为JSON创建模型，其中键是一个值 [英] Creating model for JSON where key is a value

问题描述

JSON :

{  
    "rows" :
    [
        {
            "_id": "5cdc0ede5c3dcb04bdb3a972",
            "emp_code": 187,
            "log_id": 361711,
            "punch_time": "2019-05-07T04:00:33.000Z",
            "pin_type": 1,
            "status": 4,
            "__v": 0
        },
        {
            "_id": "5cdc40de5c3dcb04bdb3a972",
            "emp_code": 111,
            "log_id": 361701,
            "punch_time": "2019-05-07T04:00:35.000Z",
            "pin_type": 101,
            "status": 4,
            "__v": 0
        }
    ],
    "pin_type_text": {
        "1": "In Fingerprint",
        "4": "In Card",
        "101": "Out Fingerprint",
        "104": "Out Card"
    }
}  

每行 pin_type 的值是指 pin_type_text 中与其键对应的记录.

The value of pin_type in each row refers to the record in pin_type_text mapped with it's key.

我正在使用 AlamofireObjectMapper 创建模型，这是PinTypeText模型:

I am using AlamofireObjectMapper for creating models, and here is the PinTypeText model :

class PinTypeText : Mappable {

    var inFingerprint: String?
    var inCard: String?
    var outFingerprint: String?
    var outCard: String?

    required init?(map: Map) {

    }

    func mapping(map: Map) {
        self.inFingerprint <- map["1"]
        self.inCard <- map["4"]
        self.outFingerprint <- map["101"]
        self.outCard <- map["104"]
    }  
}  

问题:假设将来pin_type值- 1、4、101、104 在后端发生变化，如何在不更改模型的情况下处理这种情况? .按照这种模型结构，每当后端模型更改时，我都需要更改模型类

Issue : Suppose in future, the pin_type values - 1, 4, 101, 104 change in the backend, how can I handle such a case without changing my model. As per this model structure, I need to change my model class every time the backend model changes

推荐答案

在这里，您可以使用Codable作为解决方案，

Here is how you can use Codable as a solution,

1..创建一个模型Row，该模型将包含json的rows数组中的一行的数据，即

1. Create a model Row that will contain the data of a single row in rows array of json, i.e.

class Row: Decodable {
    var id: String?
    var pinType: String?
    var pinId: Int?

    enum CodingKeys: String, CodingKey {
        case id = "_id"
        case pinId = "pin_type"
    }
}

在上面的模型中，我使用了2个不同的属性-pinType and pinId.

In the above model, I've used 2 different properties - pinType and pinId.

1. pinId将在row

pinType将包含与pinId相对应的实际值.稍后我们将填充该值.

pinType will contain the actual value corresponding to pinId. We'll fill this value later.

此外，我只使用了row的一小部分键.您可以根据需要添加更多内容.

Also, I've only used a small set of keys of the row. You can add more as required.

2..接下来，创建另一个模型Response，该模型将包含Row的array，即

2. Next create another model Response that will contain an array of Row, i.e.

class Response: Decodable {
    var rows: [Row]?

    enum CodingKeys: String, CodingKey {
        case rows, pin_type_text
    }

    required init(from decoder: Decoder) throws {
        let values = try decoder.container(keyedBy: CodingKeys.self)
        rows = try values.decodeIfPresent([Row].self, forKey: .rows)
        let pinTypeText = try values.decodeIfPresent([String:String].self, forKey: .pin_type_text)
        rows?.forEach({ (row) in
            if let pinId = row.pinId {
                row.pinType = pinTypeText?[String(pinId)]
            }
        })
    }
}

在上述模型中，

json中的

1. rows数组被解析为[Row].

pinTypeText dictionary被解析为[String:String]类型.

[Row]以使用pinId和pinTypeText dictionary在每个row中填充pinType.

[Row] is enumerated to fill pinType in each row using pinId and pinTypeText dictionary.

使用时，需要使用Row对象的pinType属性.

When using, you need to use pinType property of a Row object.

response?.rows?.forEach({ print($0.pinType) }) //This line will print - "In Fingerprint" and "Out Fingerprint"

如果遇到实施此方法的问题，请告诉我.

Let me know in case you face issue implementing this approach.

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