case? – VARAK Jul 19 '13 at 08:14
Worked it out. Thanks for the help :) – VARAK Jul 19 '13 at 08:17
No problem :) But maybe it helps other ones if you could update your post with the working code :) – Jan Peter Jul 19 '13 at 08:51

score 1 · Answer 2 · answered Jul 19 '13 at 08:18

1

try this :

XmlType(TypeName="ClassName")]
public class ClassName_0
{
    //stuff...
}

answered Jul 19 '13 at 08:18

Ashok Damani

3,896
4
30
48

score 0 · Accepted Answer · answered Jul 19 '13 at 08:16

0

Worked it out, finally, with the help of Jan Peter. XmlRoot was the wrong attribute to put on the class. It was supposed to be XmlType. With XmlType the desired effect is achieved.

answered Jul 19 '13 at 08:16

VARAK

835
10
27

score 0 · Answer 4 · answered Sep 16 '16 at 06:51

You make a root of document tree and this root will contain list of any object.

[XmlRootAttribute("myDocument")]
public class myDocument
{
   [XmlArrayAttribute]
   publict ClassName[] ArrayOfClassName {get;set;}
}

[XmlType(TypeName="ClassName")]
public class ClassName 
{
   public string stuff {get;set;}
}

C# Changing the element names of items in a list when serializing/deserializing XML

4 Answers4