我正在使用返回JSON数据的Web服务,并且在很多情况下,服务会在一个对象中返回几个属性,我想将它们分组到C#端的类中。考虑一个类结构,如:
class Person
{
public Address Address { get; set; }
public string Name { get; set; }
}
class Address
{
public string StreetAddress { get; set; }
public string City { get; set; }
public string ZipCode { get; set; }
}
和JSON数据一样:
{ "Name" : "Pilchie",
"StreetAddress" : "1234 Random St",
"City" : "Nowheretown",
"Zip" : "12345"
}
是否可以归因于我的Person
和Address
类,以便它们序列化/反序列化为此格式?
答案 0 :(得分:2)
我认为你不能让JSON.NET一次性完成这一切 - 你必须手动创建Person对象。但是,您可以在不创建单独的DTO类的情况下完成此操作。例如:
var jsonText = "{ \"Name\" : \"Pilchie\"," +
"\"StreetAddress\" : \"1234 Random St\"," +
"\"City\" : \"Nowheretown\"," +
"\"Zip\" : \"12345\"" +
"}";
JObject jsonObject = (JObject) JsonConvert.DeserializeObject(jsonText);
var person =
new Person
{
Address = new Address
{
City = (String) jsonObject["City"],
StreetAddress = (String) jsonObject["StreetAddress"],
ZipCode = (string) jsonObject["Zip"]
},
Name = (string) jsonObject["Name"]
};
序列化:
JsonConvert.SerializeObject(
new
{
person.Name,
person.Address.StreetAddress,
person.Address.City,
Zip = person.Address.ZipCode
});
答案 1 :(得分:0)
我认为,对您正在使用的数据进行反序列化的最简单方法是创建一个与JSON数据格式匹配的简单DTO对象。然后,您可以使用AutoMapper或类似的库轻松地将数据映射到新结构。
答案 2 :(得分:0)
这完全取决于您将如何使用数据。如果您只想访问地址属性,可以执行以下操作:
class Person
{
public string Name { get; set; }
public string StreetAddress { get; set; }
public string City { get; set; }
public string ZipCode { get; set; }
[ScriptIgnore]
public Address Address { get {return new Address(){StreetAddress = this.StreetAddress,
City = this.City,
ZipCode = this.ZipCode} } }
}
答案 3 :(得分:0)
var person = JsonConvert.DeserializeObject<Person>(json);
class Person
{
[JsonProperty("StreetAddress")]
private string _StreetAddress { get; set; }
[JsonProperty("City")]
private string _City { get; set; }
[JsonProperty("Zip")]
private string _ZipCode { get; set; }
public string Name { get; set; }
public Address Address
{
get
{
return new Address() { City = _City, StreetAddress = _StreetAddress, ZipCode = _ZipCode };
}
}
}
class Address
{
public string StreetAddress { get; set; }
public string City { get; set; }
public string ZipCode { get; set; }
}