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

let str = "one two"
let strSplit = str.characters.split(" ").map(String.init) // returns ["one", "two"]

Xcode 7.2(7C68)

其他回答

只有拆分才是正确的答案,这里有两个以上空格的差异。

雨燕5

var temp = "Hello world     ni hao"
let arr  = temp.components(separatedBy: .whitespacesAndNewlines)
// ["Hello", "world", "", "", "", "", "ni", "hao"]
let arr2 = temp.components(separatedBy: " ")
// ["Hello", "world", "", "", "", "", "ni", "hao"]
let arr3 = temp.split(whereSeparator: {$0 == " "})
// ["Hello", "world", "ni", "hao"]

Swift 4、Xcode 10和iOS 12更新100%工作

let fullName = "First Last"    
let fullNameArr = fullName.components(separatedBy: " ")
let firstName = fullNameArr[0] //First
let lastName = fullNameArr[1] //Last

有关更多信息,请参阅此处的Apple文档。

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]

您可以使用此公共函数并添加任何要分隔的字符串

func separateByString(String wholeString: String, byChar char:String) -> [String] {

    let resultArray = wholeString.components(separatedBy: char)
    return resultArray
}

var fullName: String = "First Last"
let array = separateByString(String: fullName, byChar: " ")
var firstName: String = array[0]
var lastName: String = array[1]
print(firstName)
print(lastName)

作为WMios答案的替代方案,您还可以使用componentsSeparatedByCharactersSet,这在分隔符(空格、逗号等)较多的情况下非常方便。

使用您的特定输入:

let separators = NSCharacterSet(charactersInString: " ")
var fullName: String = "First Last";
var words = fullName.componentsSeparatedByCharactersInSet(separators)

// words contains ["First", "Last"]

使用多个分隔符:

let separators = NSCharacterSet(charactersInString: " ,")
var fullName: String = "Last, First Middle";
var words = fullName.componentsSeparatedByCharactersInSet(separators)

// words contains ["Last", "First", "Middle"]