xcode 无法在 Swift 中使用 NSCoder 解码 Int
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Fail to decode Int with NSCoder in Swift
提问by wqyfavor
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)
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OK）
codes
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")
}
}
回答by Rob
The problem is that seq2is not an Int, but rather a Int?optional. It cannot be represented as an Objective-C integer.
问题是这seq2不是一个Int，而是一个Int?可选的。它不能表示为 Objective-C 整数。
You can use decodeObject:
您可以使用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()
}
or change it so it is not optional:
或更改它，使其不是可选的：
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)>" }
}
回答by Alessandro Ornano
Use the encodeIntegermethod to encode integers instead:
使用该encodeInteger方法对整数进行编码：
func encode(with aCoder: NSCoder) {
...
aCoder.encodeInteger(seq2, forKey: "seq2")
}
回答by Adolfo
Take care with nil values
注意 nil 值
func encode(with aCoder: NSCoder){
aCoder.encode(seq, forKey: "seq")
if let second_seq = self.seq2
{
aCoder.encode(second_seq, forKey: "seq2")
}
aCoder.encode(id, forKey: "id")
aCoder.encode(value, forKey: "value")
}
seq2will be nilif the key is not present on decode operations
seq2将是nil如果该键不存在于解码操作
