Swift 3 Xcode 8 - SwiftValue encodeWithCoder - 无法识别的选择器发送到实例
Swift 3 Xcode 8 - SwiftValue encodeWithCoder - unrecognized selector sent to instance
我的自定义对象使用以下方法符合 NSCoding 协议
required init(coder decoder: NSCoder) {
super.init()
createdDate = decoder.decodeObject(forKey: "created_date") as? Date
userId = decoder.decodeInteger(forKey: "user_id")
}
func encode(with aCoder: NSCoder) {
aCoder.encode(createdDate, forKey: "created_date")
aCoder.encode(userId, forKey: "user_id")
}
这是 Swift 3 中 nscoding 协议的正确方法名称,但是应用程序崩溃并出现错误 SwiftValue encodeWithCoder - unrecognized selector sent to instance
明明这个方法不可用,为什么不识别?
参考文献 https://developer.apple.com/reference/foundation/nscoding
这是我制作的存档方法
func encodeObject(_ defaults:UserDefaults, object:NSCoding?, key:String) {
if (object != nil) {
let encodedObject = NSKeyedArchiver.archivedData(withRootObject: object)
defaults.set(encodedObject, forKey: key)
} else {
defaults.removeObject(forKey: key)
}
}
问题是您正在尝试归档一个可选的。替换此行:
if (object != nil) {
与:
if let object = object {
我的自定义对象使用以下方法符合 NSCoding 协议
required init(coder decoder: NSCoder) {
super.init()
createdDate = decoder.decodeObject(forKey: "created_date") as? Date
userId = decoder.decodeInteger(forKey: "user_id")
}
func encode(with aCoder: NSCoder) {
aCoder.encode(createdDate, forKey: "created_date")
aCoder.encode(userId, forKey: "user_id")
}
这是 Swift 3 中 nscoding 协议的正确方法名称,但是应用程序崩溃并出现错误 SwiftValue encodeWithCoder - unrecognized selector sent to instance
明明这个方法不可用,为什么不识别?
参考文献 https://developer.apple.com/reference/foundation/nscoding
这是我制作的存档方法
func encodeObject(_ defaults:UserDefaults, object:NSCoding?, key:String) {
if (object != nil) {
let encodedObject = NSKeyedArchiver.archivedData(withRootObject: object)
defaults.set(encodedObject, forKey: key)
} else {
defaults.removeObject(forKey: key)
}
}
问题是您正在尝试归档一个可选的。替换此行:
if (object != nil) {
与:
if let object = object {