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
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)] }
}
尝试函数式编程来删除空白:
extension String {
func whiteSpacesRemoved() -> String {
return self.filter { $0 != Character(" ") }
}
}
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."
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}
这在swift 5中很管用
var myString = " Kwame Ch ef "
myString = myString.replacingOccurrences(of: " ", with: "")
print(myString)
output: Kwame Chef
