背景
我试图使用NSCoding协议编码一个字符串样式的枚举，但我遇到错误转换和返回从字符串。
解码和编码时，我得到以下错误：
字符串不能转换为Stage
额外参数ForKey：在调用
码
enum Stage : String
{
case DisplayAll = "Display All"
case HideQuarter = "Hide Quarter"
case HideHalf = "Hide Half"
case HideTwoThirds = "Hide Two Thirds"
case HideAll = "Hide All"
}
class AppState : NSCoding,NSObject
{
var idx = 0
var stage = Stage.DisplayAll
override init() {}
required init(coder aDecoder: NSCoder) {
self.idx = aDecoder.decodeIntegerForKey( "idx" )
self.stage = aDecoder.decodeObjectForKey( "stage" ) as String // ERROR
}
func encodeWithCoder(aCoder: NSCoder) {
aCoder.encodeInteger( self.idx,forKey:"idx" )
aCoder.encodeObject( self.stage as String,forKey:"stage" ) // ERROR
}
// ...
}
您需要将枚举转换为和从原始值。在Swift 1.2(Xcode 6.3)中，这将是这样：
原文链接：/swift/321231.htmlclass AppState : NSObject,NSCoding
{
var idx = 0
var stage = Stage.DisplayAll
override init() {}
required init(coder aDecoder: NSCoder) {
self.idx = aDecoder.decodeIntegerForKey( "idx" )
self.stage = Stage(rawValue: (aDecoder.decodeObjectForKey( "stage" ) as! String)) ?? .DisplayAll
}
func encodeWithCoder(aCoder: NSCoder) {
aCoder.encodeInteger( self.idx,forKey:"idx" )
aCoder.encodeObject( self.stage.rawValue,forKey:"stage" )
}
// ...
}
Swift 1.1(Xcode 6.1)，使用as而不是！！
self.stage = Stage(rawValue: (aDecoder.decodeObjectForKey( "stage" ) as String)) ?? .DisplayAll
Swift 1.0(Xcode 6.0)使用toRaw()和fromRaw()像这样：
self.stage = Stage.fromRaw(aDecoder.decodeObjectForKey( "stage" ) as String) ?? .DisplayAll
aCoder.encodeObject( self.stage.toRaw(),forKey:"stage" )
