Question

我正在尝试从JSON解析一些数据，我已经使用其他API进行了处理，但是现在我有了另一个结构，并且遇到了错误：

线程1：致命错误：“尝试！”表达式意外引发错误： Swift.DecodingError.valueNotFound（Swift.Int， Swift.DecodingError.Context（codingPath：[_JSONKey（stringValue：“索引 1“，intValue：1），CodingKeys（stringValue：” evaluation“，intValue： nil）]，debugDescription：“预期的Int值，但发现为null。”，底层错误：nil））

JSON：

{
"restaurants": [
    {
        "id": 2,
        "res_name": "Test",
        "res_phone": "Test",
        "evaluation": 4,
        "created_at": "2019-05-12 22:05:20",
        "updated_at": "2019-05-12 22:05:20"
    },
    {
        "id": 3,
        "res_name": "Test",
        "res_phone": "Test",
        "evaluation": null,
        "created_at": "2019-05-12 21:46:59",
        "updated_at": "2019-05-12 21:46:59"
    },
    {
        "id": 4,
        "res_name": "Test",
        "res_phone": "Test",
        "evaluation": null,
        "created_at": "2019-05-13 14:49:20",
        "updated_at": "2019-05-13 14:49:20"
    }
]
}

我的结构模型：

struct Restaurant: Codable {
    let id: Int
    let resName, resPhone: String
    let evaluation: Int
    let createdAt, updatedAt: String

    enum CodingKeys: String, CodingKey {
        case id
        case resName = "res_name"
        case resPhone = "res_phone"
        case evaluation
        case createdAt = "created_at"
        case updatedAt = "updated_at"
    }
}

我的代码：

Alamofire.request("http://test.tlict.com/public/api/restaurant/show?api_token=\(UserDefaults.standard.string(forKey: "token")!)").responseJSON(completionHandler: { (response) in
    print("tokennnnnn isssssssssssssss\(UserDefaults.standard.string(forKey: "token")!) ")
    //token
    switch response.result {
    case .success(let data):
        let object = data as! [String : Any]
        let result = object["restaurants"] as! [[String : Any]]

        let dataCategory = try! JSONSerialization.data(withJSONObject: result, options: [])
        self.arrayOfRestaurant = try! JSONDecoder.init().decode([Restaurant].self, from: dataCategory)



        DispatchQueue.main.async {
            self.myTable.reloadData()
        }
    case .failure(let error): break
    print(error.localizedDescription)
    }

})

JSON解码错误（Swift.DecodingError.valueNotFound）Swift 4

0 个答案: