我还没有找到能够处理3个或更多组件名称并支持较旧iOS版本的解决方案。

struct NameComponentsSplitter {

    static func split(fullName: String) -> (String?, String?) {
        guard !fullName.isEmpty else {
            return (nil, nil)
        }
        let components = fullName.components(separatedBy: .whitespacesAndNewlines)
        let lastName = components.last
        let firstName = components.dropLast().joined(separator: " ")
        return (firstName.isEmpty ? nil : firstName, lastName)
    }
}

通过测试用例：

func testThatItHandlesTwoComponents() {
    let (firstName, lastName) = NameComponentsSplitter.split(fullName: "John Smith")
    XCTAssertEqual(firstName, "John")
    XCTAssertEqual(lastName, "Smith")
}

func testThatItHandlesMoreThanTwoComponents() {
    var (firstName, lastName) = NameComponentsSplitter.split(fullName: "John Clark Smith")
    XCTAssertEqual(firstName, "John Clark")
    XCTAssertEqual(lastName, "Smith")

    (firstName, lastName) = NameComponentsSplitter.split(fullName: "John Clark Jr. Smith")
    XCTAssertEqual(firstName, "John Clark Jr.")
    XCTAssertEqual(lastName, "Smith")
}

func testThatItHandlesEmptyInput() {
    let (firstName, lastName) = NameComponentsSplitter.split(fullName: "")
    XCTAssertEqual(firstName, nil)
    XCTAssertEqual(lastName, nil)
}

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)

在解释Don Vaughn的答案时，我喜欢使用正则表达式。我很惊讶这只是Regex的第二个答案。然而，如果我们能用一个拆分方法而不是多个方法来解决这个问题，那就太好了。

我也受到了米特拉·辛加姆（Mithra Singam）的《答案》（Answer）的启发，排除了所有标点符号和空格。然而，不得不创建一个不允许的角色列表并没有引起我的共鸣。

\w-字母或数字符号的正则表达式。没有标点符号。

let foo = "(..#   Hello,,(---- World   ".split {
    String($0).range(of: #"\w"#, options: .regularExpression) == nil
}
print(foo) // Prints "Hello World"

让我们假设你对Unicode不太满意。ASKII字母和数字怎么样？

let bar = "(..#   Hello,,(---- World   ".split {
    !($0.isASCII && ($0.isLetter || $0.isNumber))
}
print(bar) // Prints "Hello World"

Swift 4或更高版本

如果您只需要正确设置人名格式，可以使用PersonNameComponentsFormater。

PersonNameComponentsFormater类提供本地化人名组成部分的表示通过PersonNameComponents对象。使用此类创建本地化当向用户显示人名信息时。

// iOS (9.0 and later), macOS (10.11 and later), tvOS (9.0 and later), watchOS (2.0 and later)
let nameFormatter = PersonNameComponentsFormatter()

let name =  "Mr. Steven Paul Jobs Jr."
// personNameComponents requires iOS (10.0 and later)
if let nameComps  = nameFormatter.personNameComponents(from: name) {
    nameComps.namePrefix   // Mr.
    nameComps.givenName    // Steven
    nameComps.middleName   // Paul
    nameComps.familyName   // Jobs
    nameComps.nameSuffix   // Jr.

    // It can also be configured to format your names
    // Default (same as medium), short, long or abbreviated

    nameFormatter.style = .default
    nameFormatter.string(from: nameComps)   // "Steven Jobs"

    nameFormatter.style = .short
    nameFormatter.string(from: nameComps)   // "Steven"

    nameFormatter.style = .long
    nameFormatter.string(from: nameComps)   // "Mr. Steven Paul Jobs jr."

    nameFormatter.style = .abbreviated
    nameFormatter.string(from: nameComps)   // SJ

    // It can also be use to return an attributed string using annotatedString method
    nameFormatter.style = .long
    nameFormatter.annotatedString(from: nameComps)   // "Mr. Steven Paul Jobs jr."
}

编辑/更新：

Swift 5或更高版本

对于按非字母字符拆分字符串，我们可以使用新的Character属性isLetter：

let fullName = "First Last"

let components = fullName.split{ !$0.isLetter }
print(components)  // "["First", "Last"]\n"

在Swift 4.2和Xcode 10中

//This is your str
let str = "This is my String" //Here replace with your string

选项1

let items = str.components(separatedBy: " ")//Here replase space with your value and the result is Array.
//Direct single line of code
//let items = "This is my String".components(separatedBy: " ")
let str1 = items[0]
let str2 = items[1]
let str3 = items[2]
let str4 = items[3]
//OutPut
print(items.count)
print(str1)
print(str2)
print(str3)
print(str4)
print(items.first!)
print(items.last!)

选项2

let items = str.split(separator: " ")
let str1 = String(items.first!)
let str2 = String(items.last!)
//Output
print(items.count)
print(items)
print(str1)
print(str2)

选项3

let arr = str.split {$0 == " "}
print(arr)

选项4

let line = "BLANCHE:   I don't want realism. I want magic!"
print(line.split(separator: " "))
// Prints "["BLANCHE:", "I", "don\'t", "want", "realism.", "I", "want", "magic!"]"

通过Apple文档。。。。

let line = "BLANCHE:   I don't want realism. I want magic!"
print(line.split(separator: " "))
// Prints "["BLANCHE:", "I", "don\'t", "want", "realism.", "I", "want", "magic!"]"

print(line.split(separator: " ", maxSplits: 1))//This can split your string into 2 parts
// Prints "["BLANCHE:", "  I don\'t want realism. I want magic!"]"

print(line.split(separator: " ", maxSplits: 2))//This can split your string into 3 parts

print(line.split(separator: " ", omittingEmptySubsequences: false))//array contains empty strings where spaces were repeated.
// Prints "["BLANCHE:", "", "", "I", "don\'t", "want", "realism.", "I", "want", "magic!"]"

print(line.split(separator: " ", omittingEmptySubsequences: true))//array not contains empty strings where spaces were repeated.
print(line.split(separator: " ", maxSplits: 4, omittingEmptySubsequences: false))
print(line.split(separator: " ", maxSplits: 3, omittingEmptySubsequences: true))

最简单的解决方案是

let fullName = "First Last"

let components = fullName.components(separatedBy: .whitespacesAndNewlines).compactMap { $0.isEmpty ? nil : $0 }

这将处理不同类型的一行中的多个空格（空格、制表符、换行符等），并且只返回一个两元素数组，您可以更改CharacterSet以包含更多您喜欢的字符，如果您想获得切割符，可以使用正则表达式解码器，这允许您编写正则表达式，该表达式可用于将字符串直接解码到实现解码协议的自己的类/结构中。对于像这样的东西来说，这是过度的，但如果您将其用作更复杂字符串的示例，则可能更有意义。