将JSON反序列化为扁平化的类

小编典典

将JSON反序列化为扁平化的类

json

我在这里发现了同样的问题…

…但是没有正确的答案。 最好的建议之一是将嵌套对象包装到新类中，但是这种方法引入了另一个问题：乐高名称。
在我的示例中，此类的最逻辑名称是与父类相同的名称，当然这是不可能的。我的示例很简单，我只想消除父类中的“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的示例：

{
    "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"
            }
        }
    ]
}

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2020-07-27

共1个答案

小编典典

如果JSON属性名称与c＃命名约定冲突，则可以在序列化期间使用DataMember或JsonProperty批注替换其他名称。

例如，以下内容可与DataContractJsonSerializerJson.NET一起使用：

[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);
        }
    }
}

如果发现DataContract属性的选择加入性质令人讨厌，并且希望使用特定于Json.NET的属性，则以下内容等效：

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);
        }
    }
}

2020-07-27