42

我正在使用 Json.Net 进行序列化。我有一堂带字典的课：

public class Test
{
    public string X { get; set; }

    public Dictionary<string, string> Y { get; set; }
}

我可以以某种方式序列化此对象以获取以下 JSON

{
    "X" : "value",
    "key1": "value1",
    "key2": "value2"
}

其中“key1”、“key2”是字典中的键？

4

3 回答 3

59

如果您使用的是Json.Net 5.0.5或更高版本，并且您愿意将字典的类型从 更改Dictionary<string, string>为Dictionary<string, object>，那么完成您想要的一种简单方法是将[JsonExtensionData]属性添加到您的字典属性中，如下所示：

public class Test
{
    public string X { get; set; }

    [JsonExtensionData]
    public Dictionary<string, object> Y { get; set; }
}

标记字典的键和值将被序列化为父对象的一部分。好处是它也适用于反序列化：JSON 中与类成员不匹配的任何属性都将被放入字典中。

于 2014-05-21T14:26:25.513 回答
9

实现JsonConverter派生类：应使用CustomCreationConverter类作为基类来创建自定义对象。

转换器的草稿版本（可以根据需要改进错误处理）：

internal class TestObjectConverter : CustomCreationConverter<Test>
{
    #region Overrides of CustomCreationConverter<Test>

    public override Test Create(Type objectType)
    {
        return new Test
            {
                Y = new Dictionary<string, string>()
            };
    }

    public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
    {
        writer.WriteStartObject();

        // Write properties.
        var propertyInfos = value.GetType().GetProperties();
        foreach (var propertyInfo in propertyInfos)
        {
            // Skip the Y property.
            if (propertyInfo.Name == "Y")
                continue;

            writer.WritePropertyName(propertyInfo.Name);
            var propertyValue = propertyInfo.GetValue(value);
            serializer.Serialize(writer, propertyValue);
        }

        // Write dictionary key-value pairs.
        var test = (Test)value;
        foreach (var kvp in test.Y)
        {
            writer.WritePropertyName(kvp.Key);
            serializer.Serialize(writer, kvp.Value);
        }
        writer.WriteEndObject();
    }

    public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
    {
        JObject jsonObject = JObject.Load(reader);
        var jsonProperties = jsonObject.Properties().ToList();
        var outputObject = Create(objectType);

        // Property name => property info dictionary (for fast lookup).
        var propertyNames = objectType.GetProperties().ToDictionary(pi => pi.Name, pi => pi);
        foreach (var jsonProperty in jsonProperties)
        {
            // If such property exists - use it.
            PropertyInfo targetProperty;
            if (propertyNames.TryGetValue(jsonProperty.Name, out targetProperty))
            {
                var propertyValue = jsonProperty.Value.ToObject(targetProperty.PropertyType);
                targetProperty.SetValue(outputObject, propertyValue, null);
            }
            else
            {
                // Otherwise - use the dictionary.
                outputObject.Y.Add(jsonProperty.Name, jsonProperty.Value.ToObject<string>());
            }
        }

        return outputObject;
    }

    public override bool CanWrite
    {
        get { return true; }
    }

    #endregion
}

客户端代码：

var test = new Test
    {
        X = "123",
        Y = new Dictionary<string, string>
            {
                { "key1", "value1" },
                { "key2", "value2" },
                { "key3", "value3" },
            }
    };

string json = JsonConvert.SerializeObject(test, Formatting.Indented, new TestObjectConverter());
var deserializedObject = JsonConvert.DeserializeObject<Test>(json);

请注意：属性名和字典的键名之间可能存在冲突。

于 2013-02-15T12:41:18.177 回答
2

您可以创建此转换器，然后将其分配给您的财产。采取了一些建议的解决方案。

public class DictionaryToJsonObjectConverter : JsonConverter
    {
        public override bool CanConvert(Type objectType)
        {
            return typeof(IDictionary<string, string>).IsAssignableFrom(objectType);
        }

        public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
        {
            throw new NotImplementedException();
        }

        public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
        {
            writer.WriteRawValue(JsonConvert.SerializeObject(value, Formatting.Indented));
        }
    }

然后在你的 poco 类中使用它。

public class Poco
{
        [JsonProperty("myid")]
        public string Id{ get; set; }

        [JsonProperty("properties")]
        [JsonConverter(typeof(DictionaryToJsonObjectConverter))]
        public IDictionary<string, string> Properties { get; set; }
    }
于 2017-09-05T17:14:51.403 回答