如何使用 NewtonSoft 更新 JSON 对象的属性 [英] How to update a property of a JSON object using NewtonSoft

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本文介绍了如何使用 NewtonSoft 更新 JSON 对象的属性的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我有一个这样的 JSON 字符串:

I have a JSON string like this:

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

如果值为 "ML""Female",我想将 option 属性更新为 "Male" 如果值为 "FM".

I would like to update the option property to "Male" if the value is "ML" and "Female" if the value is "FM".

我已经到了这一点,但不确定如何继续:

I have got to this point, but am unsure how to proceed:

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

我不知道如何访问 option,它是 value 的子项.

I don't know how to get to the option which is a child of value.

提前致谢.任何指针都会很棒.

Thanks in advance. Any pointers would be great.

推荐答案

可以通过属性作为键来访问对象:

You can access the object by using properties as keys:

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

对于您的示例,请尝试:

For your example, try:

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();

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