KakaJson使用过程中遇到的bug

在使用kakajson过程中。遇到了将模型转换成字符串的操作。结果kakajson将模型中的字符串属性转换为了字典类型。

//模型
class JifenModel: NSObject,Convertible  {
    required override init() {
        super.init()
        self.seetime = NSDate.currentDateDayString()
    }
 
//    用户id
    var userId: String = ""
//    记录观看当天的时间
    var seetime: String = ""
    var videoTime: Int = 0
    var infomationTime: Int = 0
    var videFinsihed: Bool = false
    var infomationFinished: Bool = false
}

转换用的方法

model2.userId = UserInfoManager.getUid() ?? ""
 let dictStr = model2.kj.JSONString()
        print(dictStr)

通过不断深挖。发现代码流程，

//先通过kj_JSONObject()方法将模型转为字典
        for property in properties {
            // value filter
            guard let value = kj_JSONValue(
                from: property.get(from: ptr)~!,
                property)~! else { continue }
            
            guard let v = Values.JSONValue(value) else { continue }
            
            // key filter
            json[mt.JSONKey(from: property.name,
                            kj_JSONKey(from: property))] = v
        }
通过以上方法对模型所有的属性进行取值。并添加到一个字典中。
通过Values.JSONValue(value)方法进行类型判断。在类型判断中因为工程代码中做了一个对String扩展导致了现在的问题。
extension String: Convertible{}

    static func JSONValue(_ value: Any?) -> Any? {
        guard let v = value.kj_value else { return nil }
        if v is NSNull { return nil }
        
        switch v {
        case let num as NumberValue: return _JSONValue(from: num)
        //走到这个代码后，导致了因为String遵守了Convertible协议。而返回一个空{}
        case let model as Convertible: return model.kj_JSONObject()

        case let array as [Any]: return _JSONValue(from: array)
        case let date as Date: return date.timeIntervalSince1970
        case let dict as [String: Any]: return _JSONValue(from: dict)
        case let url as URL: return url.absoluteString
        case let set as SetValue: return _JSONValue(from: set)
        case let `enum` as ConvertibleEnum: return JSONValue(`enum`.kj_value)
        default: return v as? NSCoding
        }
    }