I am trying to parse a list of input strings from an Excel file that can have a 'currency' value, and it could be in any currency. For e.g.
$200
£300
€200
CA$300
What's the best way to parse out the currency symbol and the numeric value? I'm trying to do this with a NumberFormatter but it doesn't work for the 'euro' or the 'CAD' value.
Here is my code:
let currencyFormatter = NumberFormatter()
currencyFormatter.numberStyle = .currency
currencyFormatter.maximumFractionDigits = 2
let trimmedString = String(currencyString.filter { String($0).rangeOfCharacter(from: CharacterSet(charactersIn: "0123456789.,")) == nil }).trimmingCharacters(in: .whitespacesAndNewlines)
if trimmedString.count > 0 && Locale.current.currencySymbol != trimmedString {
// Currency symbol is *not* local currency, so lookup the locale for it
let allLocales = Locale.availableIdentifiers.map({Locale(identifier: $0)})
if let localeForSymbol = allLocales.filter({$0.currencySymbol == trimmedString}).first {
currencyFormatter.locale = localeForSymbol
}
}
if let numberValue = currencyFormatter.number(from: currencyString) {
print ("\(NSDecimalNumber(decimal: numberValue.decimalValue))")
}
What am I getting wrong here? Or is this not possible without using some regex expressions?
you could try this "...to parse out the currency symbol and the numeric value":
let currencyString = "CA$300"
let valueString = currencyString.filter {
CharacterSet(charactersIn: "0123456789.,").isSuperset(of: CharacterSet(charactersIn: String($0)))
}.trimmingCharacters(in: .whitespacesAndNewlines)
print("---> valueString: \(valueString) ")
let symbol = currencyString.replacingOccurrences(of: valueString, with: "")
print("---> symbol: \(symbol)")
#workingdog has a valid answer. An alternative, and one I always reach for, is to use regex, where 3 simple regex can be used to identify any non-numeric leading characters, the numeric characters, and any non-numeric postfix characters:
let prefixRegex = #"^[^0-9]+"#
let numRegex = #"[0-9]+"#
let postfixRegex = #"[^0-9]+$"#
These can then be used with String's .range(of: options:) using the .regularExpression option to get the element's range, and then use that to extract the relevant SubString.
As a very quick demo of this in action:
["$20", "£300", "€200", "CA$300", "798xyz", "$123abc"]
.forEach{string in
var components: [Range<String.Index>?] = []
components.append( string.range(of: prefixRegex, options: .regularExpression))
components.append( string.range(of: numRegex, options: .regularExpression))
components.append( string.range(of: postfixRegex, options: .regularExpression))
print(components
.map{ $0 != nil ? String(string[$0!]).padded(to: 8) : String(repeating: " ", count: 8) }
.joined(separator: "")
)
}
This provides an output of
$ 20
£ 300
€ 200
CA$ 300
798 xyz
$ 123 abc
.padded(to:) is a utility extension on String that pads the rear of a string to a specified length with any character.
extension String {
func padded(to paddedTo :Int, with padding: String = " ") -> Self {
while count < paddedTo {
return (self + padding).padded(to: paddedTo, with: padding)
}
return self
}
}
I have a string like this:
"te_st" and like to replace all underscores followed by a character with the uppercased version of this character.
From "te_st" --> Found (regex: "_.") --------replace with next char (+ uppercase ("s"->"S")--------> "teSt"
From "te_st" ---> to "teSt"
From "_he_l_lo" ---> to "HeLLo"
From "an_o_t_h_er_strin_g" ---> to "anOTHErStrinG"
... but I can not really get it working using Swift's NSRegularExpression like this small snipped does:
var result = "te_st" // result should be teSt
result = try! NSRegularExpression(pattern: "_*").stringByReplacingMatches(in: result, range: NSRange(0..<result.count), withTemplate: ("$1".uppercased()))
There's no regular syntax to convert a match to uppercase. The code you posted is attempting to convert the string $1 to uppercase which is of course just $1. It isn't attempting to convert the value represented by the $1 match at runtime.
