json - 将 JSON 反序列化为扁平化类

Question

我在这里发现了同样的问题...

Deserializing nested JSON structure to a flattened class with Json.NET using annotations

...但没有正确的答案。 最好的建议之一是将嵌套对象包装在一个新类中，但这种方法引入了另一个问题:乐高名称。在我的示例中，此类的最逻辑名称与父类和当然不可能。我的例子很简单，我只是想消除父类中的“语言”属性。有人可以帮我做吗？

using Newtonsoft.Json;

public partial class NamedType
{
    public string Name { get; set; }
}

public class Proficiency
{
    public string Level { get; set; }

    public string Name { get; set; }
}

public class Language
{
    public int Id { get; set; }

    public string Name { get; set; }

    //public Language Language { get; set; } //Compiler error
    //public Language Value { get; set; } //Not correct
    //public NamedType Language { get; set; } //Compiler error
    //public NamedType Value { get; set; } //Ugly, isn't?

    public Proficiency Proficiency { get; set; }
}

List<Language> languageList = JsonConvert.DeserializeObject<List<Language>>(json);

json 示例:

{
    "languages": [
        {
            "id": 1,
            "language": { "name": "Spanish" },
            "proficiency": {
                "level": "native_or_bilingual",
                "name": "Native or bilingual proficiency"
            }
        },
        {
            "id": 2,
            "language": { "name": "English" },
            "proficiency": {
                "level": "full_professional",
                "name": "Full professional proficiency"
            }
        },
        {
            "id": 3,
            "language": { "name": "Japanese" },
            "proficiency": {
                "level": "elementary",
                "name": "Elementary proficiency"
            }
        }
    ]
}

Answer 1

如果 JSON 属性名称与 C# 命名约定冲突，您可以在序列化期间使用 DataMember 或 JsonProperty 注释来替换不同的名称。

例如，以下代码适用于 DataContractJsonSerializer 和 Json.NET:

[DataContract]
public class Language
{
    [DataContract]
    class NamedType
    {
        [DataMember]
        public string name { get; set; }
    }

    [DataContract]
    class ProficiencyType
    {
        [DataMember]
        public string level { get; set; }
        [DataMember]
        public string name { get; set; }
    }

    [DataMember(Name="id")]
    public int Id { get; set; }

    [IgnoreDataMember] // Do not serialize this property
    public string Name { get; set; }

    [IgnoreDataMember]
    public string ProficiencyLevel { get; set; }

    [IgnoreDataMember]
    public string ProficiencyName { get; set; }

    [DataMember(Name="language")] // serialize this nested class property with name "language"
    [JsonProperty(ObjectCreationHandling=ObjectCreationHandling.Replace)] // When deserializing, always create a fresh instance instead of reusing the proxy class.
    NamedType LanguageName
    {
        get
        {
            return new NamedType { name = Name };
        }
        set
        {
            Name = (value == null ? null : value.name);
        }
    }

    [DataMember(Name = "proficiency")]
    [JsonProperty(ObjectCreationHandling = ObjectCreationHandling.Replace)]
    ProficiencyType Proficiency
    {
        get
        {
            return new ProficiencyType { level = ProficiencyLevel, name = ProficiencyName };
        }
        set
        {
            ProficiencyLevel = (value == null ? null : value.level);
            ProficiencyName = (value == null ? null : value.name);
        }
    }
}

如果您发现 DataContract 属性的选择加入性质很麻烦，并且更喜欢使用 Json.NET 特定的属性，那么以下内容是等效的:

public class Language
{
    class NamedType
    {
        public string name { get; set; }
    }

    class ProficiencyType
    {
        public string level { get; set; }
        public string name { get; set; }
    }

    [JsonProperty(PropertyName = "id")]
    public int Id { get; set; }

    [JsonIgnore]
    public string Name { get; set; }

    [JsonIgnore]
    public string ProficiencyLevel { get; set; }

    [JsonIgnore]
    public string ProficiencyName { get; set; }

    [JsonProperty(PropertyName = "language", ObjectCreationHandling = ObjectCreationHandling.Replace)]
    NamedType LanguageName
    {
        get
        {
            return new NamedType { name = Name };
        }
        set
        {
            Name = (value == null ? null : value.name);
        }
    }

    [JsonProperty(PropertyName = "proficiency", ObjectCreationHandling = ObjectCreationHandling.Replace)]
    ProficiencyType Proficiency
    {
        get
        {
            return new ProficiencyType { level = ProficiencyLevel, name = ProficiencyName };
        }
        set
        {
            ProficiencyLevel = (value == null ? null : value.level);
            ProficiencyName = (value == null ? null : value.name);
        }
    }
}