.NET 使用 C# 的标头从 GET API 反序列化 JSON

Question

经过几个小时的研究，我找不到转换这种 JSON 的方法：

https://api.jamendo.com/v3.0/tracks/?client_id=56d30c95&format=jsonpretty&id=982090

{
    "headers":{
        "status":"success",
        "code":0,
        "error_message":"",
        "warnings":"",
        "results_count":1
    },
    "results":[
        {
            "id":"982090",
            "name":"Seul",
            "duration":297,
            "artist_id":"350774",
            "artist_name":"DON VALDES",
            "artist_idstr":"DON_VALDES",
            "album_name":"EVOLUTION",
            "album_id":"115666",
            "license_ccurl":"http:\/\/creativecommons.org\/licenses\/by-nc-nd\/3.0\/",
            "position":10,
            "releasedate":"2012-11-23",
            "album_image":"https:\/\/imgjam2.jamendo.com\/albums\/s115\/115666\/covers\/1.200.jpg",
            "audio":"https:\/\/mp3l.jamendo.com\/?trackid=982090&format=mp31&from=app-56d30c95",
            "audiodownload":"https:\/\/mp3d.jamendo.com\/download\/track\/982090\/mp32\/",
            "prourl":"https:\/\/licensing.jamendo.com\/track\/982090",
            "shorturl":"http:\/\/jamen.do\/t\/982090",
            "shareurl":"http:\/\/www.jamendo.com\/track\/982090",
            "image":"https:\/\/imgjam2.jamendo.com\/albums\/s115\/115666\/covers\/1.200.jpg"
        }
    ]
}

这是我的方法，但我的对象“track”中的每个值都是空的，我认为这是因为它无法到达 JSON 的“结果”部分（而且我找不到如何关注这部分）或这是因为它无法到达“结果”的第一个成员（[0]）（而且我也无法专注于第一个元素：/）：

public static TrackModel getTrackById(int id)
    {
        Uri uri = new Uri(String.Format("https://api.jamendo.com/v3.0/tracks/?client_id=56d30c95&format=jsonpretty&id={0}", id));
        WebRequest webRequest = WebRequest.Create(uri);
        WebResponse response = webRequest.GetResponse();
        StreamReader streamReader = new StreamReader(response.GetResponseStream());
        String responseData = streamReader.ReadToEnd();

        var track = JsonConvert.DeserializeObject<TrackModel>(responseData).;

        return track;
    }

希望你能帮助我，

非常感谢！

Answer 1

将您的TrackModel 分解为两个属性，并为Result 创建一个单独的类。

public class TrackModel
{
    [JsonProperty(PropertyName = "headers")]
    public Headers Headers { get; set; }
    [JsonProperty(PropertyName = "results")]
    public Result[] Results { get; set; }
}

public class Headers
{
    [JsonProperty(PropertyName = "status")]
    public string Status { get; set; }
    [JsonProperty(PropertyName = "code")]
    public string Code { get; set; }
    [JsonProperty(PropertyName = "error_message")]
    public string Errormessage { get; set; }
    [JsonProperty(PropertyName = "warnings")]
    public string Warnings { get; set; }
    [JsonProperty(PropertyName = "results_count")]
    public string ResultsCount { get; set; }
}

public class Result
{
    public string Id { get; set; }
    public string Name { get; set; }
    public int Duration { get; set; }
    [JsonProperty(PropertyName = "artist_id")]
    public string ArtistId { get; set; }
    [JsonProperty(PropertyName = "artist_name")]
    public string ArtistName { get; set; }
    [JsonProperty(PropertyName = "artist_idstr")]
    public string ArtistIdstr { get; set; }
    [JsonProperty(PropertyName = "album_name")]
    public string AlbumName { get; set; }
    [JsonProperty(PropertyName = "album_id")]
    public string AlbumId { get; set; }
    [JsonProperty(PropertyName = "license_ccurl")]
    public string LicenseCcurl { get; set; }
    public int Position { get; set; }
    public string Releasedate { get; set; }
    [JsonProperty(PropertyName = "album_image")]
    public string AlbumImage { get; set; }
    public string Audio { get; set; }
    public string Audiodownload { get; set; }
    public string Prourl { get; set; }
    public string Shorturl { get; set; }
    public string Shareurl { get; set; }
    public string Image { get; set; }
}

在你原来的方法中使用

var track = JsonConvert.DeserializeObject<TrackModel>(responseData).results[0].album_id;

Answer 2

您尝试反序列化子对象。那是行不通的，因为 JSON 序列化程序不知道 TrackModel 从哪里开始以及在哪里结束。添加一个包含结果列表的“父模型”。例如：

public class TopModel {
    public ICollection<TrackModel> Results {get; set; }
 }

然后使用那个模型：

var model = JsonConvert.DeserializeObject<TrackModel>(responseData).;

顺便说一句：我也会在你的 C# 文件中使用 Pascal 大小写。 JSON 序列化程序将负责大小写映射。

顺便说一句 2：在您的流周围添加一些“使用”。

Answer 3

您需要创建类似于 JSON 响应的 TrackModel。

public class TrackModel 
{
    [JsonProperty(PropertyName="headers")] 
    public Headers headers {get;set;}


    [JsonProperty(PropertyName="results")] 
    public List<Results> results {get;set;}
}


public class Headers {
    public string status {get;set;}
    public string code {get;set;}
    public string error_message {get;set;}
//Rest of the properties
}

public class Results {
    public int id {get;set;}
    public string name {get;set;}
// Rest of the properties
}

这里是小提琴：https://dotnetfiddle.net/Vr27pB

Answer 4

如果您尝试将该字符串反序列化为您编写的类定义的对象，它将永远无法工作。因为您的 TrackModel 不匹配字符串，而只匹配它的内部部分。

像这样创建一个包装类：

public class TrackModelWrapper
{
    public Headers Headers { get; set; }
    public IEnumerable<TrackModel> Results { get; set; }
}

public class Headers
{
    public string Status { get; set; }
    public int Code { get; set; }
    public string Error_message { get; set; }
    public string Warnings { get; set; }

    public int Results_count { get; set; }
}

public class TrackModel
{
    public string id { get; set; }
    public string name { get; set; }
    public int duration { get; set; }
    public string artist_id { get; set; }
    public string artist_name { get; set; }
    public string artist_idstr { get; set; }
    public string album_name { get; set; }
    public string album_id { get; set; }
    public string license_ccurl { get; set; }
    public int position { get; set; }
    public string releasedate { get; set; }
    public string album_image { get; set; }
    public string audio { get; set; }
    public string audiodownload { get; set; }
    public string prourl { get; set; }
    public string shorturl { get; set; }
    public string shareurl { get; set; }
    public string image { get; set; }
}

那就叫它：var track = JsonConvert.DeserializeObject<TrackModelWrapper>(responseData);