Here's another approach using a regular expression to find the _ followed by a lowercase letter. Those are enumerated and replaced with the uppercase letter.
extension String {
func toCamelCase() -> String {
let expr = try! NSRegularExpression(pattern: "_([a-z])")
var res = self
for match in expr.matches(in: self, range: NSRange(0..<res.count)).reversed() {
let range = Range(match.range, in: self)!
let letterRange = Range(match.range(at: 1), in: self)!
res.replaceSubrange(range, with: self[letterRange].uppercased())
}
return res
}
}
print("te_st".toCamelCase())
print("_he_l_lo".toCamelCase())
print("an_o_t_h_er_strin_g".toCamelCase())
This outputs:
teSt
HeLLo
anOTHErStrinG
Here is one implementation using NSRegularExpression. I use group match to get the character after _ and capitalize it and replace the string.
func capitalizeLetterAfterUnderscore(string: String) -> String {
var capitalizedString = string
guard let regularExpression = try? NSRegularExpression(pattern: "_(.)") else {
return capitalizedString
}
let matches = regularExpression.matches(in: string,
options: .reportCompletion,
range: NSMakeRange(0, string.count))
for match in matches {
let groupRange = match.range(at: 1)
let index = groupRange.location
let characterIndex = string.index(string.startIndex,
offsetBy: index)
let range = characterIndex ... characterIndex
let capitalizedCharacter = String(capitalizedString[characterIndex]).capitalized
capitalizedString = capitalizedString.replacingCharacters(in: range,
with: capitalizedCharacter)
}
capitalizedString = capitalizedString.replacingOccurrences(of: "_", with: "")
return capitalizedString
}
capitalizeLetterAfterUnderscore(string: "an_o_t_h_er_strin_g") // anOTHErStrinG
And here is other one without using regular expression. I made extension for method which could also be reused.
extension String {
func indexes(of character: String) -> [Index] {
precondition(character.count == 1, "character should be single letter string")
return enumerated().reduce([]) { (partial, component) in
let currentIndex = index(startIndex,
offsetBy: component.offset)
return String(self[currentIndex]) == character
? partial + [currentIndex]
: partial
}
}
func capitalizeLetter(after indexes: [Index]) -> String {
var modifiedString = self
for currentIndex in indexes {
guard let letterIndex = index(currentIndex,
offsetBy: 1,
limitedBy: endIndex)
else { continue }
let range = letterIndex ... letterIndex
modifiedString = modifiedString.replacingCharacters(in: range,
with: self[range].capitalized)
}
return modifiedString
}
}
let string = "an_o_t_h_er_strin_g"
let newString = string.capitalizeLetter(after: string.indexes(of: "_"))
.replacingOccurrences(of: "_",with: "")
You can use string range(of:, options:, range:) method with .regularExpression options to match the occurrences of "_[a-z]" and replace the subranges iterating the ranges found at reversed order by the character at the index after the range lowerbound uppercased:
let string = "an_o_t_h_er_strin_g"
let regex = "_[a-z]"
var start = string.startIndex
var ranges:[Range<String.Index>] = []
while let range = string.range(of: regex, options: .regularExpression, range: start..<string.endIndex) {
start = range.upperBound
ranges.append(range)
}
var finalString = string
for range in ranges.reversed() {
finalString.replaceSubrange(range, with: String(string[string.index(after: range.lowerBound)]).uppercased())
}
print(finalString) // "anOTHErStrinG\n"
The problem is that it is converting the string "$1" to upper case (which is, unsurprisingly unchanged, just "$1") and using "$1" as the template. If you want to use regex, you will have to enumerate through matches yourself.
The alternative is to split the string by _ characters and uppercase the first character of every substring (except the first) and joining it back together using reduce:
let input = "te_st"
let output = input.components(separatedBy: "_").enumerated().reduce("") { $0 + ($1.0 == 0 ? $1.1 : $1.1.uppercasedFirst()) }
Or, if your goal isn't to write code as cryptic as most regex, we can make that a tad more legible:
let output = input
.components(separatedBy: "_")
.enumerated()
.reduce("") { result, current in
if current.offset == 0 {
return current.element // because you don’t want the first component capitalized
} else {
return result + current.element.uppercasedFirst()
}
}
Resulting in:
teSt
Note, that uses this extension for capitalizing the first character:
extension String {
func uppercasedFirst(with locale: Locale? = nil) -> String {
guard count > 0 else { return self }
return String(self[startIndex]).uppercased(with: locale) + self[index(after: startIndex)...]
