KeyDecodingStrategy .convertFromSnakeCase 不起作用

Question

我有一个名为 ServiceHealthApi 的适配器类，它具有以下功能：

final class ServiceHealthApi {
    let mockApi = "https://staging.myapp.com/health"

    func getHealth() -> Single<ServiceHealthResponseModel> {
        let url = URL(string: mockApi)
        guard let validUrl = url else { return .never() }

        var urlRequest = URLRequest(url: validUrl)
        urlRequest.httpMethod = "GET"
        let headers = [
            "Content-Type" : "application/json; charset=utf-8"
        ]
        urlRequest.allHTTPHeaderFields = headers
        return URLSession.shared.rx.data(request: urlRequest)
            .take(1)
            .map {
                let decoder = JSONDecoder()
                decoder.keyDecodingStrategy = .convertFromSnakeCase
                return try JSONDecoder().decode(ServiceHealthResponseModel.self, from: $0) }
            .asSingle()

    }
}

struct HealthResponseModel: Decodable {
    struct DataResponse: Decodable {
        let serviceName: String
        let serviceStatus: String
        let serviceOperational: Bool
    }

    struct Meta: Decodable {
        let statusCode: Int
        let statusMessage: String
    }

    let data: [DataResponse]
    let meta: Meta
}

JSON应该被解析的字符串是这样的：

{
    "data": [
        {
            "service_name": "web",
            "service_status": "UP",
            "service_operational": true
        },
        {
            "service_name": "orm",
            "service_status": "UP",
            "service_operational": true
        }
    ],
    "meta": {
        "status_code": 200,
        "status_message": "OK"
    }
}

现在，当我尝试运行我的集成测试时，它因以下错误而失败JSONDecoder：

错误 keyNotFound(CodingKeys(stringValue: "serviceName", intValue: nil), Swift.DecodingError.Context(codingPath: [CodingKeys(stringValue: "data", intValue: nil), _JSONKey(stringValue: "Index 0", intValue: 0 )], debugDescription: "没有与键 CodingKeys(stringValue: \"serviceName\", intValue: nil) (\"serviceName\") 关联的值，underlyingError: nil))

有趣的是，如果我禁用了 .convertFromSnakeCase，并且只在响应模型中使用驼峰命名法作为变量名，它就可以正常工作。我知道我可能可以使用编码密钥，但我只是想知道，为什么我的实现不起作用？

提前致谢。

PS：我试过JSON直接解析字符串而不调用API并且它确实有效。

Answer 1

你正在创造一个JSONDecoder没有任何策略的第二个

代替

let decoder = JSONDecoder()
decoder.keyDecodingStrategy = .convertFromSnakeCase
return try JSONDecoder().decode(ServiceHealthResponseModel.self, from: $0) }

和

let decoder = JSONDecoder()
decoder.keyDecodingStrategy = .convertFromSnakeCase
return try decoder.decode(ServiceHealthResponseModel.self, from: $0) }