从技术上讲，这不是对原始问题的回答，但由于这里的许多帖子都给出了删除所有空白的答案，这里是一个更新的、更简洁的版本:

let stringWithouTAnyWhitespace = string.filter {!$0.isWhitespace}

另一个答案是，有时输入字符串在单词之间可以包含多个空格。如果你需要标准化，单词之间只有一个空格，试试这个(Swift 4/5)

let inputString = "  a very     strange   text !    "
let validInput = inputString.components(separatedBy:.whitespacesAndNewlines).filter { $0.count > 0 }.joined(separator: " ")

print(validInput) // "a very strange text !"

如果有人从字符串中删除额外的空格，例如= "这是演示文本，请在单词之间删除额外的空格。"

你可以在Swift 4中使用这个函数。

func removeSpace(_ string: String) -> String{
    var str: String = String(string[string.startIndex])
    for (index,value) in string.enumerated(){
        if index > 0{
            let indexBefore = string.index(before: String.Index.init(encodedOffset: index))
            if value == " " && string[indexBefore] == " "{
            }else{
                str.append(value)
            }
        }
    }
    return str
}

结果是

"This is the demo text remove extra space between the words."

extension String {

    var removingWhitespaceAndNewLines: String {
        return removing(.whitespacesAndNewlines)
    }

    func removing(_ forbiddenCharacters: CharacterSet) -> String {
        return String(unicodeScalars.filter({ !forbiddenCharacters.contains($0) }))
    }
}

快速解决方案:

用法:

let txt = "        hello world     "
let txt1 = txt.trimStart() // "hello world     "
let txt2 = txt.trimEnd()   // "        hello world"

用法二:

let txt = "rr rrr rrhello world r r r r r r"
let txt1 = txt.trimStart(["r", " "]) // "hello world r r r r r r"
let txt2 = txt.trimEnd(["r", " "])   // "rr rrr rrhello world"

如果你需要移除字符串中的所有空白:

txt.replace(of: " ", to: "")

public extension String {
    func trimStart(_ char: Character) -> String {
        return trimStart([char])
    }
    
    func trimStart(_ symbols: [Character] = [" ", "\t", "\r", "\n"]) -> String {
        var startIndex = 0
        
        for char in self {
            if symbols.contains(char) {
                startIndex += 1
            }
            else {
                break
            }
        }
        
        if startIndex == 0 {
            return self
        }
        
        return String( self.substring(from: startIndex) )
    }
    
    func trimEnd(_ char: Character) -> String {
        return trimEnd([char])
    }
    
    func trimEnd(_ symbols: [Character] = [" ", "\t", "\r", "\n"]) -> String {
        var endIndex = self.count - 1
        
        for i in (0...endIndex).reversed() {
            if symbols.contains( self[i] ) {
                endIndex -= 1
            }
            else {
                break
            }
        }
        
        if endIndex == self.count {
            return self
        }
        
        return String( self.substring(to: endIndex + 1) )
    }
}

/////////////////////////
/// ACCESS TO CHAR BY INDEX
////////////////////////
extension StringProtocol {
    subscript(offset: Int) -> Character { self[index(startIndex, offsetBy: offset)] }
    subscript(range: Range<Int>) -> SubSequence {
        let startIndex = index(self.startIndex, offsetBy: range.lowerBound)
        return self[startIndex..<index(startIndex, offsetBy: range.count)]
    }
    subscript(range: ClosedRange<Int>) -> SubSequence {
        let startIndex = index(self.startIndex, offsetBy: range.lowerBound)
        return self[startIndex..<index(startIndex, offsetBy: range.count)]
    }
    subscript(range: PartialRangeFrom<Int>) -> SubSequence { self[index(startIndex, offsetBy: range.lowerBound)...] }
    subscript(range: PartialRangeThrough<Int>) -> SubSequence { self[...index(startIndex, offsetBy: range.upperBound)] }
    subscript(range: PartialRangeUpTo<Int>) -> SubSequence { self[..<index(startIndex, offsetBy: range.upperBound)] }
}

Swift 3版本

  //This function trim only white space:
   func trim() -> String
        {
            return self.trimmingCharacters(in: CharacterSet.whitespaces)
        }
    //This function trim whitespeaces and new line that you enter:
     func trimWhiteSpaceAndNewLine() -> String
        {
            return self.trimmingCharacters(in: CharacterSet.whitespacesAndNewlines)
        }