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

Posted 2021-03-17 yufeiyunsui

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

d.data.replace(/LON:/g, "lng:")

string json = "{\\"Name\\" : \\"Jack\\", \\"Age\\" : 34, \\"Colleagues\\" : [{\\"Name\\" : \\"Tom\\" , \\"Age\\":44},{\\"Name\\" : \\"Abel\\",\\"Age\\":29}] }";
   
//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()