json到对象C＃(映射复杂的API响应对C＃对象)

Question

我能够处理简单的json序列化和挑选化，但是这种API响应似乎很复杂，我正在寻求有关解决此问题的理想方法的建议。

我正在尝试调用MVC应用程序的API。目标是将API数据映射到建模。API端点是https://www.alphavantage.co/query?function= time_series_intraday＆amp; symbol = msft＆amp; interval = 1min＆amp; papikey = myapikey

这里的麻烦是：

1. JSON数据键中有空白空间。
2. 当我尝试在Visual Studio中做粘贴时，它给了我很长的时间每个日期条目的类别列表，因为此API呼叫返回日期的单独信息集。

为了解决第1点中解释的问题，我在课堂中使用了[JsonProperty("1. Information")]。在我的代码中。

        public async Task TSI()
        {
            HttpClient client = new HttpClient();
            //Uri uri = new Uri("http://date.jsontest.com/");
            Uri uri = new Uri("https://www.alphavantage.co/query?function=TIME_SERIES_INTRADAY&symbol=MSFT&interval=5min&apikey=demo");
            client.DefaultRequestHeaders.Accept.Clear();
            client.DefaultRequestHeaders.Accept.Add(new MediaTypeWithQualityHeaderValue("application/json"));
            HttpResponseMessage response = await client.GetAsync(uri);
            if (response.IsSuccessStatusCode)
            {
                dynamic result = await response.Content.ReadAsAsync<object>();
                IEnumerable<dynamic> dObj = JsonConvert.DeserializeObject<dynamic>(result.ToString());
                IEnumerable<dynamic> t1 = dObj.FirstOrDefault();
                IEnumerable<dynamic> t2 = dObj.LastOrDefault();
                dynamic MetaData = t1.FirstOrDefault();
                Rootobject ro = new Rootobject();
                ro.MetaData = MetaData;
            }

ps：我对拨打API并进行处理相对较新。

我能够打电话给

date.jsontest.com

并将API数据映射到建模（我使用糊状特殊创建的模型）

//API response
    {
        "time": "12:53:22 PM",
        "milliseconds_since_epoch": 1504875202754,
        "date": "09-08-2017"
    }
//C# code to map to API data
    public class sampleObject
    {
        public string time { get; set; }
        public long milliseconds_since_epoch { get; set; }
        public string date { get; set; }
    }

我的rootObject看起来像这样：

public class Rootobject
    {
        [JsonProperty("Meta Data")]
        public MetaData MetaData { get; set; }
        [JsonProperty("Time Series (1min)")]
        public TimeSeries1Min TimeSeries1min { get; set; }
    }
    public class MetaData
    {
        [JsonProperty("1. Information")]
        public string _1Information { get; set; }
        [JsonProperty("2. Symbol")]
        public string _2Symbol { get; set; }
        [JsonProperty("3. Last Refreshed")]
        public string _3LastRefreshed { get; set; }
        [JsonProperty("4. Interval")]
        public string _4Interval { get; set; }
        [JsonProperty("5. Output Size")]
        public string _5OutputSize { get; set; }
        [JsonProperty("6. Time Zone")]
        public string _6TimeZone { get; set; }
    }
// I have so many of these sub-classes for dates, which again is an issue
    public class TimeSeries1Min
    {
            public _20170907160000 _20170907160000 { get; set; }
            public _20170907155900 _20170907155900 { get; set; }
....
....}

public class _20170907160000
    {
        public string _1open { get; set; }
        public string _2high { get; set; }
        public string _3low { get; set; }
        public string _4close { get; set; }
        public string _5volume { get; set; }
    }
    public class _20170907155900
    {
        public string _1open { get; set; }
        public string _2high { get; set; }
        public string _3low { get; set; }
        public string _4close { get; set; }
        public string _5volume { get; set; }
    }

Answer 1

很难从此JSON创建模型，但是您可以将这些数据转换为字典

var jObj = JObject.Parse(json);
var metadata = jObj["Meta Data"].ToObject<Dictionary<string, string>>();
var timeseries = jObj["Time Series (1min)"].ToObject<Dictionary<string, Dictionary<string, string>>>();

Answer 2

以下代码应执行您想要的

       if (response.IsSuccessStatusCode)
        {
            var result = await response.Content.ReadAsStringAsync();
            var obj = JsonConvert.DeserializeObject<Rootobject>(result);
            //No idea what you want to do with this line as there is no MetaData property on the root object
            obj.MetaData = MetaData;
        }