Convert from XML response to list

Question

XElement Xml = null;
var apiResponse = response.Content.ReadAsStringAsync().Result;
Xml = Newtonsoft.Json.JsonConvert.DeserializeObject<XElement>(apiResponse);

XML response from the above code:

I'm having errors with Images XML part while converting it to List, I tried so many options, please provide suggestion below:

<root>
  <Column1>
    <ID>2702</ID>
    <Desc>Failed</Desc>
    <Address>Florida</Address>
    <Date>2019-04-30T23:10:36.79</Date>
    <Images>
      <Image>
        <File>1-RRamos.PNG</File>
      </Image>
      <Image>
        <File>RRamos.PNG</File>
      </Image>
      <Image>
        <File>3-RRamos.PNG</File>
      </Image>
    </Images>
  </Column1>
</root>

Trying to convert from XML to List as below:

public class objClass
{
    public string ID { get; set; }
    public string Desc { get; set; }
    public string Address { get; set; }
    public DateTime? Date { get; set; }

    //public string[] ImageFileNames { get; set; }

    public List<Images> FileName { get; set; }
}

public class Images
{
    public string File { get; set; }
}

List<objClass> list = Xml.Elements("ID").Select(sv => new objClass()
{
    ID = (string)sv.Element("ID"),
    Desc = (string)sv.Element("Desc"),
    Address = (string)sv.Element("Address"),
    Date = (DateTime?)sv.Element("Date"),
    //**,Images = (List)sv.Element("Images")**
}).ToList();

Yong Shun · Answer 1 · 2023-02-20T06:48:49.250

You can't use the Newtonsoft.Json library to deserialize from an XML.

Solution 1: Convert XML to list via XPath

Parse XML string to XDocument.
With XPath: "//root/Column1", select the <Column1> element.

using System.Xml;
using System.Xml.Linq;
using System.Xml.Serialization;
using System.Xml.XPath;

var apiResponse = await response.Content.ReadAsStringAsync();
XDocument @Xml = XDocument.Parse(apiResponse);
        
List<ObjClass> list = @Xml.XPathSelectElements("//root/Column1")
    .Select(sv => new ObjClass()
            {
                ID = (string)sv.Element("ID"),
                Desc = (string)sv.Element("Desc"),
                Address = (string)sv.Element("Address"),
                Date = (DateTime?)sv.Element("Date"),
                Images = sv.Element("Images")
                    .Elements("Image")
                    .Select(x => new Image
                            {
                                File = (string)x.Element("File")
                            })
                    .ToList()   
            })
    .ToList();

Solution 2: Deserialize XML

This answer will be a bit complex but work the same as Solution 1.

Write the apiResponse value into MemoryStream.
Deserialize the MemoryStream via XmlSerializer as Root.
Extract root.Column and add into list.

[XmlRoot(ElementName = "root")]
public class Root
{
    [XmlElement("Column 1")]
    public ObjClass Column { get; set; }
}

public class ObjClass
{
    public string ID { get; set; }

    public string Desc { get; set; }

    public string Address { get; set; }

    public DateTime? Date { get; set; }

    [XmlArray]
    [XmlArrayItem(typeof(Image), ElementName = "Image")]
    public List<Image> Images { get; set; }
}

public class Image
{
    public string File { get; set; }
}

using System.Xml;
using System.Xml.Serialization;
using System.IO;

var apiResponse = await response.Content.ReadAsStringAsync();

using var stream = new MemoryStream();
using var writer = new StreamWriter(stream);
writer.Write(apiResponse);
writer.Flush();
stream.Position = 0;
        
XmlSerializer xmlSerializer = new XmlSerializer(typeof(Root));      
        
Root root = (Root)xmlSerializer.Deserialize(stream);
List<ObjClass> list = new List<ObjClass>();
list.Add(root.Column);

Concern:

Use await instead of Task<T>.Result as Task<T>.Result will block the calling thread until it (the task) is completed. With await, the task is waited to be completed asynchronously. Reference: What is the difference between await Task and Task.Result?

[Demo Solution 1 @ .NET Fiddle](https://dotnetfiddle.net/41xGRk) — Yong Shun, Feb 20 '23 at 05:32
[Demo Solution 2 @ .NET Fiddle](https://dotnetfiddle.net/lbBT07) — Yong Shun, Feb 20 '23 at 05:32
You missed [LINQ to XML](https://learn.microsoft.com/dotnet/standard/linq/linq-xml-overview). — Paulo Morgado, Feb 20 '23 at 06:56

score 0 · Accepted Answer · answered Feb 20 '23 at 07:07

public class objClass
{
    public string ID { get; set; }
    public string Desc { get; set; }
    public string Address { get; set; }
    public DateTime? Date { get; set; }
    public string[] ImageFileNames { get; set; }
}

var list = Xml.Elements("root").Elements("Column1")
.Select(sv => new objClass()
{
    ID = (string)sv.Element("ID"),
    Desc = (string)sv.Element("Desc"),
    Address = (string)sv.Element("Address"),
    Date = (DateTime?)sv.Element("Date"),
    ImageFileNames = sv.Element("Images")
        .Elements("Image")
        .Select(i => (string)i.Element("File"))
        .ToArray(),
})
.ToList();

Convert from XML response to list

2 Answers2