问题
I have the need to parse some unknown data which should just be a numeric value, but may contain whitespace or other non-alphanumeric characters.
Is there a new way of doing this in Swift? All I can find online seems to be the old C way of doing things.
I am looking at stringByTrimmingCharactersInSet - as I am sure my inputs will only have whitespace/special characters at the start or end of the string. Are there any built in character sets I can use for this? Or do I need to create my own?
I was hoping there would be something like stringFromCharactersInSet() which would allow me to specify only valid characters to keep
回答1:
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.
回答2:
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.
回答3:
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()
}
}
回答4:
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
}
回答5:
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
回答6:
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) }
}
}
edit/update:
Swift 5 • Xcode 10.2
In Swift5 we can use a new Character property called isNumber:
extension RangeReplaceableCollection where Self: StringProtocol {
var digits: Self {
return filter({ $0.isNumber })
}
}
extension RangeReplaceableCollection where Self: StringProtocol {
mutating func removeAllNonNumeric() {
removeAll { !$0.isNumber }
}
}
Playground testing:
"abc12345".digits // "12345"
var str = "123abc0"
str.removeAllNonNumeric()
print(str) //"1230"
回答7:
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))
回答8:
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)
回答9:
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"
回答10:
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
回答11:
Swift 3
extension String {
var keepNumericsOnly: String {
return self.components(separatedBy: CharacterSet(charactersIn: "0123456789").inverted).joined(separator: "")
}
}
回答12:
Swift 4.2
let numericString = string.filter { (char) -> Bool in
return char.isNumber
}
回答13:
Swift 4.0 version
extension String {
var numbers: String {
return String(describing: filter { String($0).rangeOfCharacter(from: CharacterSet(charactersIn: "0123456789")) != nil })
}
}
回答14:
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
回答15:
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"
回答16:
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
}
}
回答17:
Swift 4.2
let digitChars = yourString.components(separatedBy:
CharacterSet.decimalDigits.inverted).joined(separator: "")
来源：https://stackoverflow.com/questions/36594179/remove-all-non-numeric-characters-from-a-string-in-swift