我有以下Json响应:
{
"Customers": [
{
"Customer": {
"Address": {
"City": "Stockholm",
"PostalCode": "10123"
},
"Classifications": [
"LoyaltyProgram",
"Returning",
"VeryImportant"
],
"FirstName": "Peter",
"LastName": "Centers",
"Passport": {
"Expiration": "2019-01-14",
"Number": "1564931321655"
},
},
"FirstName": "Peter",
"LastName": "Centers",
"Reservation": {
"AdultCount": 2,
"AssignedSpaceId": "03f59360-8644-4e29-927a-ad85a6514466",
},
"RoomNumber": "302"
},
]
}
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我为每位客户提供以下课程:
public class CustomerDto
{
public string FirstName { get; set; }
public string LastName { get; set; }
public List<string> Classifications { get; set; }
public PassportDto Passport { get; set; }
public AddressDto Address { get; set; }
}
public class AddressDto
{
public string City { get; set; }
public string PostalCode { get; set; }
}
public class PassportDto
{
public string Expiration { get; set; }
public string Number { get; set; }
}
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从这里我使用Json.Net和我的一个方法(摘录)中的以下代码,其中下面的客户是响应:
var jsonCustomers = JObject.Parse(customers)["Customers"].Children().ToList();
IList<CustomerDto> customerList = new List<CustomerDto>();
foreach (var item in jsonCustomers) {
customerList.Add(item.ToObject<CustomerDto>());
}
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除了Address和Passport之外,CustomerDto中的所有值都被填充,这些值为null,我无法弄清楚原因.
您可以创建一个包含客户列表的类:
public class CustomerList
{
public IList<CustomerDto> Customers { get; set; }
}
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并且反序列化对象只需调用:
CustomerList jsonCustomers = JsonConvert.DeserializeObject<CustomerList>(customers);
IList<CustomerDto> customerList = jsonCustomers.Customers;
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编辑:没有注意到客户是json数组中对象列表的嵌套属性,因此需要另一个包装的类CustomerDto.@mwwills发布了完整的答案.
添加两个新类:
public class CustomersWrapper
{
public IEnumerable<CustomerWrapper> Customers { get; set; }
}
public class CustomerWrapper
{
public CustomerDto Customer { get; set; }
}
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然后将所有现有代码替换为:
var results = JsonConvert.DeserializeObject<CustomersWrapper>(input);
var customerList = results.Customers.Select(z => z.Customer).ToList();
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这将确保对层次结构中的所有对象进行标准反序列化。
由于JSON的结构奇特,因此需要使用此功能。/sf/answers/3176905651/是相同的基本问题(可能是值得一读) -基本上你应该没有真正的Customers和Customer你的JSON。没有这些,您将不需要我指定的两个包装器类。
您可能还希望避免在JSON中指定FirstName和LastName两次(不必要)。