}
}
If you want to do sort of dynamic conversion with NSRegularExpression, you can subclass NSRegularExpression and override replacementString(for:in:offset:template:):
class ToCamelRegularExpression: NSRegularExpression {
override func replacementString(for result: NSTextCheckingResult, in string: String, offset: Int, template templ: String) -> String {
if let range = Range(result.range(at: 1), in: string) {
return string[range].uppercased()
} else {
return super.replacementString(for: result, in: string, offset: 0, template: templ)
}
}
}
func toCamelCase(_ input: String) -> String { //Make this a String extension if you prefer...
let regex = try! ToCamelRegularExpression(pattern: "_(.)")
return regex.stringByReplacingMatches(in: input, options: [], range: NSRange(0..<input.utf16.count), withTemplate: "$1")
}
print(toCamelCase("te_st")) //-> teSt
print(toCamelCase("_he_l_lo")) //-> HeLLo
print(toCamelCase("an_o_t_h_er_strin_g")) //-> anOTHErStrinG
a string such as ! !! yuahl! ! , I want delete ! and , when I code like this
for index in InputName.characters.indices {
if String(InputName[index]) == "" || InputName.substringToIndex(index) == "!" {
InputName.removeAtIndex(index)
}
}
have this error " fatal error: subscript: subRange extends past String end ", how should I do? THX :D
Swift 5+
let myString = "aaaaaaaabbbb"
let replaced = myString.replacingOccurrences(of: "bbbb", with: "") // "aaaaaaaa"
If you need to remove characters only on both ends, you can use 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
If you need to remove characters also in the middle, "reconstruct from the filtered output" would be a little bit more efficient than repeating single character removal.
var tempUSView = String.UnicodeScalarView()
tempUSView.appendContentsOf(s2.unicodeScalars.lazy.filter{!delCharSet.longCharacterIsMember($0.value)})
let s2DelAll = String(tempUSView)
print(s2DelAll) //->anotherStringaString
If you don't mind generating many intermediate Strings and Arrays, this single liner can generate the expected output:
let s2DelAll2 = s2.componentsSeparatedByCharactersInSet(delCharSet).joinWithSeparator("")
print(s2DelAll2) //->anotherStringaString
I find that the filter method is a good way to go for this sort of thing:
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"
Swift 3
In Swift 3, the syntax is a bit nicer. As a result of the Great Swiftification of the old APIs, the factory method is now called trimmingCharacters(in:). Also, you can construct the CharacterSet as a Set of single-character Strings:
let string = "! !! yuahl! !"
string.trimmingCharacters(in: [" ", "!"]) // "yuahl"
If you have characters in the middle of the string you would like to remove as well, you can use components(separatedBy:).joined():
let string = "! !! yu !ahl! !"
string.components(separatedBy: ["!", " "]).joined() // "yuahl"
H/T #OOPer for the Swift 2 version
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())
}
Note:
By adding this function in String extension, you can delete the specific character of string at last.
for index in InputName.characters.indices.reversed() {
if String(InputName[index]) == "" || InputName.substringToIndex(index) == "!" {
InputName.removeAtIndex(index)
}
}
Also you can add such very helpful extension :
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)
}
}
you can use this:
for example if you want to remove "%" the percent from 10%
if let i = text.firstIndex(of: "%") {
text.remove(at: i) //10
}
I am looking for a way to replace characters in a Swift String.
Example: "This is my string"
I would like to replace " " with "+" to get "This+is+my+string".
How can I achieve this?
This answer has been updated for Swift 4 & 5. If you're still using Swift 1, 2 or 3 see the revision history.
You have a couple of options. You can do as #jaumard suggested and use replacingOccurrences()
let aString = "This is my string"
let newString = aString.replacingOccurrences(of: " ", with: "+", options: .literal, range: nil)
And as noted by #cprcrack below, the options and range parameters are optional, so if you don't want to specify string comparison options or a range to do the replacement within, you only need the following.
let aString = "This is my string"
let newString = aString.replacingOccurrences(of: " ", with: "+")
Or, if the data is in a specific format like this, where you're just replacing separation characters, you can use components() to break the string into and array, and then you can use the join() function to put them back to together with a specified separator.
let toArray = aString.components(separatedBy: " ")
let backToString = toArray.joined(separator: "+")
Or if you're looking for a more Swifty solution that doesn't utilize API from NSString, you could use this.
