c# - 在 C# 中将通用集合转换为 JSON

Question

我在将数据添加到集合中时遇到了一个场景。我需要将一些数据添加到一个集合中，然后将其转换为所需格式的 json。

问题是我没有得到我需要的 json 输出，只有字典集合给了我我需要的输出但是字典不允许其中有重复的键，我需要添加重复的数据。

我尝试了不同的集合，但无法获得所需的输出。

请查看以下代码片段并建议我一个合适的解决方案。

//with distinct emails
var dict = new Dictionary<string, object>();
dict.Add("user1@company.com", new { id = 1, first = "FirstName", last = "LastName" });
dict.Add("user2@company.com", new { id = 2, first = "FirstName", last = "LastName" });
dict.Add("user3@company.com", new { id = 3, first = "FirstName", last = "LastName" });
dict.Add("user4@company.com", new { id = 4, first = "FirstName", last = "LastName" });
string dictJson = new JavaScriptSerializer().Serialize(dict);
//json result (requires output)
//{"user1@company.com":{"id":1,"first":"FirstName","last":"LastName"},"user2@company.com":{"id":2,"first":"FirstName","last":"LastName"},"user3@company.com":{"id":3,"first":"FirstName","last":"LastName"},"user4@company.com":{"id":4,"first":"FirstName","last":"LastName"}}

//Snippet - I: with duplicate emails
var list = new List<KeyValuePair<string, object>>();
list.Add(new KeyValuePair<string, object>("user1@company.com", new { id = 1, first = "FirstName", last = "LastName" }));
list.Add(new KeyValuePair<string, object>("user1@company.com", new { id = 2, first = "FirstName", last = "LastName" }));
list.Add(new KeyValuePair<string, object>("user2@company.com", new { id = 3, first = "FirstName", last = "LastName" }));
list.Add(new KeyValuePair<string, object>("user2@company.com", new { id = 4, first = "FirstName", last = "LastName" }));
string listJson = new JavaScriptSerializer().Serialize(list);
//json result
//[{"Key":"user1@company.com","Value":{"id":1,"first":"FirstName","last":"LastName"}},{"Key":"user1@company.com","Value":{"id":2,"first":"FirstName","last":"LastName"}},{"Key":"user2@company.com","Value":{"id":3,"first":"FirstName","last":"LastName"}},{"Key":"user2@company.com","Value":{"id":4,"first":"FirstName","last":"LastName"}}]

//Snippet - II: with duplicate emails
var tupleList = new List<Tuple<string, CustomClass>>();
tupleList.Add(Tuple.Create("user1@company.com", new CustomClass { id = 1, first = "FirstName", last = "LastName" }));
tupleList.Add(Tuple.Create("user1@company.com", new CustomClass { id = 2, first = "FirstName", last = "LastName" }));
tupleList.Add(Tuple.Create("user2@company.com", new CustomClass { id = 3, first = "FirstName", last = "LastName" }));
tupleList.Add(Tuple.Create("user2@company.com", new CustomClass { id = 4, first = "FirstName", last = "LastName" }));
string tupleListJson = new JavaScriptSerializer().Serialize(tupleList);
//json result
//[{"Item1":"user1@company.com","Item2":{"id":1,"first":"FirstName","last":"LastName"}},{"Item1":"user1@company.com","Item2":{"id":2,"first":"FirstName","last":"LastName"}},{"Item1":"user2@company.com","Item2":{"id":3,"first":"FirstName","last":"LastName"}},{"Item1":"user2@company.com","Item2":{"id":4,"first":"FirstName","last":"LastName"}}]

//Snippet - III: with duplicate emails
var genericList = new List<MainClass>();
genericList.Add(new MainClass { email = "user1@company.com", details = new CustomClass { id = 1, first = "FirstName", last = "LastName" } });
genericList.Add(new MainClass { email = "user1@company.com", details = new CustomClass { id = 2, first = "FirstName", last = "LastName" } });
genericList.Add(new MainClass { email = "user2@company.com", details = new CustomClass { id = 3, first = "FirstName", last = "LastName" } });
genericList.Add(new MainClass { email = "user2@company.com", details = new CustomClass { id = 4, first = "FirstName", last = "LastName" } });
string genericListJson = new JavaScriptSerializer().Serialize(genericList);
//json result
//[{"email":"user1@company.com","details":{"id":1,"first":"FirstName","last":"LastName"}},{"email":"user1@company.com","details":{"id":2,"first":"FirstName","last":"LastName"}},{"email":"user2@company.com","details":{"id":3,"first":"FirstName","last":"LastName"}},{"email":"user2@company.com","details":{"id":4,"first":"FirstName","last":"LastName"}}]

我不想要 json 结果中的键名。我只需要电子邮件作为键和对象作为它的值(value)。 像这样

{"user1@company.com":{"id":1,"first":"FirstName","last":"LastName"}}

Answer 1

首先，使用 List<KeyValuePair<string, object>>因为您需要允许重复。

var list = new List<KeyValuePair<string, object>>();
list.Add(new KeyValuePair<string, object>("foo", new { id = 1, first = "FirstName", last = "LastName" }));
list.Add(new KeyValuePair<string, object>("foo", new { id = 1, first = "FirstName", last = "LastName" }));
list.Add(new KeyValuePair<string, object>("foo", new { id = 1, first = "FirstName", last = "LastName" }));

用 JsonConvert 序列化.

JsonSerializerSettings settings = new JsonSerializerSettings { Converters = new[] { new MyConverter() } };
string json = JsonConvert.SerializeObject(list, settings);

并使用 custom converter inspired from this answer :

public class MyConverter : JsonConverter
{

    public override void WriteJson(JsonWriter writer, object value, Newtonsoft.Json.JsonSerializer serializer)
    {
        List<KeyValuePair<string, object>> list = value as List<KeyValuePair<string, object>>;
        writer.WriteStartArray();
        foreach (var item in list)
        {
            writer.WriteStartObject();
            writer.WritePropertyName(item.Key);
            // Needed because of the dynamic object.
            var jsonValue = JsonConvert.SerializeObject(item.Value);
            writer.WriteValue(jsonValue);
            writer.WriteEndObject();
        }
        writer.WriteEndArray();
    }

    public override object ReadJson(JsonReader reader, Type objectType, object existingValue, Newtonsoft.Json.JsonSerializer serializer)
    {
        throw new NotImplementedException();
    }

    public override bool CanConvert(Type objectType)
    {
        return objectType == typeof(List<KeyValuePair<string, object>>);
    }
}

输出:

[
    {"foo":"{\"id\":1,\"first\":\"FirstName\",\"last\":\"LastName\"}"},
    {"foo":"{\"id\":1,\"first\":\"FirstName\",\"last\":\"LastName\"}"},
    {"foo":"{\"id\":1,\"first\":\"FirstName\",\"last\":\"LastName\"}"}
]