我有以下JSON和我使用Json.NET(Newtonsoft.Json):C#解析JSON问题
{
"total_items": "62",
"page_number": "6",
"page_size": "10",
"page_count": "7",
"cars": {
"car": [
{
"car_name": "Honda",
"engines": {
"engine": [ <-- HONDA has multiple engines, so this is an array
{
"name": "1.2L"
},
{
"name": "1.8L"
}
]
},
"country": "Japan"
"image": {
"thumb": {
"url": "http://image_path/Honda.jpg" <-- Image provided
}
}
},
{
"car_name": "Ford",
"engines": {
"engine": { <-- FORD has single engine, so this is an object
"name": "2.2L"
}
},
"country": "Japan"
"image": null <-- image is null
},
{
"car_name": "VW",
"engines": null, <-- VW has no engines, so this is null
"country": "Germany"
"image": null <-- image is null
}
]
}
}
而且我有以下Car对象:
class Car
{
public Car() { }
public string Name { get; set; }
public string Country { get; set; }
public List<String> EngineNames { get; set; }
}
我需要处理所有以上3种情况(HONDA阵列,FORD对象阵列,VW空白阵列)。如果它不为空,则获取所有引擎名称。因此,例如上面,我EngineNames列表中3辆汽车将是:
Honda.EngineNames = {"1.2L", "1.8L"} // array in JSON
Ford.EngineNames = {"2.2L"} //object in JSON
VW.EngineNames = null //null in JSON
我需要解析上面的JSON得到车资料。我解析car_name和国家,但我不知道如何通过处理上述3种情况来解析所有引擎名称。
private Cars GetCars(string json)
{
dynamic data = (JObject)JsonConvert.DeserializeObject(json);
foreach (dynamic d in data.cars.car)
{
Car c = new Car();
c.Name = (string)d.SelectToken("car_name");
c.Country = (string)d.SelectToken("country");
// PROBLEM: This works fine for array or null in JSON above (HONDA and VW), but it errors on JSON object (in case of FORD)
// When handling FORD, I get error "'Newtonsoft.Json.Linq.JProperty' does not contain a definition for 'name'"
c.EngineNames = (d.engines != null ? ((IEnumerable)d.engines.engine).Cast<dynamic>().Select(e => (string)e.name) : null);
CarList.Add(c);
}
return CarList;
}
为什么不将单个项目引擎放入一个项目的数组中?像:''引擎“:[{”name“:”2.2L“}]' – Steve
@Steve JSON提供给我,我没有做到。这是怎么回事,我需要解析它。谢谢 – pixel
可能的重复[如何使用JSON.net处理同一个属性的单个项目和数组](http://stackoverflow.com/questions/18994685/how-to-handle-both-a-single- item-and-an-array-for-the-same-property-using-json-n) – Rob