Question

我有一个JSON数组，包含具有不同属性的不同类型的对象。其中一个属性称为“类型”，它确定数组项的类型。以下是我的数据示例：

   [{
        type : "comment",
        text : "xxxx"
    }, {
        type : "code",
        tokens : [{
                type : "ref",
                data : "m"
            }, {
                type : "operator",
                data : "e"
            }
        ]
    }, {
        type : "for",
        boundLocal : {
            type : "local",
            name : "i",
            kind : "Number"
        },
        upperBound : {
            type : "ref",
            tokens : [{
                    type : "operator",
                    data : "3"
                }, {
                    type : "operator",
                    data : "0"
                }
            ]
        },
        body : [{
                type : "code",
                tokens : [{
                        type : "ref",
                        data : "x"
                    }
                ]
            }, {
                type : "code",
                tokens : [{
                        type : "ref",
                        data : "y"
                    }
                }
                ]
        ]
    ]

要将这些对象映射到我的.Net实现，我定义了一组类：一个基类和几个子类（具有复杂的层次结构，具有4个“代”）。以下是这些类的一个小例子：

public abstract class TExpression
{
    [JsonProperty("type")]
    public string Type { get; set; }
}

public class TComment : TExpression
{
    [JsonProperty("text")]
    public string Text { get; set; }
}   

public class TTokenSequence : TExpression
{
    [JsonProperty("tokens")]
    public List<TToken> Tokens { get; set; }
}

我想要达到的目的是能够将此数组反序列化为协变通用列表，声明为：

List<TExpression> myexpressions = JsonConvert.DeserializeObject<List<TExpression>>(aststring);

此列表应包含从TExpression继承的相应子类的实例，因此我稍后可以在代码中使用以下代码：

foreach(TExpression t in myexpressions)
{
    if (t is TComment) dosomething;
    if (t is TTokenSequence) dosomethingelse;
}

如何使用JSON.NET访问它？

Answer 1

以下是使用CustomCreationConverter的示例。

public class JsonItemConverter :  Newtonsoft.Json.Converters.CustomCreationConverter<Item>
{
    public override Item Create(Type objectType)
    {
        throw new NotImplementedException();
    }

    public Item Create(Type objectType, JObject jObject)
    {
        var type = (string)jObject.Property("valueType");
        switch (type)
        {
            case "int":
                return new IntItem();
            case "string":
                return new StringItem();
        }

        throw new ApplicationException(String.Format("The given vehicle type {0} is not supported!", type));
    }

    public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
    {
        // Load JObject from stream
        JObject jObject = JObject.Load(reader);

        // Create target object based on JObject
        var target = Create(objectType, jObject);

        // Populate the object properties
        serializer.Populate(jObject.CreateReader(), target);

        return target;
    }
}

public abstract class Item
{
    public string ValueType { get; set; }

    [JsonProperty("valueTypeId")]
    public int ValueTypeId { get; set; }

    [JsonProperty("name")]
    public string Name { get; set; }

    public new virtual string ToString() { return "Base object, we dont' want base created ValueType=" + this.ValueType + "; " + "name: " + Name; }
}

public class StringItem : Item
{
    [JsonProperty("value")]
    public string Value { get; set; }

    [JsonProperty("numberChars")]
    public int NumberCharacters { get; set; }

    public override string ToString() { return "StringItem object ValueType=" + this.ValueType + ", Value=" + this.Value + "; " + "Num Chars= " + NumberCharacters; }

}

public class IntItem : Item
{
    [JsonProperty("value")]
    public int Value { get; set; }

    public override string ToString() { return "IntItem object ValueType=" + this.ValueType + ", Value=" + this.Value; }
}

class Program
{
    static void Main(string[] args)
    {
        // json string
        var json = "[{\"value\":5,\"valueType\":\"int\",\"valueTypeId\":1,\"name\":\"numberOfDups\"},{\"value\":\"some thing\",\"valueType\":\"string\",\"valueTypeId\":1,\"name\":\"a\",\"numberChars\":11},{\"value\":2,\"valueType\":\"int\",\"valueTypeId\":2,\"name\":\"b\"}]";

        // The above is deserialized into a list of Items, instead of a hetrogenous list of
        // IntItem and StringItem
        var result = JsonConvert.DeserializeObject<List<Item>>(json, new JsonItemConverter());

        foreach (var r in result)
        {
            // r is an instance of Item not StringItem or IntItem
            Console.WriteLine("got " + r.ToString());
        }
    }
}

Answer 2

也可以使用我的JsonSubTypes库以声明方式实现它：

[JsonConverter(typeof(JsonSubtypes), "valueType")]
[JsonSubtypes.KnownSubType(typeof(IntItem), "int")]
[JsonSubtypes.KnownSubType(typeof(StringItem), "string")]
public abstract class Item
{
    public string ValueType { get; set; }

    [JsonProperty("valueTypeId")]
    public int ValueTypeId { get; set; }

    [JsonProperty("name")]
    public string Name { get; set; }

    public override string ToString()
    {
        return "Base object, we dont' want base created ValueType=" + this.ValueType + "; " + "name: " + Name;
    }
}

public class StringItem : Item
{
    [JsonProperty("value")]
    public string Value { get; set; }

    [JsonProperty("numberChars")]
    public int NumberCharacters { get; set; }

    public override string ToString()
    {
        return "StringItem object ValueType=" + this.ValueType + ", Value=" + this.Value + "; " +
               "Num Chars= " + NumberCharacters;
    }
}

public class IntItem : Item
{
    [JsonProperty("value")]
    public int Value { get; set; }

    public override string ToString()
    {
        return "IntItem object ValueType=" + this.ValueType + ", Value=" + this.Value;
    }
}

[TestMethod]
public void Demo()
{
    // json string
    var json =
        "[{\"value\":5,\"valueType\":\"int\",\"valueTypeId\":1,\"name\":\"numberOfDups\"}," +
        "{\"value\":\"some thing\",\"valueType\":\"string\",\"valueTypeId\":1,\"name\":\"a\",\"numberChars\":11}," +
        "{\"value\":2,\"valueType\":\"int\",\"valueTypeId\":2,\"name\":\"b\"}]";

    var result = JsonConvert.DeserializeObject<List<Item>>(json);

    Assert.AreEqual("IntItem object ValueType=int, Value=5", result[0].ToString());
    Assert.AreEqual("StringItem object ValueType=string, Value=some thing; Num Chars= 11", result[1].ToString());
    Assert.AreEqual("IntItem object ValueType=int, Value=2", result[2].ToString());
}

Answer 3

CustomCreationConverter应该能够解决这个问题。

将异构JSON数组反序列化为协变List＆lt;＆gt;使用JSON.NET

3 个答案: