Serializing Lists of Classes to XML
Just to check, have you marked Bar as [Serializable]?
Also, you need a parameter-less ctor on Bar, to deserialize
Hmm, I used:
public partial class Form1 : Form
{
public Form1()
{
InitializeComponent();
}
private void button1_Click(object sender, EventArgs e)
{
Foo f = new Foo();
f.BarList = new List<Bar>();
f.BarList.Add(new Bar { Property1 = "abc", Property2 = "def" });
XmlSerializer ser = new XmlSerializer(typeof(Foo));
using (FileStream fs = new FileStream(@"c:\sertest.xml", FileMode.Create))
{
ser.Serialize(fs, f);
}
}
}
public class Foo
{
[XmlArray("BarList"), XmlArrayItem(typeof(Bar), ElementName = "Bar")]
public List<Bar> BarList { get; set; }
}
[XmlRoot("Foo")]
public class Bar
{
public string Property1 { get; set; }
public string Property2 { get; set; }
}
And that produced:
<?xml version="1.0"?>
<Foo xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<BarList>
<Bar>
<Property1>abc</Property1>
<Property2>def</Property2>
</Bar>
</BarList>
</Foo>
Everything looks great. As @Carl said you need to add the [Serializable] attibute to your classes, but other than that your XML creation should work find.
Foo
[Serializable]
[XmlRoot("Foo")]
public class Foo
{
[XmlArray("BarList"), XmlArrayItem(typeof(Bar), ElementName = "Bar")]
public List<Bar> BarList { get; set; }
}
Bar
[Serializable]
public class Bar
{
public string Property1 { get; set; }
public string Property2 { get; set; }
}
Code to test
Foo f = new Foo();
f.BarList = new List<Bar>();
f.BarList.Add(new Bar() { Property1 = "s", Property2 = "2" });
f.BarList.Add(new Bar() { Property1 = "s", Property2 = "2" });
FileStream fs = new FileStream("c:\\test.xml", FileMode.OpenOrCreate);
System.Xml.Serialization.XmlSerializer s = new System.Xml.Serialization.XmlSerializer(typeof(Foo));
s.Serialize(fs, f);
Output
<?xml version="1.0" ?>
<Foo xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<BarList>
<Bar>
<Property1>s</Property1>
<Property2>2</Property2>
</Bar>
<Bar>
<Property1>s</Property1>
<Property2>2</Property2>
</Bar>
</BarList>
</Foo>
var xmlfromLINQ = new XElement("BarList",
from c in BarList
select new XElement("Bar",
new XElement("Property1", c.Property1),
new XElement("Property2", c.Property2)
));