JSON-десериализация одной строки в JSON

Question

Это моя строка JSON, которую я получаю из ответа HTTP.

[{"ts":"1395318389","date":"2014-03-20","time":"12:26:21","nodeid":"1229001363","lat":"53.292425","lon":"-6.43203333333333","speed":"76","dir":"242","sv":"9","volt":"4187","rssi":"-69","us":"3","type":0,"net":"27202","height":"108","hdop":"9","cellid":"EB84","dd":6105,"ttf":"0","lac":"4E24","odo":"0","gle":"0"}]

Мой класс настроен так

Friend Class SmartTrack_PostionList
    Public positionList As SmartTrack_Postion()
End Class

Friend Class SmartTrack_Postion
    Public ts As String
    Public [date] As String
    Public time As String
    Public nodeid As String
    Public lat As String
    Public lon As String
    Public speed As String
    Public dir As String
    Public sv As String
    Public volt As String
    Public rssi As String
    Public us As String
    Public type As String
    Public net As String
    Public height As String
    Public hdop As String
    Public cellid As String
    Public dd As String
    Public ttf As String
    Public lac As String
    Public odo As String
    Public gle As String
End Class

Я пробовал следующие строки кода, но ни одна не работает

Dim ser As New System.Web.Script.Serialization.JavaScriptSerializer
Dim getPositionList As SmartTrack_PostionList = DirectCast(ser.Deserialize(Of SmartTrack_PostionList)(getResult.ToString), SmartTrack_PostionList)

а также

Dim ser As New System.Web.Script.Serialization.JavaScriptSerializer
Dim getPositionList As SmartTrack_Postion= DirectCast(ser.Deserialize(Of SmartTrack_Postion)(getResult.ToString), SmartTrack_Postion)

Это сработало, но я думаю, что это плохое кодирование.

Dim ser As New System.Web.Script.Serialization.JavaScriptSerializer
getResult = getResult.Replace("[", "")
getResult = getResult.Replace("]", "")
Dim dict As Dictionary(Of String, String) = ser.Deserialize(Of Dictionary(Of String, String))(getResult)

Я был бы признателен за любую помощь в этом, спасибо.

Джим

Answer 1

Вам придется десериализовать вашу строку JSON в List(Of SmartTrack_Position), лайк

Dim ser = New System.Web.Script.Serialization.JavaScriptSerializer
Dim json = getResult.ToString()
Dim getPositionList = ser.Deserialize(Of List(Of SmartTrack_Postion))(json)

Answer 2

Используйте этот класс

 public class RootObject
{
public string ts { get; set; }
public string date { get; set; }
public string time { get; set; }
public string nodeid { get; set; }
public string lat { get; set; }
public string lon { get; set; }
public string speed { get; set; }
public string dir { get; set; }
public string sv { get; set; }
public string volt { get; set; }
public string rssi { get; set; }
public string us { get; set; }
public int type { get; set; }
public string net { get; set; }
public string height { get; set; }
public string hdop { get; set; }
public string cellid { get; set; }
public int dd { get; set; }
public string ttf { get; set; }
public string lac { get; set; }
public string odo { get; set; }
public string gle { get; set; }
}

Тогда используйте json.net, как это

   RootObject Json= JsonConvert.DeserializeObject<RootObject>(returnJson);