我需要一种方法来删除字符串的第一个字符,这是一个空格。我正在寻找一个方法,甚至是一个扩展的字符串类型,我可以用来削减字符串的字符。
当前回答
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)] }
}
其他回答
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 "
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}
这个String扩展删除了字符串中的所有空格,而不仅仅是尾随空格…
extension String {
func replace(string:String, replacement:String) -> String {
return self.replacingOccurrences(of: string, with: replacement, options: NSString.CompareOptions.literal, range: nil)
}
func removeWhitespace() -> String {
return self.replace(string: " ", replacement: "")
}
}
例子:
let string = "The quick brown dog jumps over the foxy lady."
let result = string.removeWhitespace() // Thequickbrowndogjumpsoverthefoxylady.
尝试函数式编程来删除空白:
extension String {
func whiteSpacesRemoved() -> String {
return self.filter { $0 != Character(" ") }
}
}
羽毛球猫的答案的Swift 3版本
extension String {
func replace(_ string:String, replacement:String) -> String {
return self.replacingOccurrences(of: string, with: replacement, options: NSString.CompareOptions.literal, range: nil)
}
func removeWhitespace() -> String {
return self.replace(" ", replacement: "")
}
}