如何使用NewtonSoft更新JSON对象的属性

Question

我有一个像这样的JSON字符串:

{
    "code": "GENDER",
    "value": { "option": "ML" }
}

我想更新option属性,"Male"如果值是"ML","Female"如果值是"FM".

我已达到这一点,但我不确定如何继续:

JArray contentobject = (JArray)JsonConvert.DeserializeObject(contentJSON);  
JObject voicgObj = contentobject.Children().FirstOrDefault(ce =>   ce["code"].ToString() == "GENDER") as JObject;
JProperty voicgProp = voicgObj.Property("value");

我不知道如何到达option哪个孩子value.

提前致谢.任何指针都会很棒.

Answer 1

您可以使用属性作为键来访问该对象:

JObject obj = JObject.Parse(json);
string gender = (string)obj["value"]["option"];

对于您的示例,请尝试:

JObject obj = JObject.Parse(json);
var val = obj["value"];
string option = (string)val["option"];

if (option == "ML")
   val["option"] = "Male";

if (option == "FM")
   val["option"] = "Female";

string result = obj.ToString();