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

对于swift 2,XCode 7.1:

let complete_string:String = "Hello world"
let string_arr =  complete_string.characters.split {$0 == " "}.map(String.init)
let hello:String = string_arr[0]
let world:String = string_arr[1]

其他回答

Xcode 8.0/Swift 3

let fullName = "First Last"
var fullNameArr = fullName.components(separatedBy: " ")

var firstname = fullNameArr[0] // First
var lastname = fullNameArr[1] // Last

漫长的道路:

var fullName: String = "First Last"
fullName += " " // this will help to see the last word

var newElement = "" //Empty String
var fullNameArr = [String]() //Empty Array

for Character in fullName.characters {
    if Character == " " {
        fullNameArr.append(newElement)
        newElement = ""
    } else {
        newElement += "\(Character)"
    }
}


var firsName = fullNameArr[0] // First
var lastName = fullNameArr[1] // Last

这在Beta 5中再次发生了变化。哭泣!它现在是CollectionType上的一个方法

旧版本:

var fullName = "First Last"
var fullNameArr = split(fullName) {$0 == " "}

新建:

var fullName = "First Last"
var fullNameArr = fullName.split {$0 == " "}

苹果发行说明

Swift 5.2的更新和最简单的方式

let paragraph = "Bob hit a ball, the hit BALL flew far after it was hit. Hello! Hie, How r u?"

let words = paragraph.components(separatedBy: [",", " ", "!",".","?"])

这张照片,

[“Bob”,“hit”,“a”,“ball”,“the”,“击球”,“ball”,“fly”,“far”,“after”,“it”,“was”,“hit”,“Hello”,“Hie”,“How”,“u”,“”]

但是,如果要过滤空字符串,

let words = paragraph.components(separatedBy: [",", " ", "!",".","?"]).filter({!$0.isEmpty})

输出

[“Bob”,“hit”,“a”,“ball”,“the”,“命中”,“ball”,“fly”,“far”,“after”,“it”,“was”,“hit”,“Hello”,”Hie“,”How“,”r“,”u“]

但请确保,Foundation已导入。

这里是我刚刚构建的一个算法,它将通过数组中的任何字符分割字符串,如果有任何希望保留具有分割字符的子字符串,可以将swall参数设置为true。

Xcode 7.3-Swift 2.2:

extension String {

    func splitBy(characters: [Character], swallow: Bool = false) -> [String] {

        var substring = ""
        var array = [String]()
        var index = 0

        for character in self.characters {

            if let lastCharacter = substring.characters.last {

                // swallow same characters
                if lastCharacter == character {

                    substring.append(character)

                } else {

                    var shouldSplit = false

                    // check if we need to split already
                    for splitCharacter in characters {
                        // slit if the last character is from split characters or the current one
                        if character == splitCharacter || lastCharacter == splitCharacter {

                            shouldSplit = true
                            break
                        }
                    }

                    if shouldSplit {

                        array.append(substring)
                        substring = String(character)

                    } else /* swallow characters that do not equal any of the split characters */ {

                        substring.append(character)
                    }
                }
            } else /* should be the first iteration */ {

                substring.append(character)
            }

            index += 1

            // add last substring to the array
            if index == self.characters.count {

                array.append(substring)
            }
        }

        return array.filter {

            if swallow {

                return true

            } else {

                for splitCharacter in characters {

                    if $0.characters.contains(splitCharacter) {

                        return false
                    }
                }
                return true
            }
        }
    }
}

例子:

"test text".splitBy([" "]) // ["test", "text"]
"test++text--".splitBy(["+", "-"], swallow: true) // ["test", "++" "text", "--"]

斯威夫特4

let words = "these words will be elements in an array".components(separatedBy: " ")