Swift4 获取String子字符串

string的长度可以直接用count了
有了prefix()和suffix()获取头尾的相应范围的子串

string.substring(to: <#T##String.Index#>)
string.substring(from: <#T##String.Index#>)
string.substring(with: <#T##Range<String.Index>#>)
这三个方方法统一为
string[<#T##Range<String.Index>#>]
即
string[index1..<index2] 左闭右开
或者
string[index1...index2] 全闭合

当然如果是从头开始你可以把
string[str.startIndex..<index]
简写成
string[..<index]

let str = "Do any additional setup after loading the view, typically from a nib."

print(str.count)
//别试了 69

//方法一：这个够简单
let sub1 = str.prefix(10)

//方法二：这个延续上一个版本
let index = str.index(str.startIndex, offsetBy: 10)

//但是下面这句被苹果废了哦
//let sub2 = str.substring(to: index1)

//改成这样了，哈哈，果然越来越精简了
let sub2 = str[..<index]

//input: "Do any add"

//这个和上面一样
let sub3 = str.suffix(10)

//这个也是一样
let index2 = str.index(str.endIndex, offsetBy: -10)

//这个同样被废弃了
//let sub4 = str.substring(from: index2)

//改成这样才多哦
let sub4 = str[index2..<str.endIndex]

//input: "rom a nib."

let index3 = str.index(str.startIndex, offsetBy: 3) //为啥是3...因为...你懂得
let index4 = str.index(str.startIndex, offsetBy: 6)

//过期了哦
//let str3 = str.substring(with: index3..<index4)

let sub4 = str[index3..<index4]

//input: "any"

extension String {
   //获取子字符串
    func substingInRange(_ r: Range<Int>) -> String? {
        if r.lowerBound < 0 || r.upperBound > self.count {
            return nil
        }
        let startIndex = self.index(self.startIndex, offsetBy:r.lowerBound)
        let endIndex   = self.index(self.startIndex, offsetBy:r.upperBound)
        return String(self[startIndex..<endIndex])
    }
}

let str4 = str.substingInRange(3..<6)
print("string from 4 - 6 : \(str4)") 
 //string from 4 - 6 : any