xml creation in C#

Question

I want final xml as follows...

<Programs>
  <Program><id>4</id><isRead>true</isRead><isWrite>false</isWrite></Program>
  <Program><id>8</id><isRead>true</isRead><isWrite>true</isWrite></Program>
</programs>

now following is the code written

XmlDocument xmlDoc = new XmlDocument();
XmlNode rootNode = xmlDoc.CreateNode(XmlNodeType.Element,"programs",null);
xmlDoc.AppendChild(rootNode);


foreach (dynamic item in access)
{
  XmlNode myXmlNode = JsonConvert.DeserializeXmlNode(item.ToString(), "program");
  rootNode.AppendChild(myXmlNode); //error
}

where in myXmlNode.InnerXml, I am already getting following

<Program><id>4</id><isRead>true</isRead><isWrite>false</isWrite></Program>

And so, running the loop for all children to add in which is parent. But I am getting error for line //error as marked above. Error is:

The specified node cannot be inserted as the valid child of this node, because the specified node is the wrong type.

for the pretty printing part (multi-line and indentation) see https://stackoverflow.com/a/1123731/1132334 — Cee McSharpface, Jan 25 '18 at 12:22
`JsonConvert.DeserializeXmlNode` is interesting choice. Have you been [here](https://stackoverflow.com/q/11492705/1997232)? — Sinatr, Jan 25 '18 at 12:22
@ghetal if you're looking to deserialize you can do this using the inverse of what I suggested in my answer below — ianrathbone, Jan 26 '18 at 10:49

lucky · Accepted Answer · 2018-01-25T17:23:53.737

2

You are trying to insert different type of xml node. You could use ImportNode to apply it.

foreach (dynamic item in access)
{
    XmlNode myXmlNode = JsonConvert.DeserializeXmlNode(item.ToString(), "program");
    rootNode.AppendChild(rootNode.OwnerDocument.ImportNode(myXmlNode.FirstChild,true));
}

edited Jan 25 '18 at 17:23

answered Jan 25 '18 at 12:21

lucky

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still gives error: Cannot import nodes of type 'Document'. – ghetal Jan 25 '18 at 15:40
You can't import a document type as node. I updated my answer. – lucky Jan 25 '18 at 17:24
1

thanks.. with minimal changes I could achieve what I wanted – ghetal Jan 26 '18 at 11:26

ianrathbone · Answer 2 · 2018-01-25T12:46:57.007

I would recommend you create a model for what you are looking to create in XML, then serialize a list of the model to XML.

The reason being that you will end up with something far more maintainable.

Your model would be something like this:

public class Program
{
    public int Id { get; set; }
    public bool IsRead { get; set; }
    public bool IsWrite { get; set; }
}

And you can serialize it by following this article:

https://support.microsoft.com/en-gb/help/815813/how-to-serialize-an-object-to-xml-by-using-visual-c

Oystein · Answer 3 · 2018-01-25T12:45:31.040

I've used the XmlSerializer library myself a few times.

public class Program
{
    public int Id { get; set; }
    public bool IsRead { get; set; }
    public bool IsWrite { get; set; }
}

public class XmlTemplate
{
    public List<Program> Programs { get; set; }
}

class MainProgram
{
    static void Main(string[] args)
    {
        var xmlTemplate = new XmlTemplate();
        xmlTemplate.Programs = new List<Program>();
        xmlTemplate.Programs.Add(new Program() { Id = 123, IsRead = true, IsWrite = false });
        xmlTemplate.Programs.Add(new Program() { Id = 456, IsRead = false, IsWrite = true });

        var serializer = new XmlSerializer(typeof(XmlTemplate));
        var stream = new StringWriter();

        serializer.Serialize(stream, xmlTemplate);

        Console.WriteLine(stream.ToString());
        Console.ReadKey();

    }
}

This outputs the XML in the following format:

<XmlTemplate xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <Programs>
    <Program>
      <Id>123</Id>
      <IsRead>true</IsRead>
      <IsWrite>false</IsWrite>
    </Program>
    <Program>
      <Id>456</Id>
      <IsRead>false</IsRead>
      <IsWrite>true</IsWrite>
    </Program>
  </Programs>
</XmlTemplate>

have reached nearer to solution, so want to see if this way can it be achieved or not — ghetal, Jan 25 '18 at 15:43
It can. By using XmlSerializer and a template class, you can both serialize from object to XML and deserialize from XML to object. — Oystein, Jan 25 '18 at 15:55

xml creation in C#

3 Answers3