在 JsonConverter.ReadJson() 中有效地获取完整的 json 字符串

public static JRaw Create(JsonReader reader){使用 (StringWriter sw = new StringWriter(CultureInfo.InvariantCulture))使用 (JsonTextWriter jsonWriter = new JsonTextWriter(sw)){jsonWriter.WriteToken(reader);返回新的 JRaw(sw.ToString());}}

生成的 JRaw 只是将该字符串封装在其 值.(当然，不能保证生成的 JSON 代表一个对象，只能代表格式正确的 JSON.)

请注意，JsonTextReader 会自动识别和解析 常用格式为DateTime对象，也将浮点值解析为double.如果您需要最字面"的 JSON 字符串，您可能希望抑制 DateTime 识别和/或将浮点值解析为 decimal.以下以 JRaw.Create() 为模型的扩展方法完成了这项工作:

public static string ReadOuterJson(this JsonReader reader, Formatting formatting = Formatting.None, DateParseHandling? dateParseHandling = null, FloatParseHandling? floatParseHandling = null){//如果您希望空 JSON 值返回空字符串，请删除此行:if (reader.TokenType == JsonToken.Null)返回空值；var oldDateParseHandling = reader.DateParseHandling;var oldFloatParseHandling = reader.FloatParseHandling;尝试{if (dateParseHandling != null)reader.DateParseHandling = dateParseHandling.Value;if (floatParseHandling != null)reader.FloatParseHandling = floatParseHandling.Value;使用 (var sw = new StringWriter(CultureInfo.InvariantCulture))使用 (var jsonWriter = new JsonTextWriter(sw) { Formatting =formatting }){jsonWriter.WriteToken(reader);返回 sw.ToString();}}最后{reader.DateParseHandling = oldDateParseHandling;reader.FloatParseHandling = oldFloatParseHandling;}}

然后这样称呼它，例如:

var json = reader.ReadOuterJson(dateParseHandling: DateParseHandling.None);

有关为什么需要这样做的详细信息，请参阅:

• Json.NET 在反序列化时解释和修改 ISO 日期JObject #862.

• JObject.Parse 修改浮点值的结尾.

How can I efficiently get full json string in JsonConverter.ReadJson() ?

I can do:

public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer) {
      var json = JObject.Load(reader).ToString(Formatting.None);

However that seems very inefficient as I serialize and deserialize for no reason

Any better way ?

ReadJson() must fully parse the JSON being read so that the JSON is confirmed to be well-formed and the JsonReader is correctly positioned at the end of the current value upon exit. However, it is not necessary to load the entire JSON into an intermediate JObject hierarchy simply to re-convert it to a JSON string. Instead, you may be able to get better performance by using JRaw.Create():

var json = JRaw.Create(reader).ToString();

As can be seen in the reference source, this method streams directly from the incoming JsonReader to a StringWriter - without loading into an intermediate JToken hierarchy and re-serializing - by using JsonWriter.WriteToken(JsonReader):

public static JRaw Create(JsonReader reader)
{
    using (StringWriter sw = new StringWriter(CultureInfo.InvariantCulture))
    using (JsonTextWriter jsonWriter = new JsonTextWriter(sw))
    {
        jsonWriter.WriteToken(reader);

        return new JRaw(sw.ToString());
    }
}

The resulting JRaw simply encapsulates that string in its Value. (Of course, there is no guarantee that the resulting JSON represents an object, only that it represents well-formed JSON.)

Note that JsonTextReader will automatically recognize and parse dates and times in common formats as DateTime objects, and also parse floating point values as double. If you need the "most literal" JSON string you may want to suppress DateTime recognition and/or parse floating point values as decimal. The following extension method, modeled on JRaw.Create(), does the job:

public static string ReadOuterJson(this JsonReader reader, Formatting formatting = Formatting.None, DateParseHandling? dateParseHandling = null, FloatParseHandling? floatParseHandling = null)
{
    // If you would prefer a null JSON value to return an empty string, remove this line:
    if (reader.TokenType == JsonToken.Null)
        return null;
    var oldDateParseHandling = reader.DateParseHandling;
    var oldFloatParseHandling = reader.FloatParseHandling;
    try
    {
        if (dateParseHandling != null)
            reader.DateParseHandling = dateParseHandling.Value;
        if (floatParseHandling != null)
            reader.FloatParseHandling = floatParseHandling.Value;
        using (var sw = new StringWriter(CultureInfo.InvariantCulture))
        using (var jsonWriter = new JsonTextWriter(sw) { Formatting = formatting })
        {
            jsonWriter.WriteToken(reader);
            return sw.ToString();
        }
    }
    finally
    {
        reader.DateParseHandling = oldDateParseHandling;
        reader.FloatParseHandling = oldFloatParseHandling;
    }
}

And then call it like, e.g.:

var json = reader.ReadOuterJson(dateParseHandling: DateParseHandling.None);

For details on why this may be necessary, see:

• Json.NET interprets and modifies ISO dates when deserializing to JObject #862.

• JObject.Parse modifies end of floating point values.

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