let aString = "Some search text"
let replaced = String(aString.map {
$0 == " " ? "+" : $0
})
You can use this:
let s = "This is my string"
let modified = s.replace(" ", withString:"+")
If you add this extension method anywhere in your code:
extension String
{
func replace(target: String, withString: String) -> String
{
return self.stringByReplacingOccurrencesOfString(target, withString: withString, options: NSStringCompareOptions.LiteralSearch, range: nil)
}
}
Swift 3:
extension String
{
func replace(target: String, withString: String) -> String
{
return self.replacingOccurrences(of: target, with: withString, options: NSString.CompareOptions.literal, range: nil)
}
}
Swift 3, Swift 4, Swift 5 Solution
let exampleString = "Example string"
//Solution suggested above in Swift 3.0
let stringToArray = exampleString.components(separatedBy: " ")
let stringFromArray = stringToArray.joined(separator: "+")
//Swiftiest solution
let swiftyString = exampleString.replacingOccurrences(of: " ", with: "+")
Did you test this :
var test = "This is my string"
let replaced = test.stringByReplacingOccurrencesOfString(" ", withString: "+", options: nil, range: nil)
var str = "This is my string"
print(str.replacingOccurrences(of: " ", with: "+"))
Output is
This+is+my+string
Swift 5.5
I am using this extension:
extension String {
func replaceCharacters(characters: String, toSeparator: String) -> String {
let characterSet = CharacterSet(charactersIn: characters)
let components = components(separatedBy: characterSet)
let result = components.joined(separator: toSeparator)
return result
}
func wipeCharacters(characters: String) -> String {
return self.replaceCharacters(characters: characters, toSeparator: "")
}
}
Usage:
"<34353 43434>".replaceCharacters(characters: "< >", toSeparator:"+") // +34353+43434+
"<34353 43434>".wipeCharacters(characters: "< >") // 3435343434
Swift 4:
let abc = "Hello world"
let result = abc.replacingOccurrences(of: " ", with: "_",
options: NSString.CompareOptions.literal, range:nil)
print(result :\(result))
Output:
result : Hello_world
A Swift 3 solution along the lines of Sunkas's:
extension String {
mutating func replace(_ originalString:String, with newString:String) {
self = self.replacingOccurrences(of: originalString, with: newString)
}
}
Use:
var string = "foo!"
string.replace("!", with: "?")
print(string)
Output:
foo?
A category that modifies an existing mutable String:
extension String
{
mutating func replace(originalString:String, withString newString:String)
{
let replacedString = self.stringByReplacingOccurrencesOfString(originalString, withString: newString, options: nil, range: nil)
self = replacedString
}
}
Use:
name.replace(" ", withString: "+")
Swift 3 solution based on Ramis' answer:
extension String {
func withReplacedCharacters(_ characters: String, by separator: String) -> String {
let characterSet = CharacterSet(charactersIn: characters)
return components(separatedBy: characterSet).joined(separator: separator)
}
}
Tried to come up with an appropriate function name according to Swift 3 naming convention.
Less happened to me, I just want to change (a word or character) in the String
So I've use the Dictionary
extension String{
func replace(_ dictionary: [String: String]) -> String{
var result = String()
var i = -1
for (of , with): (String, String)in dictionary{
i += 1
if i<1{
result = self.replacingOccurrences(of: of, with: with)
}else{
result = result.replacingOccurrences(of: of, with: with)
}
}
return result
}
}
usage
let mobile = "+1 (800) 444-9999"
let dictionary = ["+": "00", " ": "", "(": "", ")": "", "-": ""]
let mobileResult = mobile.replace(dictionary)
print(mobileResult) // 001800444999
Xcode 11 • Swift 5.1
The mutating method of StringProtocol replacingOccurrences can be implemented as follow:
extension RangeReplaceableCollection where Self: StringProtocol {
mutating func replaceOccurrences<Target: StringProtocol, Replacement: StringProtocol>(of target: Target, with replacement: Replacement, options: String.CompareOptions = [], range searchRange: Range<String.Index>? = nil) {
self = .init(replacingOccurrences(of: target, with: replacement, options: options, range: searchRange))
}
}
var name = "This is my string"
name.replaceOccurrences(of: " ", with: "+")
print(name) // "This+is+my+string\n"
var str = "This is my string"
str = str.replacingOccurrences(of: " ", with: "+")
print(str)
This is easy in swift 4.2. just use replacingOccurrences(of: " ", with: "_") for replace
var myStr = "This is my string"
let replaced = myStr.replacingOccurrences(of: " ", with: "_")
print(replaced)
Since Swift 2, String does no longer conform to SequenceType. In other words, you can not iterate through a string with a for...in loop.
The simple and easy way is to convert String to Array to get the benefit of the index just like that:
let input = Array(str)
I remember when I tried to index into String without using any conversion. I was really frustrated that I couldn’t come up with or reach a desired result, and was about to give up.
But I ended up creating my own workaround solution, and here is the full code of the extension:
extension String {
subscript (_ index: Int) -> String {
get {
String(self[self.index(startIndex, offsetBy: index)])
}
set {
remove(at: self.index(self.startIndex, offsetBy: index))
insert(Character(newValue), at: self.index(self.startIndex, offsetBy: index))
}
}
}
Now that you can read and replace a single character from string using its index just like you originally wanted to:
var str = "cat"
for i in 0..<str.count {
if str[i] == "c" {
str[i] = "h"
}
}
print(str)
It’s simple and useful way to use it and get through Swift’s String access model.
