我有一个JSON-String(下面)。我想将它反序列化为C#对象并将数据作为列表获取。我的问题是,数据在JSON数组中不可用。我如何准备我的解决方案,我可以在此结构中获取数据:
对于传感器1 - n我只需要列表中的GUID。背后的数据无关紧要。
在代码中,我用" GUID"
替换了所有GUID{
"object": {
"GUID": {
"type": "sensor",
"owner": "GUID",
"time": 1482499665,
"description": "Temperatursensor 1",
"sdevice": "00003639",
"model": "SOLUCON Industry Temperature",
"tag": [
"GUID",
"GUID"
]
},
"GUID": {
"type": "sensor",
"owner": "GUID",
"time": 1482499758,
"description": "Wassersensor 1",
"sdevice": "000056d9",
"model": "SOLUCON Industry Water",
"tag": [
"GUID",
"GUID"
]
},
"GUID": {
"type": "sensor",
"owner": "GUID",
"time": 1482499797,
"description": "Rauchmelder 1",
"sdevice": "00008519",
"model": "TG551A",
"tag": [
"GUID",
"GUID"
]
},
"GUID": {
"type": "sensor",
"owner": "GUID",
"time": 1483888365,
"description": "SOLUCON Industry Multi 2",
"sdevice": "0000d409",
"model": "SOLUCON Industry Multi",
"tag": [
"GUID",
"GUID"
]
}
},
"status": "ok"
}
答案 0 :(得分:3)
您可以像这样使用Newtonsoft.Json包:
var jsonString = ...
var result = JsonConvert.DeserializeObject<IDictionary<string, object>>(jsonString);
var obj = (JObject)result["object"];
foreach (var prop in obj.Properties())
{
Console.WriteLine(prop.Name);
}
这将打印object
节点的所有GUID属性。
如果你想获得额外的对象,你可以定义一个模型:
public class Item
{
public string Type { get; set; }
public Guid Owner { get; set; }
public string Description { get; set; }
public IList<string> Tag { get; set; }
...
}
然后你就可以得到这样的传感器:
foreach (var prop in obj.Properties())
{
Console.WriteLine(prop.Name);
Sensor sensor = prop.Value.ToObject<Sensor>();
}
答案 1 :(得分:1)
您提供的结构看起来像字典,因此它应反序列化为:
class Data
{
public Dictionary<Guid, Sensor> Object {get;set;}
}
通过此,您可以提取包含Guid列表的键列表(data.Object.Keys
)。