Question

编辑：我想我应该提到我无法控制JSON，而且我知道通常我的C＃对象应该与JSON匹配。我的问题不是“为什么不反序列化？”。我知道为什么不是这样。我在问是否有一种方法可以按照我的要求反序列化JSON。

我正在使用Newtonsoft.Json。

我有一个包含1个对象的JSON字符串。我需要将该对象反序列化为带有嵌套对象的C＃对象。

所以可以说我的JSON看起来像这样。

pick_car = raw_input("So whats your favourite car brand then?").lower()
for brand_name in cars:
    if brand_name in pick_car:
        print 'okay so you like ' + brand_name

这是我的C＃对象

{
    "id": 123,
    "userName": "fflintstone",
    "address": "345 Cave Stone Road",
    "address2": "",
    "city": "Bedrock",
    "state": "AZ",
    "zip": "",   
}

AddressModel地址属性是一个嵌套对象。该对象包含实际的地址属性。因此，我需要反序列化JSON对象，以便将id和userName添加到Customer对象，然后将地址字段添加到嵌套的Address对象。

我还无法找到newtonsoft内置的方法来完成此任务。有什么想法吗？

Answer 1

如果您是设置将要由代码使用的JSON的人，则您的JSON未设置为与对象匹配。

它应该像这样：

{
    "id": 123,
    "userName": "fflintstone",
    "address": {
        // address properties here
    }
}

否则，您需要更新C＃对象以匹配JSON，这将意味着JSON中每个项目的单独属性：

public class Customer
{
    public long Id { get; set; }

    public string UserName { get; set; }

    public string Address { get; set; }

    public string Address2 { get; set; }

    public string City { get; set; }

    public string State { get; set; }

    public string ZIP { get; set; }
}

此外，如果您同时设置了这两个名称，则可能不应将街道名称命名为“地址”。对我来说，“地址”一词指的是街道名称，号码，城市，州和邮政编码。

如果您无法控制JSON，则没有真正的方法使用Json.Net之类的方法将JSON完全反序列化为对象。您需要设置某种映射器，或直接查找属性以获得其值，以将其添加到对象中。您可以将JSON解析为JObject，然后访问所需的属性。

JObject foo = JObject.Parse(//your JSON string here);
customer.Address = (string)foo["Address"];
customer.Address2 = (string)foo["Address2];

通常，如果您不是JSON结构的控制者，那么最好使您的对象与您获得的JSON匹配。

Answer 2

您可以使用自定义JsonConverter进行此操作。

public class CustomerJsonConverter : JsonConverter
{
    public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
    {
        if (value is Customer customer)
        {
            var token = new JObject
            {
                ["id"] = customer.Id,
                ["userName"] = customer.UserName,
                ["address"] = customer.Address.Address,
                ["address2"] = customer.Address.Address2,
                ["city"] = customer.Address.City,
                ["state"] = customer.Address.State,
                ["zip"] = customer.Address.ZIP
            };

            token.WriteTo(writer);
        }
        else
        {
            throw new InvalidOperationException();
        }

    }

    public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
    {
        var obj = JToken.ReadFrom(reader);

        if (obj.Type != JTokenType.Object)
        {
            return null;
        }

        return new Customer
        {
            Id = (long) obj["id"],
            UserName = (string) obj["userName"],
            Address = obj.ToObject<AddressModel>()
        };
    }

    public override bool CanRead => true;

    public override bool CanConvert(Type objectType)
    {
        return objectType == typeof(Customer);
    }
}

在JsonConverterAttribute类上带有Customer。

[JsonConverter(typeof(CustomerJsonConverter))]
public class Customer
{
    public long Id { get; set; }

    public string UserName { get; set; }

    public AddressModel Address { get; set; }
}

public class AddressModel
{
    public string Address { get; set; }

    public string Address2 { get; set; }

    public string City { get; set; }

    public string State { get; set; }

    public string ZIP { get; set; }
}

并这样利用：

var customer = JsonConvert.DeserializeObject<Customer>(customerJson);

