反序列化 JSON 对象时的拆分字段

Question

我在JSON中有这个（数据更改以保护有罪的人）：

    {"members" : [{"Name":"Mick Jagger","Status":"ACTIVE","ExpireDate":"11/19/2015","TermType":"MONTH","State":"NY","Address2":"","Address1":"123 Anystreet","Type":"PREMIUM","EmailAddress":"[e-mail]","Zip":"12345","Country":"US","City":"New York","Term":"12"},
{"Name":"Keith Richards","Status":"ACTIVE","ExpireDate":"11/19/2015","TermType":"ANNUAL","State":"CT","Address2":"","Address1":"5150 Main St","Type":"PREMIUM","EmailAddress":"[e-mail]","Zip":"45678","Country":"US","City":"New York","Term":"1"}]}

我有这个课程：

public class Member
{
    public string Company_Num { get; set; }
    public string FirstName { get; set; }
    public string LastName { get; set; }
    public string Status { get; set; }
    public string  ExpireDate { get; set; }
    public string TermType { get; set; }
    public string State { get; set; }
    public string Address2 { get; set; }
    public string Address1 { get; set; }
    public string Type { get; set; }
    public string EmailAddress { get; set; }
    public string Zip { get; set; }
    public string Country { get; set; }
    public string City { get; set; }
    public string Term { get; set; }
}
public class MemberList
{
    public List<Member> members { get; set; }
}

当我将其反序列化为数据集时，它运行良好。但。。。

当我生成 MemberList 类时，我忘记了 JSON 中的"Name"需要拆分为 FirstName 和 LastName。有没有办法反序列化 JSON 对象并同时拆分"名称"？（所以名字是"基思"，姓氏是"理查兹"等等。

这是我将其反序列化为成员列表时的外观：

public static MemberList memList()
    {            
        WebClient atv = new WebClient();
        var data = atv.DownloadString("https://www.somewebsvc.com/memberships");
        MemberList m = Newtonsoft.Json.JsonConvert.DeserializeObject<MemberList>(data);
        return m;
    }

Answer 1

一个可能的解决方案是在模型中创建字段，这些字段将在获取它们时返回所需的值：

public class Member
{
     //other properties
     private string _lastName;
     public string LastName {get 
     {
       if (_lastName == null)
           _lastName = Name.Split(new [] {' '}, 2)[1];
       return _lastName;
     }}
     private string _firstName;
     public string FirstName {get 
     {
        if (_firstName== null)
           _firstName= Name.Split(new [] {' '}, 2)[0];
       return _firstName;
     }}
}

Answer 2

这是一个在你提出问题后立即弄清楚的情况。

我向成员添加了一个 Name 属性，然后在代码返回列表之前添加了一个 foreach 循环。所以现在它看起来像这样：

public static MemberList memList()
{            
    WebClient atv = new WebClient();
    var data = atv.DownloadString("https://www.somewebsvc.com/memberships");
    MemberList m = Newtonsoft.Json.JsonConvert.DeserializeObject<MemberList>(data);
    foreach (Member mb in m.members)
        {
            string[] names = mb.Name.Split(new char[] { ' ' }, 2);
            mb.FirstName = names[0];
            mb.LastName = names[1];
        }
    return m;
}

正如评论中指出的那样，当你必须解析"Mary Ann Smith"或"Nigel Baxter Taylor"时，这是非常丑陋的，但你几乎可以在那里插入任何字符串解析逻辑，它会起作用。客户端只想要名字的第一个单词。如果这让他们高兴，我不会争论。