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多次调用Api C#
EN

Stack Overflow用户
提问于 2020-10-13 05:45:40
回答 3查看 237关注 0票数 0

我正在尝试使用rawg api并从他们的api中获取游戏,我已经创建了要解析数据的游戏实体类

代码语言:javascript
复制
//game entity
public class Game
    {
        public int Id { get; set; }
        public string Slug { get; set; }
        public string Name { get; set; }
        public string Released { get; set; }
        public string Background_Image { get; set; }
        public int Rating { get; set; }
        public int Playtime { get; set; }
    }

现在我需要从api中获取游戏,我创建了这个方法

代码语言:javascript
复制
public static async Task<List<Game>> ApiRawgGamesRequest()
        {
            var gamesList = new List<Game>();

            for (int i = 1; i < 250; i++)
            {
                using (var msg = new HttpRequestMessage(HttpMethod.Get, new Uri($"https://api.rawg.io/api/games?page_size=40&page={i}")))
                using (var response = await _client.SendAsync(msg))
                {
                    response.EnsureSuccessStatusCode();
                    var gamesResponse = await response.Content.ReadAsAsync<Game[]>();
                    gamesList.AddRange(gamesResponse);
                }
            }

            return gamesList;
        }

我尽量不要一次添加所有的游戏,所以我使用了一个for循环来不断地将游戏添加到列表中,以潜在地减轻压力。但是我得到了一个错误

代码语言:javascript
复制
// error
Unhandled exception. System.AggregateException: One or more errors occurred. (Cannot deserialize the current JSON object (e.g. {"name":"value"}) into type 'Domain.Game[]' because the t
ype requires a JSON array (e.g. [1,2,3]) to deserialize correctly.
To fix this error either change the JSON to a JSON array (e.g. [1,2,3]) or change the deserialized type so that it is a normal .NET type (e.g. not a primitive type like integer, not a
collection type like an array or List<T>) that can be deserialized from a JSON object. JsonObjectAttribute can also be added to the type to force it to deserialize from a JSON object.
Path 'count', line 1, position 9.)

来自rawg api的响应如下所示--结果部分是我试图收集的游戏数据

代码语言:javascript
复制
// api response
{
"count": 454988,
"next": "https://api.rawg.io/api/games?page=2&page_size=40",
"previous": null,
"results": [
{
"id": 3498,
"slug": "grand-theft-auto-v",
"name": "Grand Theft Auto V",
"released": "2013-09-17",
"tba": false,
"background_image": "https://media.rawg.io/media/games/84d/84da2ac3fdfc6507807a1808595afb12.jpg",
"rating": 4.48,
},
{
"id": 4200,
"slug": "portal-2",
"name": "Portal 2",
"released": "2011-04-18",
"tba": false,
"background_image": "https://media.rawg.io/media/games/328/3283617cb7d75d67257fc58339188742.jpg",
"rating": 4.62
}
]
}

有没有更好的方法来做这件事?

EN

回答 3

Stack Overflow用户

发布于 2020-10-13 06:19:49

您的异常是由错误的JSON解析对象引起的。

在您的代码中,您将调用:

代码语言:javascript
复制
var gamesResponse = await response.Content.ReadAsAsync<Game[]>();

但是响应是对象(来自您的示例JSON):

代码语言:javascript
复制
public class GameResponse
    {
        public int Count { get; set; }
        public string Next{ get; set; }
        public string Previous { get; set; }
        public IEnumerable<Game> Results { get; set; }
    }

如果将其更改为:

代码语言:javascript
复制
var gamesResponse = await response.Content.ReadAsAsync<GameResponse>();

您将成功解析响应并通过以下方式访问游戏:

代码语言:javascript
复制
gamesResponse.Results
票数 1
EN

Stack Overflow用户

发布于 2020-10-13 06:22:53

因此,您试图只反序列化json的“result”部分,您必须使用来自响应json的所有项来构建一个类。

代码语言:javascript
复制
// Root myDeserializedClass = JsonConvert.DeserializeObject<Root>(myJsonResponse); 
    public class Result    {
        public int id { get; set; } 
        public string slug { get; set; } 
        public string name { get; set; } 
        public string released { get; set; } 
        public bool tba { get; set; } 
        public string background_image { get; set; } 
        public double rating { get; set; } 
    }

    public class Root    {
        public int count { get; set; } 
        public string next { get; set; } 
        public object previous { get; set; } 
        public List<Result> results { get; set; } 
    }

你可以随时在线使用某种Json>class生成器,使用下面的代码:https://json2csharp.com/

票数 1
EN

Stack Overflow用户

发布于 2020-10-13 06:40:27

对于结构不正确的问题,有几个很好的答案。我在这里添加的唯一一件事是注释,这是一种良好的做法,也是为了更好地控制您的变量名称。请注意,DataContract和DataMember是框架的一部分,但它们可以很好地与Json.Net一起工作,即使Json.Net提供了自己的一组注释。

代码语言:javascript
复制
    [DataContract]
    public class Payload
    {
        [DataMember(Name = "count")]
        public int Count { get; set; }

        [DataMember(Name = "next")]
        public string Next { get; set; }

        [DataMember(Name = "previous")]
        public string Previous { get; set; }

        [DataMember(Name = "results")]
        public List<GameData> Results { get; set; }
    }

    [DataContract]
    public class GameData
    {
        [DataMember(Name = "id")]
        public long Id { get; set; }

        [DataMember(Name = "slug")]
        public string Slug { get; set; }

        [DataMember(Name = "name")]
        public string Name { get; set; }

        [DataMember(Name = "rating")]
        public float Rating { get; set; }

        [DataMember(Name = "rating_top")]
        public float Rating_top { get; set; }
    }

    public static async Task<IEnumerable<Payload>> GetGameDatasAsync(int page = 1, int pageSize = 40, bool followNext = false, CancellationToken cancellationToken = default)
    {
        cancellationToken.ThrowIfCancellationRequested();

        string next = $"https://api.rawg.io/api/games?page_size={pageSize}&page={page}";
        var results = new List<Payload>();

        using (var client = new HttpClient())
            do
            {
                var result = await client.GetAsync(next, cancellationToken);

                next = string.Empty;

                if (result.IsSuccessStatusCode)
                {
                    var json = await result.Content.ReadAsStringAsync();

                    if (!string.IsNullOrWhiteSpace(json))
                    {
                        var payload = JsonConvert.DeserializeObject<Payload>(json);

                        next = payload.Next;

                        results.Add(payload);
                    }
                }

            } while (!string.IsNullOrWhiteSpace(next) && followNext);

        return results;
    }
票数 0
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/64325718

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