Swift基础之03-字符串，字典，数组，元祖等

import UIKit

---

字符串

let name: String = "zhangdanfeng"
let name01 = "zhangdanfeng01"

for char in name.characters{
    print(char)
}

let names = name + name01

let age = 30

let info = "name:\(name), age:\(age)"

let min = 2
let second = 18

let timeStr = String(format: "%02d:%02d", min, second)
let timeStr01 = String(format: "%02d:%02d", arguments: [min, second])

let urlStr = "www.baidu.com"

//let subStr = urlStr.substring(to: String.Index)//Index类型比较麻烦，所以可以把String转换成NSString进行字符串的截取

let subStr = (urlStr as NSString).substring(to: 3)
let subStr01 = (urlStr as NSString).substring(with: NSMakeRange(4, 5))
let subStr02 = (urlStr as NSString).substring(with: NSMakeRange(10, 3))

---

数组

1，数组定义

• 不可变数组

//只放字符串的不可变数组
let array: [String] = ["zhang","dan","feng"]//最常用
let array01: Array = ["zhang","dan","feng"]
//可以放任意类型的不可变数组
let array02 = ["zhang", "dan", "feng",20] as Any
let array03: Any = ["zhang", "dan", "feng",20] as Any

• 可变数组

//只放字符串的可变数组
var arrayMutable = Array()
var arrayMutable01 = [String]()
//可以放任意类型的可变数组
var arrayMutable02 = Array()
var arrayMutable03 = [Any]()

2，可变数组的基本操作

• 添加

arrayMutable.append("zhangdanfeng")
arrayMutable.append("li")
arrayMutable.append("d")

• 删除

arrayMutable.removeLast()
arrayMutable.remove(at: 2)

• 修改

arrayMutable[1] = "OK"

• 查询元素

arrayMutable[1]
let myStr = arrayMutable[1]

3，遍历数组

//常用
for var i in 0..

4，数组合并(相同类型的数组才可以合并)

let combinedArray = array + array01
//let combinedArray01 = array02 + array03//两个Any数组也不能合并

---

字典

1，定义字典：字典也是用中括号【】

• 不可变字典

let dict:Dictionary = ["name":"zhangdanfeng", "age":"20"];
let dict01 = ["name":"zhangdanfeng", "age":"20"];
//let dict02:Dictionary = ["name":"zhangdanfeng", "age":"20"];//一般这种写法会用下面那一句
let dict02 : [String: Any] = ["name":"zhangdanfeng", "age":"20"];
let dict03 = ["name":"zhangdanfeng", "age":20] as Any;

• 可变字典

var dictMutable = Dictionary()
var dictMutable01 = [String:Any]()
var dictMutable02 = [String:Any]()
var dictMutable03 = [String:AnyObject]()

2，对可变字典的基本操作

• 添加元素

dictMutable["name"] = "zhangdanfeng" as NSObject?
dictMutable["age"] = 20 as NSObject?//这是因为swift中的字符串不是NSObject类型，不是NSString，而是String类型
dictMutable["num"] = 55 as NSObject?
dictMutable

• 删除元素

dictMutable.removeValue(forKey: "name")
dictMutable

• 修改元素

dictMutable["age"] = 24 as NSObject?
dictMutable

• 获取元素

let age = dictMutable["age"]

3，遍历字典

for key in dictMutable.keys{
    print(key)//注意是无序的
}

for value in dictMutable.values {
    print(value)
}

for (key, value) in dictMutable {
    print("key:\(key), value:\(value)")
}

4，合并字典

//即使类型一致也不能直接相加合并，想合并只能遍历添加合并
var myDict01 = ["name":"zhangdanfeng", "age":24, "num":100] as [String : Any]
var myDict02 = ["height":1.69];
for (key, value) in myDict02{
    print(key, value)
    myDict01[key] = value
}

---

元组

• 元组:一般可以用于作为方法的返回值(之后还会提到)

• 为什么用元祖：

• 从下面可以看到，如果字典或者数组保存的类型是any的时候，取出的类型也是any，那么在使用的时候需要转换，甚至有时候强制转换出错的时候就会造成程序崩溃

• 元祖的优点是取出值的类型就是其保存的类型，而且在取出的时候还会有提示

//先看一下数组如何保存
let array = ["zhangdanfeng", 18, 1.69] as [Any]//一般我们也不会这么写，而是用下面这句
let array:  [Any] = ["zhangdanfeng", 18, 1.69]
array[0]//因为存储的时候是any，所以取出的也是any类型，需要转换

 //然后是字典
let dict : [String : Any]= ["name":"zhangdanfeng", "age":24]
dict["name"]//因为存储的时候是any，所以取出的也是any类型，需要转换

//方式一：常规方式
let info = ("zhangdanfeng", 18, 1.69)
info.0//String类型
info.1//Int类型
info.1//Double类型

• 可以给元组起别名

//方式二：类zi'dian
let info01 = (name:"zhangdanfeng",age:20)
info01.name
info01.0

let (name, age, height) = ("zhang", 20, 2)
name
age
height