或者，您可以简单地使用一个中间映射模型。

public class CustomerFlattened
{
    public long Id { get; set; }

    public string UserName { get; set; }

    public string Address { get; set; }

    public string Address2 { get; set; }

    public string City { get; set; }

    public string State { get; set; }

    public string ZIP { get; set; }

    public Customer ToCustomer()
    {
        return new Customer
        {
            Id = Id,
            UserName = UserName,
            Address = new AddressModel
            {
                Address = Address,
                Address2 = Address2,
                City = City,
                State = State,
                ZIP = ZIP
            }
        };
    }

    public static CustomerFlattened FromCustomer(Customer customer)
    {
        return new CustomerFlattened
        {
            Id = customer.Id,
            UserName = customer.UserName,
            Address = customer.Address.Address,
            Address2 = customer.Address.Address2,
            City = customer.Address.City,
            State = customer.Address.State,
            ZIP = customer.Address.ZIP
        };
    }
}

并这样利用：

var customer =
    JsonConvert.Deserialize<CustomerFlattened>(
        jsonOriginal
    )
    .ToCustomer();

var customerFlattened = CustomerFlattened.FromCustomer(customer);

var jsonConverted = JsonConvert.Serialize(customerFlattened );

Answer 3

如果您愿意使用其他工具，则可以创建一个与JSON匹配的DTO，然后使用诸如Automapper之类的映射器将DTO reverse flatten插入对象模型。

Answer 4

由于JSON和对象模型不匹配，因此您必须反序列化为临时对象并自己映射字段。在此示例中，我将反序列化为一个用作模板的匿名对象。

schema = return visitSchema(schema, {
    [VisitSchemaKind.OBJECT_TYPE](type: GraphQLObjectType) {
      if (type.getFields().relatedNews) {
        // console.log(type.getFields());
        return new GraphQLObjectType({
          name: type.name,
          fields: {
            ...fromPairs(Object.values(type.getFields()).map(({name, type, args, resolve, description}) => [name, {type, args: {}, resolve, description}])),
            relatedNews: {
              type: schema.getType('NewsArticle'),
              resolve(obj) {
                console.log('foo');
              }
            }
          }
        });
      }
      return type;
    },
  });

输出：

WITH RawDebug AS 
(
  SELECT 
  STRUCT(DebugReason,DebugData) AS Debug
  FROM `devicedata.Debug.T*`
  WHERE _TABLE_SUFFIX="20180624"
),
RawGPS AS (        
       SELECT
          STRUCT<DebugReason INT64, DebugData STRING>(NULL, NULL) AS Debug
        FROM
          `devicedata.Gps.T*` AS g
         WHERE _TABLE_SUFFIX="20180624"
)
SELECT Debug
FROM RawDebug
UNION ALL
SELECT Debug
FROM RawGPS

Working example on DotNetFiddle

Answer 5

首先对于要使用AdressModel类创建的对象，您需要更改JSON文档，我已更改为：

  {   "id": 123,   "userName": "fflintstone",   "Address": {
"address": "345 Cave Stone Road",
"address2": "",
"city": "Bedrock",
"state": "AZ",
"zip": ""   } }

然后针对模型，我创建了这两个模型：

public class Customer
{
    public long id { get; set; }
    public string userName { get; set; }
    public AddressModel Address { get; set; }
}

public class AddressModel
{
    public string address { get; set; }
    public string address2 { get; set; }
    public string city { get; set; }
    public string state { get; set; }
    public string zip { get; set; }
}

现在要反序列化Json文档，您可以执行以下操作：

Customer jsonConverted = new Customer();

        using (StreamReader r = new StreamReader(HostingEnvironment.ApplicationPhysicalPath + @"\infoFile.json"))
        {
            var json = r.ReadToEnd();
            jsonConverted = JsonConvert.DeserializeObject<Customer>(json);
        }

（我正在测试MVC项目）

将JSON对象反序列化为嵌套的C＃对象

5 个答案: