本文介绍了无法在 Swift 中使用 NSCoder 解码 Int的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我在 Swift 3.0 中使用 Xcode8 Beta.我试图基于 NSObject 编码一个简单的对象,但我无法解码 Int 或 NSInteger 类型.(编码过程没问题)
I am using Xcode8 Beta with Swift 3.0. I tried to encode a simple object base on NSObject, but I cannot decode Int or NSInteger type. (The encoding process is OK)
代码
class Model : NSObject, NSCoding {
var seq: NSNumber?
var seq2: Int? // problem with seq2, NSInteger is not ok, either
var id: String?
var value: String?
override init() {
super.init()
}
required init?(coder aDecoder: NSCoder){
self.seq = aDecoder.decodeObject(forKey: "seq") as? NSNumber
self.seq2 = aDecoder.decodeInteger(forKey: "seq2")
self.id = aDecoder.decodeObject(forKey: "id") as? String
self.value = aDecoder.decodeObject(forKey: "value") as? String
}
func encode(with aCoder: NSCoder){
aCoder.encode(seq, forKey: "seq")
aCoder.encode(seq2, forKey: "seq2")
aCoder.encode(id, forKey: "id")
aCoder.encode(value, forKey: "value")
}
}
推荐答案
问题在于 seq2
不是 Int
,而是 Int?
可选.它不能表示为 Objective-C 整数.
The problem is that seq2
is not an Int
, but rather a Int?
optional. It cannot be represented as an Objective-C integer.
你可以使用decodeObject
:
required init?(coder aDecoder: NSCoder){
self.seq = aDecoder.decodeObject(forKey: "seq") as? NSNumber
self.seq2 = aDecoder.decodeObject(forKey: "seq2") as? Int
self.id = aDecoder.decodeObject(forKey: "id") as? String
self.value = aDecoder.decodeObject(forKey: "value") as? String
super.init()
}
或更改它使其不是可选的:
or change it so it is not optional:
class Model : NSObject, NSCoding {
var seq: NSNumber?
var seq2: Int
var id: String?
var value: String?
init(seq: NSNumber, seq2: Int, id: String, value: String) {
self.seq = seq
self.seq2 = seq2
self.id = id
self.value = value
super.init()
}
required init?(coder aDecoder: NSCoder) {
self.seq = aDecoder.decodeObject(forKey: "seq") as? NSNumber
self.seq2 = aDecoder.decodeInteger(forKey: "seq2")
self.id = aDecoder.decodeObject(forKey: "id") as? String
self.value = aDecoder.decodeObject(forKey: "value") as? String
super.init()
}
func encode(with aCoder: NSCoder) {
aCoder.encode(seq, forKey: "seq")
aCoder.encode(seq2, forKey: "seq2")
aCoder.encode(id, forKey: "id")
aCoder.encode(value, forKey: "value")
}
override var description: String { return "<Model; seq=\(seq); seq2=\(seq2); id=\(id); value=\(value)>" }
}
这篇关于无法在 Swift 中使用 NSCoder 解码 Int的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!