Question

我有这个JSON：

{ “JOE”：{ “ID”：7， “年龄”： “23”}， “比利”：{ “ID”：8， “年龄”： “29”}}

我有this solution for a more simple JSON structure posted by RajN。

Dim j1 As String = "{ "JOE"":""0.90000000"",""JOE"":""3.30000000"",""MONROE"":""1.20000000""}"
Dim dict = JsonConvert.DeserializeObject(Of Dictionary(Of String, String))(j1)
For Each kvp In dict
    Console.WriteLine(kvp.Key & " - " + kvp.Value)
Next

我正在寻找如何处理新的JSON数据。提前致谢

Answer 1

解决方案与@RajN的the other solution you linked几乎相同，除了使用Dictionary(Of String, String)，您需要使用Dictionary(Of String, T)，其中T是您定义的类保留id和age。

所以，定义一个类：

Public Class PersonData
    Public Property id As Integer
    Public Property age As String
End Class

然后反序列化：

Dim json As String = "{""JOE"":{""id"":7,""age"":""23""},""BILLY"":{""id"":8,""age"":""29""}}"

Dim dict = JsonConvert.DeserializeObject(Of Dictionary(Of String, PersonData))(json)

For Each kvp In dict
    Console.WriteLine("name: " & kvp.Key)
    Console.WriteLine("id: " & kvp.Value.id)
    Console.WriteLine("age: " & kvp.Value.age)
    Console.WriteLine()
Next

小提琴：https://dotnetfiddle.net/HMv7Om

有意义吗？

无需一个多个属性反序列化JSON

1 个答案: