Reputation: 541
I am serialising a List of US States into XML, and while I can control the names of most of the output elements with attributes, the root node is always called "ArrayOfStates". Is there a way to change this so it is just "States"?
Code:
public class Program
{
[XmlArray("States")]
public static List<State> States;
public static void Main(string[] args)
{
PopulateListOfStates();
var xml = new XmlSerializer(typeof(List<State>));
xml.Serialize(new XmlTextWriter(@"C:\output.xml",Encoding.Default), States);
}
}
public struct State
{
[XmlAttribute]
public string Name;
[XmlArray("Neighbours")]
[XmlArrayItem("Neighbour")]
public List<string> Neighbours;
}
Output:
<?xml version="1.0" encoding="Windows-1252"?>
<ArrayOfState xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<State Name="AL">
<Neighbours>
<Neighbour>FL</Neighbour>
<Neighbour>GA</Neighbour>
<Neighbour>MS</Neighbour>
<Neighbour>TN</Neighbour>
</Neighbours>
</State>
<State Name="FL">
<Neighbours>
<Neighbour>AL</Neighbour>
<Neighbour>GA</Neighbour>
</Neighbours>
</State>
<State Name="GA">
<Neighbours>
<Neighbour>AL</Neighbour>
<Neighbour>FL</Neighbour>
<Neighbour>NC</Neighbour>
<Neighbour>SC</Neighbour>
<Neighbour>TN</Neighbour>
</Neighbours>
</State>
...
</ArrayOfState>
As an aside, is it also possible to have the contents of the "Neighbour" elements as attributes of those elements (i.e. <Neighbour name="XX"/>
)?
Upvotes: 2
Views: 563
Reputation: 123
[Serializable]
public class Worksheet
{
[XmlRoot(ElementName = "XML")]
public class XML
{
[XmlArray("States")]
public List<State> States { get; set; }
}
public class State
{
[XmlAttribute]
public string Name { get; set; }
[XmlArray("Neighbours")]
[XmlArrayItem("Neighbour")]
public List<Neighbour> Neighbours { get; set; }
}
public class Neighbour
{
[XmlAttribute]
public string Name { get; set; }
}
}
public static void Main(string[] args)
{
Worksheet.XML xml = PopulateListOfStates();
XmlSerializer serializer = new XmlSerializer(typeof(Worksheet.XML));
using (StreamWriter writer = new StreamWriter(@"C:\output.xml", false))
{
serializer.Serialize(writer, xml);
}
}
Upvotes: 2