Swift字典

字典初始化

基本语法：

[key 1: value 1, key 2: value 2, key 3: value

3]

 var   airports:    Dictionary<String,       String>    = ["TYO": "Tokyo", "DUB":"Dublin"]

 字典追加元素

 var   airports:    Dictionary<String,       String>    = ["TYO": "Tokyo", "DUB":"Dublin"] airports["LHR"] = "London"

 println("The dictionary of airports contains

(airports.count) items.")

字典删除元素

通过 removeValueForKey(key)方法删除

 var airports: Dictionary<String, String> =

["TYO": "Tokyo", "DUB": "Dublin"]

if         let          removedValue                   =

airports.removeValueForKey("DUB") {

println("The removed airport's nameis (removedValue).")

} else {

println("The airports dictionary doesnot contain a value forDUB.")

}

字典长度

使用 count 属性。

 var   airports:    Dictionary<String,       String>    = ["TYO": "Tokyo", "DUB":"Dublin"]

 println("The dictionary of airports contains

(airports.count) items.")

字典遍历

1.遍历方法

 var airports = ["TYO": "Tokyo", "DUB": "Dublin"]

 for (airportCode,airportName) in airports {

 println("(airportCode): (airportName)")

 }

２.遍历键和值

for airportCode in airports.keys {

 println("Airport code: (airportCode)")

 }

 for airportName in airports.values {

 println("Airport name: (airportName)")

 }

获得键或值的数组：

let airportCodes= Array(airports.keys)

 // airportCodes is ["TYO", "LHR"]

let airportNames = Array(airports.values)

  // airportNamesis ["Tokyo","London Heathrow"]

Swift交流讨论论坛论坛：http://www.cocoagame.net

欢迎加入Swift技术交流群：362298485