Now that you’ll feel it’s smooth sailing next time when you can loop through the string just as it is, not casting it into Array.
Try it out, and see if it can help!
I've implemented this very simple func:
func convap (text : String) -> String {
return text.stringByReplacingOccurrencesOfString("'", withString: "''")
}
So you can write:
let sqlQuery = "INSERT INTO myTable (Field1, Field2) VALUES ('\(convap(value1))','\(convap(value2)')
I think Regex is the most flexible and solid way:
var str = "This is my string"
let regex = try! NSRegularExpression(pattern: " ", options: [])
let output = regex.stringByReplacingMatchesInString(
str,
options: [],
range: NSRange(location: 0, length: str.characters.count),
withTemplate: "+"
)
// output: "This+is+my+string"
Swift extension:
extension String {
func stringByReplacing(replaceStrings set: [String], with: String) -> String {
var stringObject = self
for string in set {
stringObject = self.stringByReplacingOccurrencesOfString(string, withString: with)
}
return stringObject
}
}
Go on and use it like let replacedString = yorString.stringByReplacing(replaceStrings: [" ","?","."], with: "+")
The speed of the function is something that i can hardly be proud of, but you can pass an array of String in one pass to make more than one replacement.
Here is the example for Swift 3:
var stringToReplace = "This my string"
if let range = stringToReplace.range(of: "my") {
stringToReplace?.replaceSubrange(range, with: "your")
}
Here's an extension for an in-place occurrences replace method on String, that doesn't no an unnecessary copy and do everything in place:
extension String {
mutating func replaceOccurrences<Target: StringProtocol, Replacement: StringProtocol>(of target: Target, with replacement: Replacement, options: String.CompareOptions = [], locale: Locale? = nil) {
var range: Range<Index>?
repeat {
range = self.range(of: target, options: options, range: range.map { self.index($0.lowerBound, offsetBy: replacement.count)..<self.endIndex }, locale: locale)
if let range = range {
self.replaceSubrange(range, with: replacement)
}
} while range != nil
}
}
(The method signature also mimics the signature of the built-in String.replacingOccurrences() method)
May be used in the following way:
var string = "this is a string"
string.replaceOccurrences(of: " ", with: "_")
print(string) // "this_is_a_string"
If you don't want to use the Objective-C NSString methods, you can just use split and join:
var string = "This is my string"
string = join("+", split(string, isSeparator: { $0 == " " }))
split(string, isSeparator: { $0 == " " }) returns an array of strings (["This", "is", "my", "string"]).
join joins these elements with a +, resulting in the desired output: "This+is+my+string".
you can test this:
let newString = test.stringByReplacingOccurrencesOfString(" ", withString: "+", options: nil, range: nil)
Swift 5.5
But this might work in earlier versions.
I'm frequently replacing because I want to replace "any whitespace or -" with a _ or something like that. This extension on string lets me do that.
extension String {
func removingCharacters(_ characters:CharacterSet) -> Self {
Self(self.unicodeScalars.filter {
!characters.contains($0)
})
}
func removingCharacters(in string:String) -> Self {
Self(self.unicodeScalars.filter {
!CharacterSet(charactersIn:string).contains($0)
})
}
func replacingCharacters(_ characters:CharacterSet, with newChar:Character) -> Self {
String(self.compactMap( {
CharacterSet(charactersIn: "\($0.1)").isSubset(of: characters)
? newChar : $0.1
}))
}
func replacingCharacters(in string:String, with newChar:Character) -> Self {
String(self.compactMap( {
CharacterSet(charactersIn: "\($0)").isSubset(of: CharacterSet(charactersIn:string))
? newChar : $0
}))
}
}
usage:
print("hello \n my name\t is Joe".removingCharacters(.whitespacesAndNewlines))
print("hello \n my name\t is Joe".removingCharacters(in: " \t\n"))
print("ban annan anann ana".replacingCharacters(.whitespacesAndNewlines, with: "_"))
print("ban-annan anann ana".replacingCharacters(in: " -", with: "_"))
Obviously for a single character the .replacingOccurrences(of: " ", with: "+") is better.
I have not done a performance comparison to the
let toArray = aString.components(separatedBy: characterSet)
let backToString = toArray.joined(separator: "+")
style done in Ramis's extension. I'd be interested if someone does.
See also replacing emoji's: https://stackoverflow.com/a/63416058/5946596