I am using Json.net for serializing and then making an JObject that looks like this:
\"RegistrationList\": [
{
\"CaseNumber\": \"120654-1330\",
I only want to display the CaseNumber, FirstName and Comment.
As always in ASP.NET MVC you could start by writing a view model that matches your requirements:
public class MyViewModel
{
public string CaseNumber { get; set; }
public string FirstName { get; set; }
public string Comment { get; set; }
}
then in your controller action you build the view model from the JObject instance you already have:
public ActionResult Index()
{
JObject json = ... the JSON shown in your question (after fixing the errors because what is shown in your question is invalid JSON)
IEnumerable<MyViewModel> model =
from item in (JArray)json["RegistrationList"]
select new MyViewModel
{
CaseNumber = item["CaseNumber"].Value<string>(),
FirstName = item["Person"]["FirstName"].Value<string>(),
Comment = item["User"]["Comment"].Value<string>(),
};
return View(model);
}
and finally in your strongly typed view you display the desired information:
@model IEnumerable<MyViewModel>
<table>
<thead>
<tr>
<th>Case number</th>
<th>First name</th>
<th>Comment</th>
</tr>
</thead>
<tbody>
@foreach (var item in Model)
{
<tr>
<td>@item.CaseNumber</td>
<td>@item.FirstName</td>
<td>@item.Comment</td>
</tr>
}
</tbody>
</table>
Several ways:
1) According documentation 'Using LINQ for JSON' you can query JObject in LINQ way
JObject o = JObject.Parse(@"{
'CPU': 'Intel',
'Drives': [
'DVD read/writer',
'500 gigabyte hard drive'
]
}");
string cpu = (string)o["CPU"];
// Intel
string firstDrive = (string)o["Drives"][0];
// DVD read/writer
IList<string> allDrives = o["Drives"].Select(t => (string)t).ToList();
// DVD read/writer
// 500 gigabyte hard drive
2) Querying JSON with SelectToken
3) Use custom helper extention method for querying by specified path like this:
public static class JsonHelpers
{
public static JToken QueryJson(this object jsonObject, params string[] jsonPath)
{
const string separator = " -> ";
if (jsonObject == null)
throw new Exception(string.Format("Can not perform JSON query '{0}' as the object is null.",
string.Join(separator, jsonPath ?? new string[0])));
var json = (jsonObject as JToken) ?? JObject.FromObject(jsonObject);
var token = json;
var currentPath = "";
if (jsonPath != null)
foreach (var level in jsonPath)
{
currentPath += level + separator;
token = token[level];
if (token == null) break;
}
if (token == null)
throw new Exception(string.Format("Can not find path '{0}' in JSON object: {1}", currentPath, json));
return token;
}
}
I think you want to get the JSON string like below:
{
'RegistrationList': [
{
'CaseNumber': '120654-1330',
'Priority': 5,
'PersonId': 7,
'Person': {
'FirstName': '0',
'LastName': '',
},
'UserId': 7,
'User': {
'Id': 7,
'CreatedTime': '2013-07-05T13:09:57.87',
'Comment': ''
}
},
{
'CaseNumber': '120654-1330',
'Priority': 5,
'PersonId': 7,
'Person': {
'FirstName': '0',
'LastName': '',
},
'UserId': 7,
'User': {
'Id': 7,
'CreatedTime': '2013-07-05T13:09:57.87',
'Comment': ''
}
},
]
}
If so, you can get below code working for your question:
string json = @"{
'RegistrationList': [
{
'CaseNumber': '120654-1330',
'Priority': 5,
'PersonId': 7,
'Person': {
'FirstName': '0',
'LastName': '',
},
'UserId': 7,
'User': {
'Id': 7,
'CreatedTime': '2013-07-05T13:09:57.87',
'Comment': ''
}
},
{
'CaseNumber': '120654-1330',
'Priority': 5,
'PersonId': 7,
'Person': {
'FirstName': '0',
'LastName': '',
},
'UserId': 7,
'User': {
'Id': 7,
'CreatedTime': '2013-07-05T13:09:57.87',
'Comment': ''
}
},
]
}";
JObject o = JObject.Parse(json);
JArray list = (JArray)o["RegistrationList"];
List<Tuple<string, string, string>> rList = new List<Tuple<string, string, string>>();
foreach (var r in list)
{
Tuple<string, string, string> temp = new Tuple<string, string, string>(r["CaseNumber"].Value<string>(), r["Person"]["FirstName"].Value<string>(), r["User"]["Comment"].Value<string>());
rList.Add(temp);
Console.WriteLine(temp);
}
var serializer = new JavaScriptSerializer(); object modelData = serializer.DeserializeObject(jsonstring);