c# - 如何使用包含 "raw"JSON 的字符串成员序列化和反序列化类型，而无需在此过程中转义 JSON

Question

我想将 JSON 反序列化为对象，但我不想反序列化嵌套的 JSON，嵌套的 JSON 应该转换为 JSON 列表(请检查“我的预期输出”以获得清晰的想法)。 ..

//假设我有以下 JSON 数据，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...

Answer 1

您的问题是，如何使用包含“原始”JSON 的 string 成员对类型进行序列化和反序列化，而无需在此过程中转义 JSON？

这可以通过 custom JsonConverter 来完成使用 JsonWriter.WriteRawValue() 读写原始 JSON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);
    }
}

然后按如下方式将其应用于您的类型:

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; }
}

样本 fiddle .

请注意，原始 JSON 字符串必须代表有效的 JSON。如果没有，则创建的 JSON 将不可读。如果你想保证 JSON 文字是有效的，你可以在内部保持 JSON 处于解析状态:

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。