Question

我试图将json转换为数据库的平面文件。为了做到这一点，我必须在根节点中获取一些值，然后在特定子节点中获取一个键，然后每个键的子节点值。

for each，JObject和JToken

之间JProperty和投射问题是一场噩梦

源JSON

{
  "id": "030010014B",
  "type": "street",
  "housenumbers": {
    "13": {
      "lat": 46.085617,
      "lon": 3.460492
    },
    "1bis": {
      "lat": 46.08686,
      "lon": 3.459992
    },
    "5": {
      "lat": 46.086469,
      "lon": 3.460912
    }
  }
}

预期结果

// <id value>;<key of housenumber #>;<lon value>;<lat value>
030010014B;13;3.460492;46.085617
030010014B;1bis;3.459992;46.08686
030010014B;5;3.460912;46.086469

当前代码错误地失败

Dim tmpJson As JObject = JObject.Parse(<json sample as string>)
Dim house As JToken
Dim house_p As JToken
If tmpJson.Property("housenumbers") IsNot Nothing Then
For Each houses As JObject In tmpJson.Property("housenumbers").Children 'only one
    For Each house In houses.Descendants
        fs.Append(tmpJson.Property("id").Value.ToString & ";")
        Dim hh As JProperty = house 'failing here
        fs.Append(hh.Name & ";") 'no way to get a key without jproperty 
        For Each house_p In house.Children()
            fs.AppendLine(house_p.Value(Of String)("lon") & ";" & house_p.Value(Of String)("lat"))
        Next
    Next
Next
End If
...

我失败的例子受到围绕SO的C＃答案的启发，但价值的关键是＆＃34;值。似乎很少见。

Answer 1

您可以将它们作为对象的集合并迭代：

Public Class Location
    <JsonProperty("lat")>
    Public Property Latitude As Single
    <JsonProperty("lon")>
    Public Property Longitude As Single
End Class

然后将内部部分反序列化为sa Dictionary：

Dim jstr = from whereever
Dim jobj = JObject.Parse(jstr)

Dim id As String = jobj("id").ToString

Dim numbers = JsonConvert.DeserializeObject(Of Dictionary(Of String, Location))(jobj("housenumbers").ToString())

For Each kvp In numbers
    Console.WriteLine("key:{0}, lat: {1}, long: {2}",
                      kvp.Key, kvp.Value.Latitude,
                      kvp.Value.Longitude)
Next

键：13，纬度：46.08562，长：3.460492
  key：1bis，lat：46.08686，long：3.459992
  key：5，lat：46.08647，long：3.460912

030010014B将出现在您的Id变量中

如何通过JSON.net获得关键和关键儿童的价值

1 个答案: