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eeboy
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XML Deserialization

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I am attempting to deserialize an XML file that looks like so:

<?xml version="1.0"?>
<cameras>
  <camera name="z1485" id="0">
      <mode name="Video" cmd="96">
        <resolution name="1280x720" value="0"/>
        <resolution name="640x480" value="1"/>
        <resolution name="320x240" value="2"/>
      </mode>
  </camera>
  <camera name="a950" id="2">
      <mode name="Video" cmd="96">
        <resolution name="1280x720" value="0"/>
        <resolution name="640x480" value="1"/>
      </mode>
  </camera>
</cameras>


Using the following code:

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml;
using System.Xml.Serialization;
using System.IO;


namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args)
        {

            List<Camera> Cameras = new List<Camera>();

            XmlSerializer deserializer = new XmlSerializer(typeof(List<Camera>));
            TextReader textReader = new StreamReader("cameras.xml");
            Cameras = (List<Camera>)deserializer.Deserialize(textReader);
            textReader.Close();
            
        }
    }

    [XmlRootAttribute("camera",Namespace="",IsNullable=false)]
    public class Camera
    {

        [XmlElement("mode")]
        List<Mode> Modes;

        public Camera() 
        {
            Modes = new List<Mode>();
        }

    }

    public class Mode
    {
        [XmlElement("resolution")]
        List<Resolution> Resolution;

        [XmlAttribute("name")]
        public string Name { get; set; }
        [XmlAttribute("id")]
        public byte Id { get; set; }

        public Mode()
        {
            Resolution = new List<Resolution>();
        }
    }


    public class Resolution
    {
        [XmlAttribute("name")]
        public string name {get; set;}
        [XmlAttribute("value")]
        public byte value { get; set; } 

    }


}


There will be many camera entries. There will be many modes per camera. There will also be more properties of a single camera mode. This is a very simple xml document for now. I want to deserialize this data into a list of cameras. The list of cameras will be made up of a list of modes (and perhaps other elements). The list of modes will be made up of various lists of elements (like resolution).

Now, when I run the above the deserialize process throws an 'InvalidOperationException' and tells me that 'There is an error in XML document (2,2).

Any thoughts?

xml deserialization
  • asked 2 years ago
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last answered 2 years ago

1 answers

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up1down
tick
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You've got one or two problems there, particularly with the root node. Try the following instead:

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml;
using System.Xml.Serialization;
using System.IO;


namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args)
        {
            List<Camera> Cameras;
            XmlRootAttribute root = new XmlRootAttribute();
            root.ElementName = "cameras";
            root.Namespace = "";
            root.IsNullable = false; 
            XmlSerializer deserializer = new XmlSerializer(typeof(List<Camera>), root);
            TextReader textReader = new StreamReader("cameras.xml");
            Cameras = (List<Camera>)deserializer.Deserialize(textReader);
            textReader.Close();
  
            // print out results
            Console.Clear();
            foreach(Camera c in Cameras)
            {
               Console.WriteLine("Camera Name = {0}, Id = {1}\n", c.Name, c.Id);  
               foreach(Mode m in c.Modes)
               {
                   Console.WriteLine("  Mode Name = {0}, Cmd = {1}\n", m.Name, m.Cmd);
                   foreach(Resolution r in m.Resolutions)
                   {
                      Console.WriteLine("    Resolution Name = {0}, Value = {1}", r.Name, r.Value);
                   }           
               }
               Console.WriteLine();
            }
            Console.ReadKey(); 
        }
    }

    [XmlType("camera")]
    public class Camera
    {
        [XmlAttribute("name")]
        public string Name { get; set; }

        [XmlAttribute("id")]
        public string Id { get; set; }  

        [XmlElement("mode")]
        public List<Mode> Modes;

        public Camera() 
        {
            Modes = new List<Mode>();
        }

    }

    public class Mode
    {
        [XmlElement("resolution")]
        public List<Resolution> Resolutions;

        [XmlAttribute("name")]
        public string Name { get; set; }

        [XmlAttribute("cmd")]
        public byte Cmd { get; set; }

        public Mode()
        {
            Resolutions = new List<Resolution>();
        }
    }


    public class Resolution
    {
        [XmlAttribute("name")]
        public string Name {get; set;}

        [XmlAttribute("value")]
        public byte Value { get; set; } 

    }

}

  • answered 2 years ago by:

    vulpes
    17279
  • edited 2 years ago
(x)
eeboy
499

Thanks Vulpes!!! You don't know what XML deserialization has just done for my life! :) If you'd have seen how I was parsing my XML file... well... I won't go there. :)

(x)
eeboy
499

The [XmlAttribute] and [XmlElement] are pretty intuitive to me but what does [XmlType("camera")] do?

(x)
vulpes
17279

Yes, once you get your classes right, XML deserialization is much easier than trying to parse the file directly with XmlReader, XmlDocument etc. As we're deserializing directly to a List<Camera), the deserializer needs a way of associating the Camera class with the camera (small 'c') elements in the Xml file. The XmlType attribute does this.

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