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
这在swift 5中很管用
var myString = " Kwame Ch ef "
myString = myString.replacingOccurrences(of: " ", with: "")
print(myString)
output: Kwame Chef
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)] }
}
class SpaceRemover
{
func SpaceRemover(str :String)->String
{
var array = Array(str)
var i = array.count
while(array.last == " ")
{
var array1 = [Character]()
for item in 0...i - 1
{
array1.append(array[item])
}
i = i - 1
array = array1
print(array1)
print(array)
}
var arraySecond = array
var j = arraySecond.count
while(arraySecond.first == " ")
{
var array2 = [Character]()
if j > 1
{
for item in 1..<j
{
array2.append(arraySecond[item])
}
}
j = j - 1
arraySecond = array2
print(array2)
print(arraySecond)
}
print(arraySecond)
return String(arraySecond)
}
}
斯威夫特3
你可以简单地使用这个方法删除字符串中的所有普通空格(不考虑所有类型的空白):
let myString = " Hello World ! "
let formattedString = myString.replacingOccurrences(of: " ", with: "")
结果将是:
HelloWorld!
对于任何想要从字符串中删除前导空格的人(正如问题标题所明确要求的那样),这里有一个答案:
假设:
let string = " Hello, World! "
要删除所有前导空格,使用以下代码:
var filtered = ""
var isLeading = true
for character in string {
if character.isWhitespace && isLeading {
continue
} else {
isLeading = false
filtered.append(character)
}
}
print(filtered) // "Hello, World! "
我相信有比这更好的代码,但它已经完成了我的工作。
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.