我正在编写一个通用方法,该方法采用Dictionary<string, object>
并使用反射来填充类型为T
的对象:
private T HydrateRequestObject<T>(Dictionary<string, object> requestParameters)
{
dynamic requestObject = Activator.CreateInstance<T>();
PropertyInfo[] requestProperties = typeof(T).GetProperties();
foreach (PropertyInfo prop in requestProperties)
{
//Fails with complex json
prop.SetValue(requestObject, requestParameters[prop.Name]);
}
return requestObject;
}
目前,它处理简单的参数和属性,如:
"params" : {
"CompanyID" : 50
}
但我也需要支持复杂的参数,如:
"params": {
"Resource": {
"FirstName": "test",
"LastName": "test",
"MiddleInitial": null,
"Suffix": null,
"Phone1": "test",
"Phone2": null,
"Phone3": null,
"Email": "test",
"AddressID": 47
},
"ResourceSaveType": 0
}
目前我得到一个ArgumentException:"类型Newtonsoft.Json.Linq.JObject
的对象不能转换为类型Resource
"是否有一种方法来遍历json设置对象中的所有属性?还是我的思路不对?
问题是没有人能确切地知道对象应该解析成什么类型——你必须明确地指定它。想象一下,例如你在字典中也有Person类型的对象(具有属性FirstName, LastName) - json库如何知道Resource应该属于Resource类?
明确指定的类型示例:
using System.Collections.Generic;
namespace ConsoleApplication1 {
public class Resource {
public string FirstName { get; set; }
public string LastName { get; set; }
}
class Program {
static void Main(string[] args) {
var foo = @"{
""$type"":""System.Collections.Generic.Dictionary`2[[System.String, mscorlib],[System.Object, mscorlib]], mscorlib"",
""Resource"":{""$type"":""ConsoleApplication1.Resource, ConsoleApplication1"",
""FirstName"":""rrr"",
""LastName"":null}}";
var bar = Newtonsoft.Json.JsonConvert.DeserializeObject<Dictionary<string, object>>(
foo,
new Newtonsoft.Json.JsonSerializerSettings {
TypeNameHandling = Newtonsoft.Json.TypeNameHandling.Objects,
});
}
}
}