JObject 操作; Linq to JSON ;将json转换为JObject

https://www.cnblogs.com/gygang/p/8954376.html

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string json = "{\"Name\" : \"Jack\", \"Age\" : 34, \"Colleagues\" : [{\"Name\" : \"Tom\" , \"Age\":44},{\"Name\" : \"Abel\",\"Age\":29}] }";    <br>//将json转换为JObject             JObject jObj = JObject.Parse(json);             //通过属性名或者索引来访问，仅仅是自己的属性名，而不是所有的             JToken ageToken =  jObj["Age"];             Console.WriteLine(ageToken.ToString());     jObj["Age"] = 35;             Console.WriteLine(jObj.ToString());      JToken colleagues = jObj["Colleagues"];             colleagues[0]["Age"] = 45;             jObj["Colleagues"] = colleagues;//修改后，再赋给对象             Console.WriteLine(jObj.ToString());    jObj.Remove("Colleagues");//跟的是属性名称             Console.WriteLine(jObj.ToString());             jObj["Age"].Parent.AddAfterSelf(new JProperty("Department", "Personnel Department"));    JObject linda = new JObject(new JProperty("Name", "Linda"), new JProperty("Age", "23"));             jObj["Colleagues"].Last.AddAfterSelf(linda);           JToken name = jObj.SelectToken("Name");     var names = jObj.SelectToken("Colleagues").Select(p => p["Name"]).ToList();            foreach (var name in names)                 var tends = from c in jObj.First.First.First.First.Children()