当你想要删除所有类型的空白时(基于这个SO答案)的正确方法是:

extension String {
    var stringByRemovingWhitespaces: String {
        let components = componentsSeparatedByCharactersInSet(.whitespaceCharacterSet())
        return components.joinWithSeparator("")
    }
}

Swift 3.0+ (3.0, 3.1, 3.2, 4.0)

extension String {
    func removingWhitespaces() -> String {
        return components(separatedBy: .whitespaces).joined()
    }
}

EDIT

这个答案是在问题是关于删除所有空白时发布的，问题被编辑为只提到前导空白。如果你只想删除前导空白，请使用以下方法:

extension String {
    func removingLeadingSpaces() -> String {
        guard let index = firstIndex(where: { !CharacterSet(charactersIn: String($0)).isSubset(of: .whitespaces) }) else {
            return self
        }
        return String(self[index...])
    }
}

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)] }
}

嗨，这可能有点晚，但值得一试。这是一个游乐场的文件。你可以让它成为一个字符串扩展名。

这是在Swift 5.3中编写的

方法1:

var str = "\n \tHello, playground       "
if let regexp = try? NSRegularExpression(pattern: "^\\s+", options: NSRegularExpression.Options.caseInsensitive) {
    let mstr = NSMutableString(string: str)
    regexp.replaceMatches(in: mstr, options: [], range: NSRange(location: 0, length: str.count), withTemplate: "")
    str = mstr as String
}

Result: "Hello, playground       "

方法2:

if let c = (str.first { !($0 == " " || $0 == "\t" || $0 == "\n") }) {
    if let nonWhiteSpaceIndex = str.firstIndex(of: c) {
        str.replaceSubrange(str.startIndex ..< nonWhiteSpaceIndex, with: "")
    }
}

Result: "Hello, playground       "

当你想要删除所有类型的空白时(基于这个SO答案)的正确方法是:

extension String {
    var stringByRemovingWhitespaces: String {
        let components = componentsSeparatedByCharactersInSet(.whitespaceCharacterSet())
        return components.joinWithSeparator("")
    }
}

Swift 3.0+ (3.0, 3.1, 3.2, 4.0)

extension String {
    func removingWhitespaces() -> String {
        return components(separatedBy: .whitespaces).joined()
    }
}

EDIT

这个答案是在问题是关于删除所有空白时发布的，问题被编辑为只提到前导空白。如果你只想删除前导空白，请使用以下方法:

extension String {
    func removingLeadingSpaces() -> String {
        guard let index = firstIndex(where: { !CharacterSet(charactersIn: String($0)).isSubset(of: .whitespaces) }) else {
            return self
        }
        return String(self[index...])
    }
}

你也可以使用正则表达式。

let trimmedString = myString.stringByReplacingOccurrencesOfString("\\s", withString: "", options: NSStringCompareOptions.RegularExpressionSearch, range: nil)

尝试函数式编程来删除空白:

extension String {
  func whiteSpacesRemoved() -> String {
    return self.filter { $0 != Character(" ") }
  }
}