Question

我正在使用JSON.Net将数据库中的数据序列化为JSON格式。但是我无法获得预期的结果。

我必须根据某些条件来创建JSON对象，例如如果datamap的数据为空，则不应将其包含在JSON中；如果不为空，则应将其包含在内。

public class DatamapKey
{
    [JsonExtensionData]
    public Dictionary<string, JToken> DatamapKeyFields = new Dictionary<string, JToken>();
}
public class DatamapKey1
{
    [JsonExtensionData]
    public Dictionary<string, JToken> DatamapKey1Fields = new Dictionary<string, JToken>();
}
public class DatamapItem
{
    [JsonExtensionData]
    public Dictionary<string, JToken> DatamapItemFields = new Dictionary<string, JToken>();
    public DatamapKey datamapKey { get; set; }
    public DatamapKey1 datamapKey1 { get; set; }
}
public class RootObject
{
    public List<DatamapItem> datamapItems { get; set; }
}

输出JSON：

{
  "datamapItems": [
    {
      "datamapKey": {
        "module": 1,
        "id": 1391
      },
      "datamapKey1": {},
      "paramName": "VE8321C",
      "min": "0",
      "max": "40"
    },
    {
      "datamapKey": {},
      "datamapKey1": {},
      "paramName": "VE8321C",
      "min": "0",
      "max": "40"
    },
    {
      "datamapKey": {
        "module": 1,
        "id": 1391
      },
      "datamapKey1": {
        "module": 1,
        "id": 1391
      },
      "paramName": "VE8321C",
      "min": "0",
      "max": "40"
    }
  ]
}

预期输出：

{
  "datamapItems": [
    {
      "paramName": "VE8321C",
      "datamapKey": {
        "module": 1,
        "id": 1391
      },
      "min": "0",
      "max": "40"
    },
    {
      "paramName": "VE8321C",
      "min": "0",
      "max": "40"
    },
    {
      "paramName": "VE8321C",
      "datamapKey": {
        "module": 1,
        "id": 1391
      },
      "datamapKey1": {
        "module": 1,
        "id": 1391
      },
      "min": "0",
      "max": "40"
    }
  ]
}

Answer 1

[JsonExtensionData]属性使标记的字典中的键/值对被序列化，就好像它们是包含该字典的类的属性一样。该属性由Json.Net专门处理，因此@Alex在注释中建议的IgnoreEmptyEnumerablesResolver对此无效。但是您可以重组类，这样就不需要datamapKey和datamapKey1的属性，这将使解析器在这些字典为空时可以对其进行处理，从而为您提供所需的输出

public class DatamapItem
{
    [JsonExtensionData]
    public Dictionary<string, JToken> DatamapItemFields { get; set; } = new Dictionary<string, JToken>();
    public Dictionary<string, JToken> datamapKey { get; set; } = new Dictionary<string, JToken>();
    public Dictionary<string, JToken> datamapKey1 { get; set; } = new Dictionary<string, JToken>();
}

public class RootObject
{
    public List<DatamapItem> datamapItems { get; set; }
}

演示小提琴：https://dotnetfiddle.net/Gw03gY

如果您不喜欢该主意或不想修改您的类结构，另一种选择是将对象实例加载到JObject中，并使用递归方法从层次结构中删除空标记。 How to omit/ignore/skip empty object literals in the produced JSON?中详细介绍了此方法。

下面是一个演示小提琴，用您现有的类结构显示了该方法：https://dotnetfiddle.net/jmNgYi

使用C＃和Json.NET根据条件序列化JSON对象

1 个答案: