我有一个类似的集合:
public class Xml
{
public string version { get; set; }
public string encoding { get; set; }
}
public class Content
{
public string Expires { get; set; }
public string MaxArrivalScope { get; set; }
}
public class Trip
{
public string ETA { get; set; }
public string TripNo { get; set; }
public string WheelchairAccess { get; set; }
}
public class Destination
{
public string Name { get; set; }
public List<Trip> Trip { get; set; }
}
public class Route
{
public string RouteNo { get; set; }
public string Name { get; set; }
public Destination Destination { get; set; }
}
public class Platform
{
public string PlatformTag { get; set; }
public string Name { get; set; }
public Route Route { get; set; }
}
public class JPRoutePositionET
{
public string xmlns { get; set; }
public string xsi { get; set; }
public string schemaLocation { get; set; }
public Content Content { get; set; }
public Platform Platform { get; set; }
}
public class RootObject
{
public Xml xml { get; set; }
public JPRoutePositionET JPRoutePositionET { get; set; }
}
}
我有这样的JSON:
{
"xml": {
"version": "1.0",
"encoding": "utf-8"
},
"JPRoutePositionET": {
"xmlns": "urn:connexionz-co-nz:jp",
"xsi": "http://www.w3.org/2001/XMLSchema-instance",
"schemaLocation": "urn:connexionz-co-nz:jp JourneyPlanner.xsd",
"Content": {
"Expires": "2017-04-09T15:59:31+12:00",
"MaxArrivalScope": "60"
},
"Platform": {
"PlatformTag": "2628",
"Name": "Awatea Rd near Awatea Gdns",
"Route": {
"RouteNo": "125",
"Name": "Redwood/Westlake",
"Destination": {
"Name": "Westlake & Halswell",
"Trip": [
{
"ETA": "4",
"TripNo": "4751",
"WheelchairAccess": "true"
},
{
"ETA": "32",
"TripNo": "4752",
"WheelchairAccess": "true"
}
]
}
}
}
}
}
为什么Newtonsoft无法正确解析Platform
类?它返回null
。我是否需要格式化JSON以删除我不想要的所有其他信息?还是我想念的东西?
答案 0 :(得分:0)
法比奥对
的评论 var result = Newtonsoft.Json.JsonConvert.DeserializeObject<RootObject>(jsonString);
的工作。
但是现在当我得到像pastebin.com/pebp178这样的JSON时,我似乎无法使用Json2CSharp来获取我需要的类,因为它没有使Destination成为List,它也会改变访问对象而不是类。这真的很奇怪。它只是抛出一个错误,说JSON无法被解析。