How can I rename class-names via Xml attributes?
Solution: Use [XmlType(TypeName="g")]
XmlRoot only works with XML root nodes as per the documentation (and what you would expect, given its name includes root)!
I was unable to get any of the other answers to work so kept digging...
Instead I found that the XmlTypeAttribute (i.e. [XmlType]
) and its TypeName property do a similar job for non-root classes/objects.
e.g.
[XmlType(TypeName="g")]class Song{ public string Artist; public string SongTitle;}
Assuming you apply it to the other classes e.g.:
[XmlType(TypeName="a")]class Artist{ .....}[XmlType(TypeName="s")]class SongTitle{ .....}
This will output the following exactly as required in the question:
<g> <a>Britney Spears</a> <s>I Did It Again</s></g>
I have used this in several production projects and found no problems with it.
Checkout the XmlRoot attribute.
Documentation can be found here:http://msdn.microsoft.com/en-us/library/system.xml.serialization.xmlrootattribute(v=VS.90).aspx
[XmlRoot(Namespace = "www.contoso.com", ElementName = "MyGroupName", DataType = "string", IsNullable=true)]public class Group
UPDATE:Just tried and it works perfectly on VS 2008.This code:
[XmlRoot(ElementName = "sgr")]public class SongGroup{ public SongGroup() { this.Songs = new List<Song>(); }[XmlElement(ElementName = "sgs")] public List<Song> Songs { get; set; }}[XmlRoot(ElementName = "g")]public class Song{ [XmlElement("a")] public string Artist { get; set; } [XmlElement("s")] public string SongTitle { get; set; }}
Outputs:
<?xml version="1.0" encoding="utf-8"?><sgr xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema"> <sgs> <a>A1</a> <s>S1</s> </sgs> <sgs> <a>A2</a> <s>S2</s> </sgs></sgr>
If this is the root element of the document, you can use [XmlRoot("g")].
Here is my updated response based on your clarification. The degree of control you are asking for is not possible without a wrapping class. This example uses a SongGroup
class to wrap the list so that you can give alternate names to the items within.
using System;using System.Collections.Generic;using System.IO;using System.Xml.Serialization;public class SongGroup{ public SongGroup() { this.Songs = new List<Song>(); } [XmlArrayItem("g", typeof(Song))] public List<Song> Songs { get; set; }}public class Song { public Song() { } [XmlElement("a")] public string Artist { get; set; } [XmlElement("s")] public string SongTitle { get; set; }} internal class Test{ private static void Main() { XmlSerializer serializer = new XmlSerializer(typeof(SongGroup)); SongGroup group = new SongGroup(); group.Songs.Add(new Song() { Artist = "A1", SongTitle = "S1" }); group.Songs.Add(new Song() { Artist = "A2", SongTitle = "S2" }); using (Stream stream = new MemoryStream()) using (StreamWriter writer = new StreamWriter(stream)) { serializer.Serialize(writer, group); stream.Seek(0, SeekOrigin.Begin); using (StreamReader reader = new StreamReader(stream)) { Console.WriteLine(reader.ReadToEnd()); } } }}
This has the side effect of generating one more inner element representing the list itself. On my system, the output looks like this:
<?xml version="1.0" encoding="utf-8"?><SongGroup xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema"> <Songs> <g> <a>A1</a> <s>S1</s> </g> <g> <a>A2</a> <s>S2</s> </g> </Songs></SongGroup>