Swift学习：8.字典

字典

字典是一种存储多个相同类型的值的容器。每个值（value）都关联唯一的键（key），键作为字典中的这个值数据的标识符。和数组中的数据项不同，字典中的数据项并没有具体顺序。

1.初始化

var games: [String:String] = [“Diablo3”:”2014:8:12”,
“Dragon Age”:”2014:10:07”]
var games = [“Diablo3”:”2014:8:12”,“DragonAge”:”2014:10:07”]
games[“LittleBigPlanet3”] = “2014:11:29"
games[“LittleBigPlanet3”] = “2014:11:30"
var nameOfIntegers = [Int:String]()
namesOfIntegers[16] = “sixteen"
namesOfIntegers = [:]

2.修改已有键值：

if let oldValue = games.updateValue(“2014:8:14”,forKey:”Diablo3”){
     println(“Diablo3的旧值：\(oldValue)”)
}

3.获取键值为可选类型：

if let releaseDate = games[“Diablo3”]{
     println(“该游戏的发布日期是\(releaseDate)”)
}else {
     println(“该游戏的发布日期不在games字典里”)
}
games[“LittleBigPlanet3”] = nil 移除键值
games.removeValueForKey(“Diablo3”) 和updateValue一样

4.字典遍历

let airports = [“TYO”:”Tokyo”,”LHR”:”London”]
for (airportCode, airportName) in airports{
     println(“\(airportCode): \(airportName)”)
}
for airportCode in airports.keys{}
for airportName in airports.values{}

5.示例代码：

var airports:Dictionary<String,String> = ["TKO":"Tokyo","CHA":"China"]

println("the airports Dictionary has \(airports.count) airport")

airports["LON"] ="London"

airports["LON"] = "London weather"

airports["CHA"] =nil

iflet oldValue = airports.updateValue("Dublin", forKey:"CHA"){

    println("the old value is\(oldValue)")

}else{

    println("there is no airport named CHA")

}

for (airportNumber,airportName)in airports{

    println("airportNumber:\(airportNumber) airportName:"+airportName)

}

for keyin airports.keys{

    

}

for valuein airports.values{

    

}

let airportCode = Array(airports.keys)

var nameOfIntergers = Dictionary<String,Int>()

nameOfIntergers = [:]