在C#中反序列化JSON的多个记录

本文关键字：记录 JSON 反序列化 | 更新日期: 2023-09-27 18:28:00

我有以下JSON

string json = @"[[{""campaignId"":201410018,""programCode"":""54321""},{""reason"":201410018,"‌​"about"":""54321""}],[{""campaignId"":201410019,""programCode"":""54322""},{""rea‌​son"":201410018,""about"":""54321""}]]"

我创建了以下类

public class JSONResponse
    {
        public number[] number{ get; set; }
        public Inf[] inf{ get; set; }
    }
    public class number
    {
        public int reason { get; set; }
        public string about { get; set; }
    }
    public class Inf
    {
        public int campaignId { get; set; }
        public string programCode { get; set; }
    }

为了反序列化，我调用下面的代码

List<List<JSONResponse>> myDeserializedObjList = JsonConvert.DeserializeObject<List<List<JSONResponse>>>(jsonstr);

但我的两个类数据仍然为空。

非常感谢您的帮助。

在C#中反序列化JSON的多个记录

因为您的JSON在同一数组中有两种不同的对象类型，所以您无法以典型且简单的方式将其反序列化为对象。除非您能够控制JSON，将结构更改为在同一数组中没有Inf和number对象的结构，否则您唯一的选择就是为什么选择Expando/dynamic对象。下面的代码将提供的JSON解析为现有的类结构。

static List<JSONResponse> Parse(string json)
{
    var responses = new List<JSONResponse>();
    //string)obj[0][0].programCode.Value;
    dynamic obj = JsonConvert.DeserializeObject<dynamic>(json);
    for (int i = 0; i < obj.Count; i++)
    {
        //responses[i] = new JSONResponse() {
        var inf = new List<Inf>();
        var numbers = new List<number>();
        for (int j = 0; j < obj[i].Count; j++)
        {
            if (obj[i][j].campaignId != null)
                inf.Add(new Inf()
                    {
                        campaignId = (int) obj[i][j].campaignId.Value,
                        programCode = obj[i][j].programCode.Value
                    });
            if (obj[i][j].reason != null)
                numbers.Add(new number()
                    {
                        about = obj[i][j].about.Value,
                        reason = (int)obj[i][j].reason.Value
                    });
        }
        responses.Add(new JSONResponse()
            {
                number = numbers.ToArray(),
                inf = inf.ToArray()
            });
    }
    return responses;
}

您的JSON没有您在类中显示的结构。json不是一个包含Inf和number类数组的响应列表，而是一个Inf和number类对的列表。如果您将json粘贴到http://jsonformatter.curiousconcept.com/.将每个对打包到一个容器类中是有意义的，但不幸的是，这些对是在数组中传输的，而不是作为不同的属性传输的，因此它们没有可以映射到类属性名称的名称。相反，您可以编写一个自定义的JsonConverter来解析数组并适当地分配字段：

public class number
{
    public int reason { get; set; }
    public string about { get; set; }
}
public class Inf
{
    public int campaignId { get; set; }
    public string programCode { get; set; }
}
public class JSONResponse
{
    public number number { get; set; }
    public Inf Inf { get; set; }
    public static List<JSONResponse> DeserializeList(string jsonstr)
    {
        return JsonConvert.DeserializeObject<List<JSONResponse>>(jsonstr, new JSONResponseConverter());
    }
}
class JSONResponseConverter : JsonConverter
{
    public override bool CanConvert(Type objectType)
    {
        return typeof(JSONResponse).IsAssignableFrom(objectType);
    }
    public override object ReadJson(JsonReader reader,
        Type objectType, object existingValue, JsonSerializer serializer)
    {
        JSONResponse response = new JSONResponse();
        var array = JArray.Load(reader);
        if (array.Count != 2)
        {
            // Or maybe throw an exception?
            Debug.WriteLine("Unexpected array length for " + array.ToString());
        }
        foreach (var entry in array)
        {
            if (entry["campaignId"] != null)
            {
                response.Inf = entry.ToObject<Inf>();
            }
            else if (entry["reason"] != null)
            {
                response.number = entry.ToObject<number>();
            }
            else
            {
                // Or maybe throw an exception?
                Debug.WriteLine("Unknown entry " + entry.ToString());
            }
        }
        return response;
    }
    public override void WriteJson(JsonWriter writer,
        object value, JsonSerializer serializer)
    {
        throw new NotImplementedException();
    }
}

假设你负责你的数据，那么你可以改变你的结构，而不是让事情变得复杂

eg而不是

@"[[{""campaignId"":201410018,""programCode"":""54321""},{""reason"":201410018,"‌​"about"":""54321""}],[{""campaignId"":201410019,""programCode"":""54322""},{""rea‌​son"":201410018,""about"":""54321""}]]"

尝试

@"[{info:{""campaignId"":201410018,""programCode"":""54321""},number:{""reason"":201410018,"‌​"about"":""54321""}},{info:{""campaignId"":201410019,""programCode"":""54322""},number:{""rea‌​son"":201410018,""about"":""54321""}}]"

并处理

public class JSONResponse
{
    public number number{ get; set; }
    public Inf info{ get; set; }
}
public class number
{
    public int reason { get; set; }
    public string about { get; set; }
}
public class Inf
{
    public int campaignId { get; set; }
    public string programCode { get; set; }
}

现在您只有一个处理的JSONResponse列表

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