Swift 字符串处理总结

1.截取前面三个字符串,要求输出结果为abc

let str ="abcdefghi"

let index = str.index(str.startIndex, offsetBy:3)//获取字符d的索引

let result = str.substring(to: index)//从起始截取到索引的所有字符串,即abc,注意不包含d

打印结果为 abc

2.截取第四个开始的所有字符串,要求输出结果为defghi

let str ="abcdefghi"

let index = str.index(str.startIndex, offsetBy:3)//获取字符d的索引

let result = str.substring(from: index)//从d的索引开始截取后面所有的字符串即defghi

打印结果为defghi

3.截取第4个开始的长度为2个字符串,要求结果输出为de

let str ="abcdefghi"

let startIndex = str.index(str.startIndex, offsetBy:3)//获取d的索引

let endIndex = str.index(startIndex, offsetBy:2)//从d的索引开始往后两个,即获取f的索引

let result = str.substring(with: startIndex..

打印结果为de

4.判断字符串是否包含某段字符

let str ="abcdefghi"

let result = str.contains("ac")

打印结果为false

let result = str.contains("abc")

打印结果为true

5.字符串大小写转换

let str ="ABCDEFGHI"

let result = str.lowercased()

输出结果为abcdefghi

let str ="abcdefghi"

let result = str.uppercased()

输出结果为ABCDEFGHI

6.指定字符串替换为其他字符串,将abc替换成m,使输出结果为mdefghi

let str ="abcdefghi"

let result = str.replacingOccurrences(of:"abc", with:"m")

输出结果为mdefghi

7.指定范围内的字符串替换为其他字符串,将cde替换成m,使输出结果为abmfghi

let str ="abcdefghi"

let startIndex = str.index(str.startIndex, offsetBy:2)

let endIndex = str.index(startIndex, offsetBy:3)

let result = str.replacingCharacters(in: startIndex..

输出结果为abmfghi

8.有时候不确定字符串的长度,需求为截取字符串末尾面开始往前5位,截取长度为2的字符串,使输出为ef

let str ="abcdefghi"

let startIndex = str.index(str.startIndex, offsetBy: str.lengthOfBytes(using: .utf8) -5)

let endIndex = str.index(startIndex, offsetBy:2)

let result = str.substring(with: startIndex..

输出结果为ef

9. 字符串转为数组,使得输出为["a", "b", "c", "d", "e", "f", "g", "h", "i"]

let str ="a.b.c.d.e.f.g.h.i"

let result = str.components(separatedBy:".")

输出为["a", "b", "c", "d", "e", "f", "g", "h", "I"]

10.Model数组格式化的时间字符串排序

self.appArr.sort(by: { (model1, model2) -> Bool in

                            return Int(model1.createTime!.dateStamp!) > Int(model2.createTime!.dateStamp!)

                        })