xmlserialize argumentexception (ArgumentException: Could not cast or convert from System.String to S-CodePudding

Below piece of code is failing and throwing ArgumentException

  static void Main(string[] args)
    {
        string xml = "<root><SourcePatient><Communication>HP:6055550120</Communication></SourcePatient></root>";

        XmlDocument doc = new XmlDocument();

        doc.LoadXml(xml);


        var serializedString = JsonConvert.SerializeXmlNode(doc, Newtonsoft.Json.Formatting.None,true);

        var deserialise = serializedString.ToObject<SampleModel>();

    }

Models are,

public class SampleModel
{
    public SourcePatientModel SourcePatient { get; set; }

}

public class SourcePatientModel
{
    public List<string> Communication { get; set; }


}

How to deserialize this? Sometimes Communication node from xml string will have multiple entries

CodePudding user response：

Your current xml is only a single entry

<Communication>HP:6055550120</Communication>

Change your xml input

<Communication><Entry>HP:6055550120</Entry></Communication>

So later when you get multiple entries, they can be processed

<Communication><Entry>HP:6055550120</Entry><Entry>HP:xxxxxxxxx</Entry></Communication>

Your class needs tweaked a bit

if a string [] is acceptable

[XmlArray(ElementName = "Communication")]
[XmlArrayItem(ElementName = "Entry")]
public string[] comm_list // Whatever name you want here
{
   get; set;
}
// if you want a list here
// also if your going to do this, realize it creates a new list every time you use it, not the best. (bad practice)
List<string> Communication
{
   get => new List<string>(comm_list );
}

otherwise it gets a little complicated

[XmlRoot(ElementName="Communication")]
public class Communication // element name by def
{
  [XmlElement(ElementName="Entry")]
  public List<string> entry { get; set; }
}

Another possibility, not sure how your multiple entries come in. If multiple entries look like the following

<Communication>HP:6055550120, HP:7055550120</Communication>

Then you cant have a direct list,

public class SourcePatientModel
{
    public string Communication { get; set; }
    // which again this creates a list everytime, its better to change your xml to match a tag name for each entry
    [XmlIgnore]
    public List<string> CommunicationValues { get => Communication.Split(',').ToList();

}

Also this is just typed up code, there may be some typos or compile errors

CodePudding user response：

First I parsed xml to the valid model, then you can convert to json if needed

using (TextReader sr = new StringReader(xml))
{
    XmlSerializer serializer = new XmlSerializer(typeof(SampleModel));
    var schema = (SampleModel)serializer.Deserialize(sr);
}

[Serializable, XmlRoot("root")]
public class SampleModel
{
    public SourcePatientModel SourcePatient { get; set; }
}

public class SourcePatientModel
{
    [XmlElement("Communication")]
    public List<string> Communication { get; set; }
}

CodePudding user response：

I modified your classes to include some XML serialization attributes. (More on how to figure those out later - the short version is that you don't have to figure it out.)

[XmlRoot(ElementName = "SourcePatient")]
public class SourcePatientModel
{
    [XmlElement(ElementName = "Communication")]
    public List<string> Communication { get; set; }
}

[XmlRoot(ElementName = "root")]
public class SampleModel
{
    [XmlElement(ElementName = "SourcePatient")]
    public SourcePatientModel SourcePatientModel { get; set; }
}

...and this code deserializes it:

var serializer = new XmlSerializer(typeof(SampleModel));
using var stringReader = new StringReader(xmlString);
SampleModel deserialized = (SampleModel) serializer.Deserialize(stringReader);

Here's a unit test to make sure a test XML string is deserialized with a list of strings as expected. A unit test is a little easier to run and repeat than a console app. Using them makes writing code a lot easier.

[Test]
public void DeserializationTest()
{
    string xmlString = @"
<root>
    <SourcePatient>
        <Communication>A</Communication>
        <Communication>B</Communication>
    </SourcePatient>
</root>";

    var serializer = new XmlSerializer(typeof(SampleModel));

    using var stringReader = new StringReader(xmlString);
    SampleModel deserialized = (SampleModel) serializer.Deserialize(stringReader);
    Assert.AreEqual(2, deserialized.SourcePatientModel.Communication.Count);
}

Here's a key takeaway: You don't need to memorize XML serialization attributes. I don't bother because I find them confusing. Instead, google "XML to Csharp" and you'll find sites like this one. I pasted your XML into that site and let it generate the classes for me. (Then I renamed them so that they matched your question.)

But be sure that you include enough sample data in your XML so that it can generate the classes for you. I made sure there were two Communication elements so it would create a List<string> in the generated class.

Sites like that might not work for extremely complicated XML, but they work for most scenarios, and it's much easier than figuring out how to write the classes ourselves.