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