Serializing a class under a different element name

Question

I have this class:

[XmlRoot(ElementName ="Lesson")]
public class LessonOld
{
    public LessonOld()
    {
        Students = new List<string>();
    }
    public string Name { get; set; }
    public DateTime FirstLessonDate { get; set; }
    public int DurationInMinutes { get; set; }
    public List<string> Students { get; set; }
}

And I'm using this code to serialize it:

TextWriter writer = new StreamWriter(Path.Combine(UserSettings, "Lessons-temp.xml"));
XmlSerializer xmlSerializer = new XmlSerializer(typeof(List<LessonOld>));
xmlSerializer.Serialize(writer, tempList);
writer.Close();

(Note that that is a List<LessonOld>)

Here is my resulting XML:

<?xml version="1.0" encoding="utf-8"?>
<ArrayOfLessonOld xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <LessonOld>
    <FirstLessonDate>0001-01-01T00:00:00</FirstLessonDate>
    <DurationInMinutes>0</DurationInMinutes>
    <Students />
  </LessonOld>
</ArrayOfLessonOld>

I would like to change it to serialize as <ArrayOfLesson and <Lesson> for the XML elements. Is this possible? (As you can see, I've already tried using [XmlRoot(ElementName ="Lesson")])

Possible duplicate of [How can I rename class-names via Xml attributes?](https://stackoverflow.com/questions/3653411/how-can-i-rename-class-names-via-xml-attributes) — maccettura, Apr 24 '18 at 20:14

Francesco B. · Accepted Answer · 2018-04-24T20:33:01.720

You're almost there. Use:

[XmlType(TypeName = "Lesson")]

instead of

[XmlRoot(ElementName = "Lesson")]

Of course you can test it easily; your code with the above change

[XmlType(TypeName = "Lesson")]
public class LessonOld
{
    public LessonOld()
    {
        Students = new List<string>();
    }
    public string Name { get; set; }
    public DateTime FirstLessonDate { get; set; }
    public int DurationInMinutes { get; set; }
    public List<string> Students { get; set; }
}

class Program
{
    static void Main(string[] args)
    {
        TextWriter writer = new StreamWriter(Path.Combine(@"C:\Users\Francesco\Desktop\nanovg", "Lessons-temp.xml"));
        XmlSerializer xmlSerializer = new XmlSerializer(typeof(List<LessonOld>));
        xmlSerializer.Serialize(writer, new List<LessonOld> { new LessonOld() { Name = "name", DurationInMinutes = 0 } });
        writer.Close();
    }
}

produces this

<?xml version="1.0" encoding="utf-8"?>
<ArrayOfLesson xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <Lesson>
    <Name>name</Name>
    <FirstLessonDate>0001-01-01T00:00:00</FirstLessonDate>
    <DurationInMinutes>0</DurationInMinutes>
    <Students />
  </Lesson>
</ArrayOfLesson>

As I've seen it, XmlRoot works fine when you want to serialize a single object. Consider this code, derived from yours:

[XmlRoot(ElementName = "Lesson")]
public class LessonOld
{
    public LessonOld()
    {
        Students = new List<string>();
    }
    public string Name { get; set; }
    public DateTime FirstLessonDate { get; set; }
    public int DurationInMinutes { get; set; }
    public List<string> Students { get; set; }
}

class Program
{
    static void Main(string[] args)
    {
        TextWriter writer = new StreamWriter(Path.Combine(@"C:\Users\Francesco\Desktop\nanovg", "Lessons-temp.xml"));
        XmlSerializer xmlSerializer = new XmlSerializer(typeof(LessonOld));
        xmlSerializer.Serialize(writer, new LessonOld() { Name = "name", DurationInMinutes = 0  });
        writer.Close();
    }
}

It will output

<?xml version="1.0" encoding="utf-8"?>
<Lesson xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <Name>name</Name>
  <FirstLessonDate>0001-01-01T00:00:00</FirstLessonDate>
  <DurationInMinutes>0</DurationInMinutes>
  <Students />
</Lesson>

@maccettura thanks for the edit:) it looks much better – Francesco B. Apr 24 '18 at 20:14 — Francesco B., Apr 24 '18 at 20:14
Ah, `XmlType` not `XmlRoot`. Thanks! – derekantrican Apr 24 '18 at 20:25 — derekantrican, Apr 24 '18 at 20:25
@derekantrican I added an example of use of `XmlRoot` – Francesco B. Apr 24 '18 at 20:33 — Francesco B., Apr 24 '18 at 20:33

Serializing a class under a different element name

1 Answers1