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

斯威夫特4

let string = "loremipsum.dolorsant.amet:"

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

print(result[0])
print(result[1])
print(result[2])
print("total: \(result.count)")

输出

loremipsum
dolorsant
amet:
total: 3

其他回答

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。

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])
    }

}

我在寻找loosy拆分,例如PHP的分解,其中空序列包含在结果数组中,这对我很有用:

"First ".split(separator: " ", maxSplits: 1, omittingEmptySubsequences: false)

输出:

["First", ""]

我发现了一个有趣的案例

方法1

var data:[String] = split( featureData ) { $0 == "\u{003B}" }

当我使用此命令从从服务器加载的数据中拆分一些符号时,它可以在模拟器中测试时拆分,并与测试设备同步,但不会在发布应用程序和Ad Hoc中拆分

我花了很多时间来跟踪这个错误,它可能会被某些Swift版本或iOS版本诅咒,或者两者都没有

这也与HTML代码无关,因为我尝试stringByRemovangPercentEncoding,但它仍然不起作用

2015年10月10日新增

在Swift 2.0中,此方法已更改为

var data:[String] = featureData.split {$0 == "\u{003B}"}

方法2

var data:[String] = featureData.componentsSeparatedByString("\u{003B}")

当我使用此命令时,它可以正确分割从服务器加载的相同数据


结论,我真的建议使用方法2

string.componentsSeparatedByString("")

或者没有闭包,你可以在Swift 2中这样做:

let fullName = "First Last"
let fullNameArr = fullName.characters.split(" ")
let firstName = String(fullNameArr[0])