假设我这里有一个字符串:

var fullName: String = "First Last"

我想在空白处拆分字符串,并将值分配给它们各自的变量

var fullNameArr = // something like: fullName.explode(" ") 

var firstName: String = fullNameArr[0]
var lastName: String? = fullnameArr[1]

此外,有时用户可能没有姓氏。


当前回答

空白问题

一般来说,人们会一次又一次地重复这个问题和糟糕的解决方案。这是一个空间吗?“”以及“\n”、“\t”或一些您从未见过的unicode空白字符,在很大程度上是不可见的。虽然你可以逃脱

弱解决方案

import Foundation
let pieces = "Mary had little lamb".componentsSeparatedByString(" ")

如果你需要在现实中动摇你的控制,请观看WWDC视频中的字符串或日期。简而言之,让苹果解决这类平凡的任务几乎总是更好的。

稳健的解决方案:使用NSCharacterSet

IMHO,正确做到这一点的方法是使用NSCharacterSet,因为正如前面所述,您的空白可能不是您所期望的,并且Apple提供了一个空白字符集。要探索提供的各种字符集,请查看Apple的NSCharacterSet开发人员文档,然后,如果不符合您的需要,才可以扩充或构建新的字符集。

NSCharacterSet空白

返回包含Unicode常规中的字符的字符集Z类和字符表(U+0009)。

let longerString: String = "This is a test of the character set splitting system"
let components = longerString.components(separatedBy: .whitespaces)
print(components)

其他回答

Swift Dev.4.0(2017年5月24日)

Swift 4(Beta版)中拆分了一个新功能。

import Foundation
let sayHello = "Hello Swift 4 2017";
let result = sayHello.split(separator: " ")
print(result)

输出:

["Hello", "Swift", "4", "2017"]

访问值:

print(result[0]) // Hello
print(result[1]) // Swift
print(result[2]) // 4
print(result[3]) // 2017

Xcode 8.1/Swift 3.0.1

这是多个分隔符与数组的方式。

import Foundation
let mathString: String = "12-37*2/5"
let numbers = mathString.components(separatedBy: ["-", "*", "/"])
print(numbers)

输出:

["12", "37", "2", "5"]

我有一个场景,其中多个控制字符可以出现在我要拆分的字符串中。我只是让苹果处理这一部分,而不是维护一系列这些。

以下功能适用于iOS 10上的Swift 3.0.1:

let myArray = myString.components(separatedBy: .controlCharacters)

将字符串拆分为数组的简单方法

var fullName:String=“First Last”;var fullNameArr=fullName.componentsSeparatedByString(“”)var firstName:String=fullNameArr[0]var lastName:String=fullNameArr[1]

根据Swift 2.2

您只需编写2行代码,就会得到拆分字符串。

let fullName = "FirstName LastName"
var splitedFullName = fullName.componentsSeparatedByString(" ")
print(splitedFullName[0])
print(splitedFullName[1]) 

享受。:)

非顶部:

对于搜索如何使用子字符串(而不是字符)拆分字符串的人来说,下面是一个有效的解决方案:

// TESTING
let str1 = "Hello user! What user's details? Here user rounded with space."
let a = str1.split(withSubstring: "user") // <-------------- HERE IS A SPLIT
print(a) // ["Hello ", "! What ", "\'s details? Here ", " rounded with space."]

// testing the result
var result = ""
for item in a {
    if !result.isEmpty {
        result += "user"
    }
    result += item
}
print(str1) // "Hello user! What user's details? Here user rounded with space."
print(result) // "Hello user! What user's details? Here user rounded with space."
print(result == str1) // true

/// Extension providing `split` and `substring` methods.
extension String {
    
    /// Split given string with substring into array
    /// - Parameters:
    ///   - string: the string
    ///   - substring: the substring to search
    /// - Returns: array of components
    func split(withSubstring substring: String) -> [String] {
        var a = [String]()
        var str = self
        while let range = str.range(of: substring) {
            let i = str.distance(from: str.startIndex, to: range.lowerBound)
            let j = str.distance(from: str.startIndex, to: range.upperBound)
            let left = str.substring(index: 0, length: i)
            let right = str.substring(index: j, length: str.length - j)
            a.append(left)
            str = right
        }
        if !str.isEmpty {
            a.append(str)
        }
        return a
    }
    
    /// the length of the string
    public var length: Int {
        return self.count
    }
    
    /// Get substring, e.g. "ABCDE".substring(index: 2, length: 3) -> "CDE"
    ///
    /// - parameter index:  the start index
    /// - parameter length: the length of the substring
    ///
    /// - returns: the substring
    public func substring(index: Int, length: Int) -> String {
        if self.length <= index {
            return ""
        }
        let leftIndex = self.index(self.startIndex, offsetBy: index)
        if self.length <= index + length {
            return String(self[leftIndex..<self.endIndex])
        }
        let rightIndex = self.index(self.endIndex, offsetBy: -(self.length - index - length))
        return String(self[leftIndex..<rightIndex])
    }

}