我正在尝试将JSON转换为列表,请任何人都可以帮助我 .
public class User
{
public String id { get; set; }
public String imageURL { get; set; }
public String search { get; set; }
public String status { get; set; }
public String username { get; set; }
}
List<User> users = JsonConvert.DeserializeObject<List<User>>(resp.Body);
JSON
{
"KfWE8S9jWJdWnAZEbOtHTtisNwO2":
{"id":"KfWE8S9jWJdWnAZEbOtHTtisNwO2"
,"imageURL":"https://firebasestorage.googleapis.com/v0/b/bchat-af5e5.appspot.com/o/uploads%2F1542785437375.jpg?alt=media&token=be1ce806-fecf-4081-9dad-f0a20e5d8489"
,"search":"rene vizconde"
,"status":"online"
,"username":"Rene Vizconde"},
"ScpDnyQCyKemXSgdo3jEvZFNxY83":
{"id":"ScpDnyQCyKemXSgdo3jEvZFNxY83"
,"imageURL":"default"
,"search":"yeli potpot"
,"status":"offline"
,"username":"Yeli Potpot"},
"cnPYOdHYWaaLDQmchELLvw85DBf1":
{"id":"cnPYOdHYWaaLDQmchELLvw85DBf1"
,"imageURL":"https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcSHilM1ke9pZePBJTobMTcktggiw-UywdqAIpf-VX9nqecKl6b4wQ"
,"search":"bards disc"
,"status":"offline"
,"username":"Bards Disc"},
"tWTbllTxaVM9WQnsNwnBgc3ixLM2":
{"id":"tWTbllTxaVM9WQnsNwnBgc3ixLM2"
,"imageURL":"default"
,"search":"renz angelo"
,"status":"offline"
,"username":"Renz Angelo"}
}
2 回答
您可以简单地将json反序列化为
Dictionary<string, User>
而不是List<User>
.Output:
你的JSON有问题 . 虽然它的结构“语法”正确,但它设计得很糟糕 .
您使用ID命名JSON的成员,然后在该成员的每个对象中引用该ID .
不要重复自己 .
由于成员的名称是动态的,因此您无法使用该根对象来反序列化JSON . (这不是
List<User>
,JSON中没有列表,但是对象)请考虑使用此结构(注意括号以创建对象数组)