从 swift 中的字符串中删除所有非数字字符

Question

我需要解析一些未知数据，这些数据应该只是一个数值，但可能包含空格或其他非字母数字字符。

在 Swift 中有新的方法吗？我在网上所能找到的似乎都是旧的 C 做事方式。

我正在查看 stringByTrimmingCharactersInSet - 因为我确定我的输入在字符串的开头或结尾只有 whitespace/special 个字符。有没有我可以使用的内置字符集？还是我需要自己创建？

我希望有类似 stringFromCharactersInSet() 的东西，它允许我只指定要保留的有效字符

Answer 1

你可以这样做...

let string = "[,myString1. \"" // string : [,myString1. " 
let characterSet = NSCharacterSet(charactersInString: "[,. \"")
let finalString = (string.componentsSeparatedByCharactersInSet(characterSet) as NSArray).componentsJoinedByString("") 
print(finalString)   
//finalString will be "myString1"

Answer 2

使用filter函数和rangeOfCharacterFromSet

的解决方案

let string = "sld [f]34é7*˜µ"

let alphaNumericCharacterSet = NSCharacterSet.alphanumericCharacterSet()
let filteredCharacters = string.characters.filter {
  return  String([=10=]).rangeOfCharacterFromSet(alphaNumericCharacterSet) != nil
}
let filteredString = String(filteredCharacters) // -> sldf34é7µ

要仅过滤数字字符，请使用

let string = "sld [f]34é7*˜µ"

let numericSet = "0123456789"
let filteredCharacters = string.characters.filter {
  return numericSet.containsString(String([=11=]))
}
let filteredString = String(filteredCharacters) // -> 347

或

let numericSet : [Character] = ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]
let filteredCharacters = string.characters.filter {
  return numericSet.contains([=12=])
}
let filteredString = String(filteredCharacters) // -> 347

Answer 3

I was hoping there would be something like stringFromCharactersInSet() which would allow me to specify only valid characters to keep.

您可以使用 trimmingCharacters 和 inverted 字符集来删除字符串开头或结尾的字符。在 Swift 3 及更高版本中：

let result = string.trimmingCharacters(in: CharacterSet(charactersIn: "0123456789.").inverted)

或者，如果您想删除字符串中任意位置的非数字字符（不仅仅是开头或结尾），您可以 filter characters，例如在 Swift 4.2.1:

let result = string.filter("0123456789.".contains)

或者，如果您想从字符串中的任意位置删除 CharacterSet 中的字符，请使用：

let result = String(string.unicodeScalars.filter(CharacterSet.whitespaces.inverted.contains))

或者，如果您只想匹配特定格式的有效字符串（例如 ####.##），您可以使用正则表达式。例如：

if let range = string.range(of: #"\d+(\.\d*)?"#, options: .regularExpression) {
    let result = string[range] // or `String(string[range])` if you need `String`
}

这些不同方法的行为略有不同，因此它完全取决于您要执行的操作。如果您想要小数或仅整数，请包括或排除小数点。有很多方法可以做到这一点。

对于较旧的 Swift 2 语法，请参阅 previous revision of this answer。

Answer 4

let result = string.stringByReplacingOccurrencesOfString("[^0-9]", withString: "", options: NSStringCompareOptions.RegularExpressionSearch, range:nil).stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceCharacterSet())

Swift 3

let result = string.replacingOccurrences( of:"[^0-9]", with: "", options: .regularExpression)

您可以投票 this answer。

Answer 5

Rob 的第一个解决方案的问题是 stringByTrimmingCharactersInSet 仅过滤字符串的结尾而不是整个字符串，如 Apple 文档中所述：

Returns a new string made by removing from both ends of the receiver characters contained in a given character set.

而是使用 componentsSeparatedByCharactersInSet 首先将所有未出现的字符集隔离到数组中，然后使用空字符串分隔符将它们连接起来：

"$34%^56()78*9££".componentsSeparatedByCharactersInSet(NSCharacterSet(charactersInString: "0123456789").invertedSet)).joinWithSeparator("")

哪个returns123456789

Answer 6

Swift 3、过滤除数字以外的所有

let myString = "dasdf3453453fsdf23455sf.2234"
let result = String(myString.characters.filter { String([=10=]).rangeOfCharacter(from: CharacterSet(charactersIn: "0123456789")) != nil })
print(result)

Answer 7

Swift 3 版本

extension String
{
    func trimmingCharactersNot(in charSet: CharacterSet) -> String
    {
        var s:String = ""
        for unicodeScalar in self.unicodeScalars
        {
            if charSet.contains(unicodeScalar)
            {
                s.append(String(unicodeScalar))
            }
        }
        return s
    }
}

Answer 8

我更喜欢 this solution，因为我喜欢扩展，而且对我来说它看起来更干净一些。解决方案转载于此：

extension String {
    var digits: String {
        return components(separatedBy: CharacterSet.decimalDigits.inverted)
            .joined()
    }
}

