从字符串中过滤非数字

Question

仅使用快速代码我无法弄清楚如何取"(555)555-5555"并仅返回数值并得到"5555555555".我需要删除所有括号,空格和短划线.我能找到的唯一例子是Objective-C,他们似乎都使用.trim()方法.似乎swift没有这个方法,但它确实有.stringByTrimmingCharacters方法,但这似乎只是修剪数据之前和之后的空格.

Answer 1

Swift 3&4:

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

Answer 2

将字符串中的非数字字符拆分为数字数组,并将它们连接回字符串:

斯威夫特1:

let stringArray = origString.componentsSeparatedByCharactersInSet(
    NSCharacterSet.decimalDigitCharacterSet().invertedSet)
let newString = NSArray(array: stringArray).componentsJoinedByString("")

斯威夫特2:

let stringArray = origString.componentsSeparatedByCharactersInSet(
    NSCharacterSet.decimalDigitCharacterSet().invertedSet)
let newString = stringArray.joinWithSeparator("")

Swift 3&4:

let newString = origString
    .components(separatedBy:CharacterSet.decimalDigits.inverted)
    .joined()

Answer 3

我喜欢正则表达式:

var s = "(555) 555-5555"
s = s.stringByReplacingOccurrencesOfString(
    "\\D", withString: "", options: .RegularExpressionSearch, 
    range: s.startIndex..<s.endIndex)

Answer 4

这是@Tapani的Swift 2.0答案作为一个方便的String扩展,(长度属性不是解决方案的一部分,但我在示例中留下它因为它也很方便):

import Foundation

extension String {

    var length : Int {
        return self.characters.count
    }

    func digitsOnly() -> String{
        let stringArray = self.componentsSeparatedByCharactersInSet(
            NSCharacterSet.decimalDigitCharacterSet().invertedSet)
        let newString = stringArray.joinWithSeparator("")

        return newString
    }

}

用法:

let phone = "(123)-123 - 1234"
print(phone.digitsOnly())

Answer 5

我有一个类似的问题，但我需要保留小数点。我调整了对此的最佳答案：

extension String {

    /// Returns a string with all non-numeric characters removed
    public var numericString: String {
        let characterSet = CharacterSet(charactersIn: "0123456789.").inverted
        return components(separatedBy: characterSet)
            .joined()
    }
}

Answer 6

细节

• Xcode 版本 10.2.1 (10E1001)，Swift 5

解决方案

import Foundation

extension String {

    private func filterCharacters(unicodeScalarsFilter closure: (UnicodeScalar) -> Bool) -> String {
        return String(String.UnicodeScalarView(unicodeScalars.filter { closure($0) }))
    }

    private func filterCharacters(definedIn charSets: [CharacterSet], unicodeScalarsFilter: (CharacterSet, UnicodeScalar) -> Bool) -> String {
        if charSets.isEmpty { return self }
        let charSet = charSets.reduce(CharacterSet()) { return $0.union($1) }
        return filterCharacters { unicodeScalarsFilter(charSet, $0) }
    }

    func removeCharacters(charSets: [CharacterSet]) -> String { return filterCharacters(definedIn: charSets) { !$0.contains($1) } }
    func removeCharacters(charSet: CharacterSet) -> String { return removeCharacters(charSets: [charSet]) }

    func onlyCharacters(charSets: [CharacterSet]) -> String { return filterCharacters(definedIn: charSets) { $0.contains($1) } }
    func onlyCharacters(charSet: CharacterSet) -> String { return onlyCharacters(charSets: [charSet]) }
}

用法

let string = "23f45gdor#@%#i425v wer 24 1+DWEJwi 3u09ru49w*()9uE2R_)$I#Q)_ U383q04+RFJO{dgnkvlj b`kefl;nwdl qsa`WKFSA,.E"
print("original string:                                 \(string)")
print("only .decimalDigits:                             \(string.onlyCharacters(charSet: .decimalDigits))")
print("only [.lowercaseLetters, .symbols]:              \(string.onlyCharacters(charSets: [.lowercaseLetters, .symbols]))")
print("remove .letters:                                 \(string.removeCharacters(charSet: .letters))")
print("remove [.decimalDigits, .lowercaseLetters]:      \(string.removeCharacters(charSets: [.decimalDigits, .lowercaseLetters]))")

结果

original string:                                 23f45gdor#@%#i425v wer 24 1+DWEJwi 3u09ru49w*()9uE2R_)$I#Q)_ U383q04+RFJO{dgnkvlj b`kefl;nwdl qsa`WKFSA,.E
only .decimalDigits:                             2345425241309499238304
only [.lowercaseLetters, .symbols]:              fgdorivwer+wiuruwu$q+dgnkvljb`keflnwdlqsa`
remove .letters:                                 2345#@%#425  24 1+ 30949*()92_)$#)_ 38304+{ `; `,.
remove [.decimalDigits, .lowercaseLetters]:      #@%#   +DWEJ *()ER_)$I#Q)_ U+RFJO{ `; `WKFSA,.E

（可选）字符串扩展

extension String {
    var onlyDigits: String { return onlyCharacters(charSets: [.decimalDigits]) }
    var onlyLetters: String { return onlyCharacters(charSets: [.letters]) }
}

（可选）字符串扩展用法

let string = "23f45gdor#@%#i425v wer 24 1+DWEJwi 3u09ru49w*()9uE2R_)$I#Q)_ U383q04+RFJO{dgnkvlj b`kefl;nwdl qsa`WKFSA,.E"
print("original string:     \(string)")
print(".onlyDigits:         \(string.onlyDigits)")
print(".onlyLetters:        \(string.onlyLetters)")

（可选）字符串扩展使用结果

original string:     23f45gdor#@%#i425v wer 24 1+DWEJwi 3u09ru49w*()9uE2R_)$I#Q)_ U383q04+RFJO{dgnkvlj b`kefl;nwdl qsa`WKFSA,.E
.onlyDigits:         2345425241309499238304
.onlyLetters:        fgdorivwerDWEJwiuruwuERIQUqRFJOdgnkvljbkeflnwdlqsaWKFSAE

Answer 7

在Swift 4中，解决方案更加出色：

import Foundation

let sourceText = "+5 (555) 555-5555"

let allowedCharset = CharacterSet
    .decimalDigits
    .union(CharacterSet(charactersIn: "+"))

let filteredText = String(sourceText.unicodeScalars.filter(allowedCharset.contains))

print(filteredText) // +55555555555

Answer 8

您需要使用 NSCharacterSet：

查看有关 Swift 和 Obj-C 实现的 NSHipster 链接： http://nshipster.com/nscharacterset/

类似的例子：

var string = "  Lorem    ipsum dolar   sit  amet. "

let components = string.componentsSeparatedByCharactersInSet(NSCharacterSet.whitespaceCharacterSet()).filter({!isEmpty($0)})

string = join(" ", components)

请参阅：标点字符集

描述：

返回包含标点符号类别中的字符的字符集。非正式地，该集合是用于分隔脚本中的语言单元的所有非空白字符的集合，例如句点、破折号、括号等。

@Tapani提出了一个很好的建议：NSCharacterSet.decimalDigitCharacterSet().invertedSet