Deserializing Json in windows phone 7 using Json.NET

Question

I have a JSON in a web server like this :

{"My Book List": [{"ID":"5","TYPE":"History","TITLE":"Ekannoborti","PRICE":"200","IMAGE":"Ekannoborti.jpg","DOWNLOAD LINK":"http://www.starhostbd.com/"}],"success":3}

To deserialize it I have done so far :

 public class Attributes
    {
        public string ID{ get; set; }
        public string TYPE { get; set; }
        public string TITLE { get; set; }
        public string  PRICE { get; set; }
        public string IMAGE { get; set; }
        public string DOWNLOADLINK { get; set; }
    }

    public class DataJsonAttributeContainer
    {
        public List<Attributes> attributes { get; set; }
        //public Attributes attributes { get; set; }
    }

    public static T DeserializeFromJson<T>(string json)
    {
        T deserializedProduct = JsonConvert.DeserializeObject<T>(json);
        return deserializedProduct;
    }

& in my code :

void webClient_DownloadStringCompleted(object sender, DownloadStringCompletedEventArgs e)
    {
        //var deserializedJSON = JsonConvert.DeserializeObject<Attributes>(e.Result);
        var container = DeserializeFromJson<DataJsonAttributeContainer>(e.Result);
        string asd = container.attributes[0].DOWNLOADLINK[0].ToString();
        //string asd = deserializedJSON.DOWNLOADLINK[0].ToString();
    }

The problem is : From debug wiindow i can see that data is assigned in e.Result but container remains null . How to solve this problem ? Please help !

With mine and pedro'd modification this code works for me. I can extract the whole link. But you need to use string asd = container.attributes[0].DOWNLOADLINK.ToString(); to obtain the whole link. In your case you obtain only the first character. — saxos
– saxos, Commented Oct 1, 2012 at 14:03

Pedro Lamas · Accepted Answer · 2012-10-01 11:50:34Z

1

Add a JsonProperty attribute to the attributes property to match the property name in the JSON, like so:

public class DataJsonAttributeContainer
{
    [JsonProperty("My Book List")]
    public List<Attributes> attributes { get; set; }
}

Also, you should add a JsonProperty attribute to the Attributes.DOWNLOADLINK property with "DOWNLOAD LINK" value in order for it to match the JSON property name.

edited Oct 1, 2012 at 11:50

answered Oct 1, 2012 at 11:10

Pedro Lamas

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3 Comments

Pedro Lamas Over a year ago

Working fine here, but I did notice an error in your sample code above: there is a missing " after the My Book List

Fahim Ahmed Over a year ago

sorry ! that was a typo due to copy paste from the actual large json file . Actual file is ok !

Pedro Lamas Over a year ago

You can check here what I did, and it is working fine: gist.github.com/3811718

saxos · Accepted Answer · 2012-10-01 11:15:42Z

0

At first glance I think the problem is at the DOWNLOADLINK property. Your server returns "DOWNLOAD LINK" but your property hasn't the space in it's name.

You should define the json representation at your property like this:

[JsonProperty(PropertyName = "DOWNLOAD LINK")]
public string DOWNLOADLINK { get; set; }

Hope this helps.

answered Oct 1, 2012 at 11:15

saxos

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Comments

nkchandra · Accepted Answer · 2012-10-01 14:41:28Z

Actually both the above answers should solve your problem, you have to club them

public class MyBookList
{
    public string ID { get; set; }
    public string TYPE { get; set; }
    public string TITLE { get; set; }
    public string PRICE { get; set; }
    public string IMAGE { get; set; }

    [JsonProperty("DOWNLOAD LINK")]
    public string DOWNLOADLINK { get; set; }
}

public class DataJsonAttributeContainer
{
    [JsonProperty("My Book List")]
    public List<MyBookList> MyBookList { get; set; }

    public int success { get; set; }
}

and also

var container = DeserializeFromJson<DataJsonAttributeContainer>(e.Result);
string asd = container.attributes[0].DOWNLOADLINK.ToString();

Try with these classes. Should work.

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