Answer 9

Swift 3

extension String {
    var keepNumericsOnly: String {
        return self.components(separatedBy: CharacterSet(charactersIn: "0123456789").inverted).joined(separator: "")
    }
}

Answer 10

Swift 4.0 版本

extension String {
    var numbers: String {
        return String(describing: filter { String([=10=]).rangeOfCharacter(from: CharacterSet(charactersIn: "0123456789")) != nil })
    }
}

Answer 11

Swift 4

String.swift

import Foundation

extension String {

    func removeCharacters(from forbiddenChars: CharacterSet) -> String {
        let passed = self.unicodeScalars.filter { !forbiddenChars.contains([=10=]) }
        return String(String.UnicodeScalarView(passed))
    }

    func removeCharacters(from: String) -> String {
        return removeCharacters(from: CharacterSet(charactersIn: from))
    }
}

ViewController.swift

let character = "1Vi234s56a78l9"
        let alphaNumericSet = character.removeCharacters(from: CharacterSet.decimalDigits.inverted)
        print(alphaNumericSet) // will print: 123456789

        let alphaNumericCharacterSet = character.removeCharacters(from: "0123456789")
        print("no digits",alphaNumericCharacterSet) // will print: Vishal

Answer 12

您可以使用范围的模式匹配运算符过滤字符串的 UnicodeScalarView，传递从 0 到 9 的 UnicodeScalar ClosedRange，并使用生成的 UnicodeScalarView 初始化新字符串：

extension String {
    private static var digits = UnicodeScalar("0")..."9"
    var digits: String {
        return String(unicodeScalars.filter(String.digits.contains))
    }
}

"abc12345".digits   // "12345"

edit/update:

Swift 4.2

extension RangeReplaceableCollection where Self: StringProtocol {
    var digits: Self {
        return filter(("0"..."9").contains)
    }
}

或作为变异方法

extension RangeReplaceableCollection where Self: StringProtocol {
    mutating func removeAllNonNumeric() {
        removeAll { !("0"..."9" ~= [=12=]) }
    }
}

Swift 5.2 • Xcode 11.4 或更高版本

在 Swift5 中我们可以使用一个名为 isWholeNumber:

的新角色属性

extension RangeReplaceableCollection where Self: StringProtocol {
    var digits: Self { filter(\.isWholeNumber) }
}

extension RangeReplaceableCollection where Self: StringProtocol {
    mutating func removeAllNonNumeric() {
        removeAll { ![=14=].isWholeNumber }
    }
}

为了也允许句点，我们可以扩展 Character 并创建一个计算属性:

extension Character {
    var isDecimalOrPeriod: Bool { "0"..."9" ~= self || self == "." }
}

extension RangeReplaceableCollection where Self: StringProtocol {
    var digitsAndPeriods: Self { filter(\.isDecimalOrPeriod) }
}

游乐场测试：

"abc12345".digits   // "12345"

var str = "123abc0"
str.removeAllNonNumeric()
print(str) //"1230"

"Testing0123456789.".digitsAndPeriods // "0123456789."

Answer 13

Swift 4

我找到了一个不错的方法来从字符串中仅获取字母数字字符集。例如：-

func getAlphaNumericValue() {

    var yourString  = "123456789!@#$%^&*()AnyThingYouWant"

    let unsafeChars = CharacterSet.alphanumerics.inverted  // Remove the .inverted to get the opposite result.  

    let cleanChars  = yourString.components(separatedBy: unsafeChars).joined(separator: "")


    print(cleanChars)  // 123456789AnyThingYouWant

}

Answer 14

Swift 4

但没有扩展或 componentsSeparatedByCharactersInSet 也不能读取。

let allowedCharSet = NSCharacterSet.letters.union(.whitespaces)
let filteredText = String(sourceText.unicodeScalars.filter(allowedCharSet.contains))

Answer 15

Swift 4.2

let digitChars  = yourString.components(separatedBy:
        CharacterSet.decimalDigits.inverted).joined(separator: "")

Answer 16

Swift 4.2

let numericString = string.filter { (char) -> Bool in
    return char.isNumber
}

Answer 17

let string = "+1*(234) fds567@-8/90-"
let onlyNumbers = string.components(separatedBy: CharacterSet.decimalDigits.inverted).joined()

print(onlyNumbers) // "1234567890"

或

extension String {

  func removeNonNumeric() -> String {
    return self.components(separatedBy: CharacterSet.decimalDigits.inverted).joined()
  }
}

let onlyNumbers = "+1*(234) fds567@-8/90-".removeNonNumeric() 
print(onlyNumbers)// "1234567890"

从 swift 中的字符串中删除所有非数字字符

Remove all non-numeric characters from a string in swift

swift

nscharacterset

Swift 4