Question

当反序列化以下json响应时，我设法获得订阅者ID，电子邮件和状态，这很容易，因为这些名称始终相同并且不依赖于列表中的特定设置。但是，如何将合并字段和在不同列表中具有不同名称的兴趣反序列化？

{ "members": [ {
    "id": "f777bbffab8d1ceca8b757df63c47cb8",
    "email_address": "urist.mcvankab+1@freddiesjokes.co",
    "unique_email_id": "882e9bec19",
    "status": "subscribed",
    "merge_fields":{
        "FNAME": "",
        "LNAME": "" },
    "interests":{
        "9143cf3bd1": true,
        "3a2a927344": false,
        "f9c8f5f0ff": false,
        "f231b09abc": true,
        "bd6e66465f": false },
//And so on...

如何更改我的类以反序列化响应？

我的会员班

public class Member
{
    public string id { get; set; }
    public string email_address { get; set; }
    public string status { get; set; }
    //public List<MergeField> mergeFields { get; set; }
    //public List<Interests> interests { get; set; }
}

MergeField引用我的旧MergeField类，其中字段是硬编码的

public class MergeField
{
    public string FNAME { get; set; }
    public string LNAME { get; set; }
    public string ADDRESS { get; set; }
    public string MERGE4 { get; set; }
}

但我想在请求列表的合并字段时使用相同的两个Mergefield（s）类来反序列化json响应。

public class MergeFields
{
    public List<Mergefield> merge_fields { get; set; }
    public string list_id { get; set; }
    public int total_items { get; set; }
}

public class Mergefield
{
    public string merge_id { get; set; }
    public string tag { get; set; }
    public string name { get; set; }
    public string list_id { get; set; }
}

我找到了这个答案How do deserialize JSON with non-standard (and varying) property names，但我无法弄清楚如何在我的场景中使用它。

Answer 1

您可以对合并字段使用Dictionary<string, object>类型，为感兴趣使用Dictionary<string, bool>类型。这是Member课程应该是什么样子

public class Member
{
    public string id { get; set; }
    public string email_address { get; set; }
    public string status { get; set; }
    public Dictionary<string, object> merge_fields { get; set; }
    public Dictionary<string, bool> interests { get; set; }
}

假设您正在使用MemberResponse类作为响应，如下所示

public class MemberResponse
{
    public List<Member> members { get; set; }
    public string list_id { get; set; }
    public int total_items { get; set; }
}

以下是获取每个合并字段和兴趣值的方法

MemberResponse memberResponse = .... // your code that calls the API

// enumerate members
foreach (var member in memberResponse.members)
{
    // enumerate merge fields
    foreach (var key in member.merge_fields.Keys)
    {
        // when key = "FNAME", value would be the first name of the member
        // when key = "LNAME", value would be the last name of the member
        string value = member.merge_fields[key].ToString();
    }

    // enumerate interests
    foreach (var key in member.interests.Keys)
    {
        // when key = "9143cf3bd1", value would be true
        // when key = "3a2a927344", value would be false
        // ... and so on
        bool value = member.interests[key];
    }

}

很遗憾没有办法使用name端点获取合并字段的merge_id和/lists/{list_id}/members，因为响应正文只包含tag的{{1}}合并字段，即FNAME，LNAME等。您需要单独调用/lists/{list_id}/merge-fields端点并比较每个合并字段的tag获取相关的name和merge_id。

在不同的MailChimp列表中反序列化具有不同名称的合并字段和兴趣

1 个答案: