JToken:获取原始/原始JSON值 [英] JToken: Get raw/original JSON value

问题描述

是否可以从JToken获取原始/原始JSON值?

Is there a way to get the raw/original JSON value from a JToken?

问题:

var data = JObject.Parse(@"{
    ""SimpleDate"":""2012-05-18T00:00:00Z"",
    ""PatternDate"":""2012-11-07T00:00:00Z""
}");

var value = data["SimpleDate"].Value<string>();

value现在是05/18/2012 00:00:00，但是我需要原始字符串2012-05-18T00:00:00Z.

The value is now 05/18/2012 00:00:00 but I need the original string 2012-05-18T00:00:00Z.

有没有办法获得这个原始值?另外，我不能更改JObject的创建方式(例如更改设置)，因为它会作为参数传递给我的课程...

Is there a way to get this original value? Also, I cannot change the way how the JObject is created (e.g. change settings), because it is passed as parameter into my class...

(参考:原始的NJsonSchema问题)

推荐答案

您无法获取原始字符串，可以识别日期字符串并将其转换为DateTime结构. com/json/help/html/T_Newtonsoft_Json_JsonReader.htm"rel =" noreferrer> JsonReader 本身.如果您这样做，可以看到以下内容:

You cannot get the original string, date strings are recognized and converted to DateTime structs inside the JsonReader itself. You can see this if you do:

Console.WriteLine(((JValue)data["SimpleDate"]).Value.GetType()); // Prints System.DateTime

但是，您可以执行以下操作，以 ISO 8601 格式提取日期:

You can, however, extract the dates in ISO 8601 format by doing:

var value = JsonConvert.SerializeObject(data["SimpleDate"]);
// value is "2012-05-18T00:00:00Z"

这将始终以JSON合适的字符串格式输出JValue.由于您的原始日期采用这种格式，因此可以满足您的需求.

This will always output a JValue in a JSON-appropriate string format. Since your original dates are in this format, this may meet your needs.

(老实说，我很惊讶 JValue.ToString() 输出日期以非ISO格式，假设 JObject.ToString() 会输出包含日期以ISO格式).

(Honestly, I'm surprised JValue.ToString() outputs dates in non-ISO format, given that JObject.ToString() does output contained dates in ISO format.)

如果您能够在阅读JObject时更改设置，则可以使用 JsonSerializerSettings.DateParseHandling = DateParseHandling.None :

If you were able to change settings while reading your JObject, you could use JsonSerializerSettings.DateParseHandling = DateParseHandling.None:

        var settings = new JsonSerializerSettings { DateParseHandling = DateParseHandling.None };
        var data = JsonConvert.DeserializeObject<JObject>(@"{
            ""SimpleDate"":""2012-05-18T00:00:00Z"",
            ""PatternDate"":""2012-11-07T00:00:00Z""
        }", settings);

        var value = data["SimpleDate"].Value<string>();

        Debug.WriteLine(value); // Outputs 2012-05-18T00:00:00Z

JObject.Parse() 不会超载，它需要，因此您需要使用 DeserializeObject .此设置最终会传播到 JsonReader.DateParseHandling .

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