如何使用包含“原始” JSON的字符串成员对类型进行序列化和反序列化，而又不会在过程中转义JSON

一尘不染

如何使用包含“原始” JSON的字符串成员对类型进行序列化和反序列化，而又不会在过程中转义JSON

json

我想反序列化JSON为对象，但我不想反序列化嵌套的JSON，嵌套的嵌套JSON应该转换为JSON列表（请检查“ 我的预期输出
”以获得清晰的主意）…

//假设我下面有JSON数据，在这里我为“地址”实体嵌套了JSON

String jsonEmployees =
"{"Employees":
[{"EmpId":1, "EmpName":"ABC", "Address":[{"AddressId":1, "Address":"Something"},{"AddressId":2, "Address":"Anything"}]},
{"EmpId":2, "EmpName":"XYZ", "Address":[{"AddressId":1, "Address":"Something"},{"AddressId":2, "Address":"Anything"}]}]
}"

public class Employee
{
    public int EmpId { get; set; }
    public string EmpName { get; set; }
    // **Note** : I'm not using List<Address> data type for Address, instead of I want list of address in JSON string
    public string Address { get; set; }
}

public class RootObject
{
    public List<Employee> Employees { get; set; }
}

var Employees = JsonConvert.DeserializeObject<RootObject>(jsonEmployees);

//我的预期输出

Employees[0].EmpId = 1;
Employees[0].EmpName = "ABC";
Employees[0].Address = "[{"AddressId":1, "Address":"Something"},{"AddressId":2, "Address":"Anything"}]";

Employees[1].EmpId = 2;
Employees[1].EmpName = "XYZ";
Employees[1].Address = "[{"AddressId":1, "Address":"Something"},{"AddressId":2, "Address":"Anything"}]";

请建议我解决此问题的最佳方法…

阅读 214

2020-07-27

共1个答案

一尘不染

您的问题是，如何 在不转义JSON的情况下 ， 如何使用string包含“原始” JSON 的成员对类型进行序列化和反序列化？

这可以通过做自定义JsonConverter读取和写入使用原始JSON
JsonWriter.WriteRawValue()和JRaw.Create()：

public class RawConverter : JsonConverter
{
    public override bool CanConvert(Type objectType)
    {
        throw new NotImplementedException();
    }

    public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
    {
        if (reader.TokenType == JsonToken.Null)
            return null;
        var raw = JRaw.Create(reader);
        return raw.ToString();
    }

    public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
    {
        var s = (string)value;
        writer.WriteRawValue(s);
    }
}

然后将其应用于您的类型，如下所示：

public class Employee
{
    public int EmpId { get; set; }
    public string EmpName { get; set; }
    // **Note** : I'm not using List<Address> data type for Address, instead of I want list of address in JSON string
    [JsonConverter(typeof(RawConverter))]
    public string Address { get; set; }
}

public class RootObject
{
    public List<Employee> Employees { get; set; }
}

样品提琴。

请注意，原始JSON字符串必须
代表有效的JSON。如果不是，则创建的JSON将不可读。如果要保证JSON文字有效，则可以在内部将JSON保持为已解析状态：

public class Employee
{
    public int EmpId { get; set; }
    public string EmpName { get; set; }

    [JsonProperty("Address")]
    JToken AddressToken { get; set; }

    [JsonIgnore]
    public string Address
    {
        get
        {
            if (AddressToken == null)
                return null;
            return AddressToken.ToString(Formatting.Indented); // Or Formatting.None if you prefer
        }
        set
        {
            if (value == null)
                AddressToken = null;
            else
                // Throw an exception if value is not valid JSON.
                AddressToken = JToken.Parse(value);
        }
    }
}

此实现不需要转换器。

2020-07-27