Deserialize XML to object C#

Question

I'm trying to serialize xml to object, but after serializing my list of transport-agreements is empty. I pretty sure this is quite easy, but i cant find my error.

I have the following xml

<transport-agreements type="array">
   <transport-agreement>
     <id type="integer">1047</id>
     <description>Standard</description>
     <products type="array">
       <product>not important</product>
     </products>
   </transport-agreement>
   <transport-agreement>
     <id type="integer">1048</id>
     <description>Standard</description>
     <products type="array">
       <product>not important</product>
    </products>
   </transport-agreement>
</transport-agreements>

And i have the following code:

var resultContent = await response.Content.ReadAsStreamAsync();
var serializer = new XmlSerializer(typeof(TransportAgreementRoot));
var transportAgreements = (TransportAgreementRoot)serializer.Deserialize(resultContent);

And my classes:

[Serializable, XmlRoot("transport-agreements")]

    public class TransportAgreementRoot
    {
        [XmlArrayItem("transport-agreement")]
        TransportAgreement[] TransportAgreements { get; set; }         
     }

    public class TransportAgreement
    {
        [XmlElement("description")]
        public string Description { get; set; }
        [XmlElement("id")]
        public int Id { get; set; }
        [XmlElement("number")]
        public string Number { get; set; }
        [XmlElement("carrier")]
        public Carrier Carrier { get; set; }
        [XmlArray("products")]
        [XmlArrayItem("product")]
        public Product[] Products { get; set; }

    }

Have you tried changing `var transportAgreements = (TransportAgreementRoot)serializer.Deserialize(resultContent);` to `(TransportAgreement[])serializer.Deserialize(resultContent)` — Ashley John, Jul 11 '16 at 09:25
pls have look at this http://stackoverflow.com/questions/364253/how-to-deserialize-xml-document — Ashley John, Jul 11 '16 at 09:32

score 3 · Answer 1 · answered Jul 11 '16 at 09:40

3

You can use some of on-line converters, which will give you full classes diagram. My fave is: this one

answered Jul 11 '16 at 09:40

Maciej S.

742
11
20

This should be a comment – dotNET Jul 11 '16 at 09:42

score 2 · Accepted Answer · answered Jul 11 '16 at 10:08

[XmlRoot("transport-agreements")]
public class TransportAgreementRoot
{
    [XmlElement("transport-agreement")]
    public TransportAgreement[] TransportAgreements { get; set; }
}

[XmlRoot("transport-agreement")]
public class TransportAgreement
{
    [XmlElement("description")]
    public string Description { get; set; }
    [XmlElement("id")]
    public int Id { get; set; }
    // other properties
}

score -1 · Answer 3 · answered Jul 11 '16 at 09:18

-1

I Changed XmlArray to XmlElement. The Array would add a extra tag which you don't have in the xml file.

   [Serializable, XmlRoot("transport-agreements")]

    public class TransportAgreementRoot
    {
        [XmlElement("transport-agreement")]
        TransportAgreement[] TransportAgreements { get; set; }
    }

answered Jul 11 '16 at 09:18

jdweng

29,955
2
14
18

Hi! The array of transportagreements is still empty after changing to element – sindrem Jul 11 '16 at 09:21

Deserialize XML to object C#

3 Answers3