我有一个JSON文件的结构:
[ {"unit_id": {"type":[string],"customer_id":[int]} },
...,
...]
我想把它转换成一个字典(或任何有用的)的结构:
Dictionary<int,Tuple<string,int>>
我正在尝试以下内容:
Dictionary<int, Tuple<string,int>> units =
JsonConvert.DeserializeObject<Dictionary<int, Tuple<string, int>>>
(File.ReadAllText(jsonFile));
失败,因为文件不能反序列化为该结构。我也试过创建一个类:
class Unit{
public int unitID;
public Tuple<string, int> details;
}
然后:
List<Unit> units = JsonConvert.DeserializeObject<List<Unit>>(File.ReadAllText(jsonFile));
它不会失败,但不会用任何值填充列表。
谢谢
对于JSON结构,您可能需要创建这样的类:
class Unit
{
public CustomerType UnitId;
}
class CustomerType
{
public int CustomId { get; set; }
public string Type { get; set; }
}
或者使用dynamic
:
Unit unit = new Unit();
unit.UnitId = new CustomerType()
{
CustomId = 1001,
Type = "Customer"
};
//generate test json string
string jsonTest = JsonConvert.SerializeObject(unit);
//convert to dynamic
var result = JsonConvert.DeserializeObject<dynamic>(jsonTest);
Console.WriteLine(result.UnitId.CustomId);
Console.WriteLine(result.UnitId.Type);
所以我设法用以下(混乱的)代码创建了我的字典:
public static Dictionary<int, Tuple<string, int>> parseJsonIntoDictionary(string jsonFile) {
Dictionary<int, Tuple<string, int>> unitDictionary = new Dictionary<int,Tuple<string,int>>();
var json = System.IO.File.ReadAllText(jsonFile);
var units = JArray.Parse(json);
foreach (JToken unit in units) {
JProperty property = ((JObject)unit).Properties().ToArray()[0];
int unitID = Convert.ToInt32(property.Name);
Tuple<string,int> tempTuple = new Tuple<string,int>(Convert.ToString(unit[property.Name]["unit_type"]),Convert.ToInt32(unit[property.Name]["customer_id"]));
unitDictionary[unitID] = tempTuple;
}
return unitDictionary;
}