Deserialize Json file

Question

{
    "Location": "St Kilda", 
    "Name": "Movie Plaza theatre", 
    "Day": ["Sunday","Monday","Tuesday","Wednesday","Tuesday","Friday","Saturday"]
}

I can't deserialize this Json. can anyone help me ?

Too many tutorials/examples/existing questions available: no effort. — user2864740, Aug 08 '15 at 05:00
you have to show some effort of your own (for example, code you tried), and tell us what does not work, and how. — hoijui, Aug 08 '15 at 05:01
For example, http://stackoverflow.com/questions/4611031/convert-json-string-to-c-sharp-object?lq=1 , http://stackoverflow.com/questions/18887739/convert-json-to-c-sharp-type?lq=1 (and any relevant web search) — user2864740, Aug 08 '15 at 05:02

Krunal Mevada · Accepted Answer · 2015-08-08T05:09:24.010

1

Create class of your json as below

public class RootObject
{
    public string Location { get; set; }
    public string Name { get; set; }
    public List<string> Day { get; set; }
}

Then write some code as below to get data from json

JObject json = JObject.Parse(your json string); 
RootObject obj = Newtonsoft.Json.JsonConvert.DeserializeObject<RootObject>(json);

Update:

Console.WriteLine(obj.Location);
Console.WriteLine(obj.Name);

foreach (var d in obj.Day)
{
    Console.WriteLine(d);
}

edited Aug 08 '15 at 05:09

answered Aug 08 '15 at 04:50

Krunal Mevada

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Yeah, it works. Can you tell me how to show the data of this Json file? coz i can only show the type of this list. – Sam Aug 08 '15 at 04:57
I am not able to get you. What you want ? – Krunal Mevada Aug 08 '15 at 05:00
i used console.writeline in foreach to show the data in Json file, i can get the data of location and name, but i only get the type of the List not the data in this List. – Sam Aug 08 '15 at 05:03
static void PrintData(List data) { foreach (var item in data) { Console.WriteLine(item.Location + item.Name + item.Day); } } – Sam Aug 08 '15 at 05:11
@Sam I update my answer please check. – Krunal Mevada Aug 08 '15 at 05:12
Thank you very much . :) – Sam Aug 08 '15 at 05:17

score 0 · Answer 2 · answered Aug 08 '15 at 05:01

I think that you can use the C# dynamic type to make things easier. This technique also makes re-factoring simpler as it does not rely on magic-strings.

1. JsonConvert.DeserializeObject Use JsonConvert.DeserializeObject<RootObject>(string json);

Create your classes on JSON 2 C#. How it read here

If any fields missing in the JSON data should simply be left NULL.

2. Deserializing JSON with Json.NET You can also read how deserializing JSON with Json.NET

3. JsonConvert.Populate You can also use JsonConvert.Populate(json,obj); i.e. json is the json string,obj is the target object

Deserialize Json file

2 Answers2