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 airportNameinairports.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

if let 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 key in airports.keys{

}

for value in airports.values{

}

let airportCode = Array(airports.keys)

var nameOfIntergers = Dictionary<String,Int>()

nameOfIntergers = [:]

相关文章

软件简介:蓝湖辅助工具,减少移动端开发中控件属性的复制和粘...
现实生活中,我们听到的声音都是时间连续的,我们称为这种信...
前言最近在B站上看到一个漂亮的仙女姐姐跳舞视频,循环看了亿...
【Android App】实战项目之仿抖音的短视频分享App(附源码和...
前言这一篇博客应该是我花时间最多的一次了,从2022年1月底至...
因为我既对接过session、cookie,也对接过JWT,今年因为工作...