Question

! !! yuahl! !之类的字符串，我希望删除!和，当我像这样编码时

for index in InputName.characters.indices {
    if String(InputName[index]) == "" || InputName.substringToIndex(index) == "!" {
        InputName.removeAtIndex(index)
    }
}

有这个错误“致命错误：下标：subRange扩展超过String结束”，我该怎么办？ THX：D

Answer 1

Swift 3 +

let myString = "aaaaaaaabbbb"
let replaced = myString.replacingOccurrences(of: "bbbb", with: "") // "aaaaaaaa"

Answer 2

我发现filter方法是一种很好的方法：

let unfiltered = "!   !! yuahl! !"

// Array of Characters to remove
let removal: [Character] = ["!"," "]

// turn the string into an Array
let unfilteredCharacters = unfiltered.characters

// return an Array without the removal Characters
let filteredCharacters = unfilteredCharacters.filter { !removal.contains($0) }

// build a String with the filtered Array
let filtered = String(filteredCharacters)

print(filtered) // => "yeah"

// combined to a single line
print(String(unfiltered.characters.filter { !removal.contains($0) })) // => "yuahl"

Answer 3

如果您只需要删除两端的字符，可以使用stringByTrimmingCharactersInSet(_:)

let delCharSet = NSCharacterSet(charactersInString: "! ")

let s1 = "! aString! !"
let s1Del = s1.stringByTrimmingCharactersInSet(delCharSet)
print(s1Del) //->aString

let s2 = "! anotherString !! aString! !"
let s2Del = s2.stringByTrimmingCharactersInSet(delCharSet)
print(s2Del) //->anotherString !! aString

如果您还需要在中间删除字符，请从过滤后的输出中重建＆＃34;比重复单个字符删除更有效率。

var tempUSView = String.UnicodeScalarView()
tempUSView.appendContentsOf(s2.unicodeScalars.lazy.filter{!delCharSet.longCharacterIsMember($0.value)})
let s2DelAll = String(tempUSView)
print(s2DelAll) //->anotherStringaString

如果您不介意生成许多中间字符串和数组，这个单线程可以生成预期的输出：

let s2DelAll2 = s2.componentsSeparatedByCharactersInSet(delCharSet).joinWithSeparator("")
print(s2DelAll2) //->anotherStringaString

Answer 4

Swift 3

在Swift 3中，语法更好一些。由于旧API的Great Swiftification，工厂方法现在称为trimmingCharacters(in:)。此外，您可以将CharacterSet构建为单个字符Set的{{1}}：

String

如果您要删除字符串中间的字符，则可以使用let string = "! !! yuahl! !" string.trimmingCharacters(in: [" ", "!"]) // "yuahl"：

components(separatedBy:).joined()

Swift 2版本的H / T @OOPer

Answer 5

func trimLast(character chars: Set<Character>) -> String {

    let str: String = String(self.reversed())

    guard let index = str.index(where: {!chars.contains($0)}) else {
        return self
    }
    return String((str[index..<str.endIndex]).reversed())
}

注意：通过在String扩展名中添加此功能，您可以最后删除字符串的特定字符。

Answer 6

for index in InputName.characters.indices.reversed() {
    if String(InputName[index]) == "" || InputName.substringToIndex(index) == "!" {
        InputName.removeAtIndex(index)
    }
}

Answer 7

您还可以添加这样非常有用的扩展程序：

import Foundation


extension String{
 func exclude(find:String) -> String {
    return stringByReplacingOccurrencesOfString(find, withString: "", options: .CaseInsensitiveSearch, range: nil)
 }
 func replaceAll(find:String, with:String) -> String {
    return stringByReplacingOccurrencesOfString(find, withString: with, options: .CaseInsensitiveSearch, range: nil)
 }
}

swift：如何删除特定字符？

7 个答案: