快速删除字符串中的所有非数字字符

You can do something like this...

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

A solution using the filter function and rangeOfCharacterFromSet

let string = "sld [f]34é7*˜µ"


let alphaNumericCharacterSet = NSCharacterSet.alphanumericCharacterSet()
let filteredCharacters = string.characters.filter {
return  String($0).rangeOfCharacterFromSet(alphaNumericCharacterSet) != nil
}
let filteredString = String(filteredCharacters) // -> sldf34é7µ

To filter for only numeric characters use

let string = "sld [f]34é7*˜µ"


let numericSet = "0123456789"
let filteredCharacters = string.characters.filter {
return numericSet.containsString(String($0))
}
let filteredString = String(filteredCharacters) // -> 347

or

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

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

You can either use trimmingCharacters with the inverted character set to remove characters from the start or the end of the string. In Swift 3 and later:

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

Or, if you want to remove non-numeric characters anywhere in the string (not just the start or end), you can filter the characters, e.g. in Swift 4.2.1:

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

Or, if you want to remove characters from a CharacterSet from anywhere in the string, use:

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

Or, if you want to only match valid strings of a certain format (e.g. ####.##), you could use regular expression. For example:

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

The behavior of these different approaches differ slightly so it just depends on precisely what you're trying to do. Include or exclude the decimal point if you want decimal numbers, or just integers. There are lots of ways to accomplish this.

For older, Swift 2 syntax, see previous revision of this answer.

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)

You can upvote this answer.

The issue with Rob's first solution is stringByTrimmingCharactersInSet only filters the ends of the string rather than throughout, as stated in Apple's documentation:

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

Instead use componentsSeparatedByCharactersInSet to first isolate all non-occurrences of the character set into arrays and subsequently join them with an empty string separator:

"$$1234%^56()78*9££".componentsSeparatedByCharactersInSet(NSCharacterSet(charactersInString: "0123456789").invertedSet)).joinWithSeparator("")

Which returns 123456789

Swift 3, filters all except numbers

let myString = "dasdf3453453fsdf23455sf.2234"
let result = String(myString.characters.filter { String($0).rangeOfCharacter(from: CharacterSet(charactersIn: "0123456789")) != nil })
print(result)

Swift 3 Version

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
}
}

I prefer this solution, because I like extensions, and it seems a bit cleaner to me. Solution reproduced here:

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

Swift 3

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

Swift 4.0 version

extension String {
var numbers: String {
return String(describing: filter { String($0).rangeOfCharacter(from: CharacterSet(charactersIn: "0123456789")) != nil })
}
}

Swift 4

String.swift

import Foundation


extension String {


func removeCharacters(from forbiddenChars: CharacterSet) -> String {
let passed = self.unicodeScalars.filter { !forbiddenChars.contains($0) }
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

You can filter the UnicodeScalarView of the string using the pattern matching operator for ranges, pass a UnicodeScalar ClosedRange from 0 to 9 and initialise a new String with the resulting UnicodeScalarView:

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)
}
}

or as a mutating method

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

Swift 5.2 • Xcode 11.4 or later

In Swift5 we can use a new Character property called isWholeNumber:

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

extension RangeReplaceableCollection where Self: StringProtocol {
mutating func removeAllNonNumeric() {
removeAll { !$0.isWholeNumber }
}
}

To allow a period as well we can extend Character and create a computed property:

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

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

Playground testing:

"abc12345".digits   // "12345"


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


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

Swift 4

I found a decent way to get only alpha numeric characters set from a string. For instance:-

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


}

Swift 4

But without extensions or componentsSeparatedByCharactersInSet which doesn't read as well.

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

Swift 4.2

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

Swift 4.2

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

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


print(onlyNumbers) // "1234567890"

or

extension String {


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


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