I am struggling to modify captured value with regex.
For example, I wanna change "Hello, he is hero" to "HEllo, HE is HEro" using Regex.
I know there are ways to change this without regex, but it is just an example to show the problem. I actually use the regex instead of just he, but I cannot provide it here. That is why using regex is required.
The code below somehow does not work. Are there any ways to make it work?
"Hello, he is hero".replacingOccurrences(
of: #"(he)"#,
with: "$1".uppercased(), // <- uppercased is not applied
options: .regularExpression
)
You need to use your regex in combination with Range (range(of:)) to find matches and then replace each found range separately
Here is a function as an extension to String that does this by using range(of:) starting from the start of the string and then moving the start index to match from forward to after the last match. The actual replacement is done inside a separate function that is passed as an argument
extension String {
func replace(regex: String, with replace: (Substring) -> String) -> String {
var string = self
var startIndex = self.startIndex
let endIndex = self.endIndex
while let range = string.range(of: regex, options: [.regularExpression] , range: startIndex..<endIndex) {
if range.isEmpty {
startIndex = string.index(startIndex, offsetBy: 1)
if startIndex >= endIndex { break }
continue
}
string.replaceSubrange(range, with: replace(string[range]))
startIndex = range.upperBound
}
return string
}
}
Example where we do an case insensitive search for words starting with "he" and replace each match with the uppercased version
let result = "Hello, he is hero. There he is".replace(regex: #"(?i)\bhe"#) {
$0.uppercased()
}
Output
HEllo, HE is HEro. There HE is
You can try NSRegularExpression. Something like:
import Foundation
var sourceStr = "Hello, he is hero"
let regex = try! NSRegularExpression(pattern: "(he)")
let matches = regex.matches(in: sourceStr, range: NSRange(sourceStr.startIndex..., in: sourceStr))
regex.enumerateMatches(in: sourceStr, range: NSRange(sourceStr.startIndex..., in: sourceStr)) { (match, _, _) in
guard let match = match else { return }
guard let range = Range(match.range, in: sourceStr) else { return }
let sub = sourceStr[range]
sourceStr = sourceStr.replacingOccurrences(of: sub, with: sub.uppercased(), options: [], range: range)
}
print(sourceStr)
this is the solution i can provide
var string = "Hello, he is hero"
let occurrence = "he"
string = string.lowercased().replacingOccurrences(
of: occurrence,
with: occurrence.uppercased(),
options: .regularExpression
)
print(string)
for the following code:
import Foundation
extension String {
var fullRange: NSRange {
return .init(self.startIndex ..< self.endIndex, in: self)
}
public subscript(range: Range<Int>) -> Self.SubSequence {
let st = self.index(self.startIndex, offsetBy: range.startIndex)
let ed = self.index(self.startIndex, offsetBy: range.endIndex)
let sub = self[st ..< ed]
return sub
}
func split(regex pattern: String) throws -> [String] {
let regex = try NSRegularExpression.init(pattern: pattern, options: [])
let fRange = self.fullRange
let match = regex.matches(in: self, options: [], range: fRange)
var list = [String]()
var start = 0
for m in match {
let r = m.range
let end = r.location
list.append(String(self[start ..< end]))
start = end + r.length
}
if start < self.count {
list.append(String(self[start ..< self.count]))
}
return list
}
}
print(try! "مرتفع جداً\nVery High".split(regex: "\n"))
the output should be :
["مرتفع جداً", "Very High"]
but instead it is:
["مرتفع جداً\n", "ery High"]
that because regex (for this case) matched the \n at the offset 10 instead of 9
is there any thing wrong in my code, or it is a bug in swift with regex !!
It's not a bug. You are trying to use Int indexes which is error-prone and strongly discouraged in an Unicode environment.
This is the equivalent of your code with the proper String.Index type and the dedicated API to convert NSRange to Range<String.Index> and vice versa. fullRange and subscript are obsolete.
I just left out the print line. startIndex and endIndex are properties of String
extension String {
func split(regex pattern: String) throws -> [String] {
let regex = try NSRegularExpression(pattern: pattern)
let matches = regex.matches(in: self, range: NSRange(startIndex..., in: self))
var list = [String]()
var start = startIndex
for match in matches {
let range = Range(match.range, in: self)!
let end = range.lowerBound
list.append(String(self[start..<end]))
start = range.upperBound
}
if start < endIndex {
list.append(String(self[start..<endIndex]))
}
return list
}
}
print(try! "مرتفع جداً\nVery High".split(regex: "\n"))
The result is ["مرتفع جداً", "Very High"]
I found the issue behind this bug?!
Swift Strings are so much weirder than any other language; since every character is 4 bytes length, then a single character (may, would, will, ..) contains 1 or 2 unicode characters (witch what happened in my case), so the solution is to subarray the unicodeScalars of the swift String instead of the string it self !!
I want to extract substrings from a string that match a regex pattern.
So I'm looking for something like this:
func matchesForRegexInText(regex: String!, text: String!) -> [String] {
???
}
So this is what I have:
func matchesForRegexInText(regex: String!, text: String!) -> [String] {
var regex = NSRegularExpression(pattern: regex,
options: nil, error: nil)
var results = regex.matchesInString(text,
options: nil, range: NSMakeRange(0, countElements(text)))
as Array<NSTextCheckingResult>
/// ???
return ...
}
The problem is, that matchesInString delivers me an array of NSTextCheckingResult, where NSTextCheckingResult.range is of type NSRange.
NSRange is incompatible with Range<String.Index>, so it prevents me of using text.substringWithRange(...)
Any idea how to achieve this simple thing in swift without too many lines of code?
Even if the matchesInString() method takes a String as the first argument,
it works internally with NSString, and the range parameter must be given
using the NSString length and not as the Swift string length. Otherwise it will
fail for "extended grapheme clusters" such as "flags".
As of Swift 4 (Xcode 9), the Swift standard
library provides functions to convert between Range<String.Index>
and NSRange.
func matches(for regex: String, in text: String) -> [String] {
do {
let regex = try NSRegularExpression(pattern: regex)
let results = regex.matches(in: text,
range: NSRange(text.startIndex..., in: text))
return results.map {
String(text[Range($0.range, in: text)!])
}
} catch let error {
print("invalid regex: \(error.localizedDescription)")
return []
}
}
Example:
let string = "🇩🇪€4€9"
let matched = matches(for: "[0-9]", in: string)
print(matched)
// ["4", "9"]
Note: The forced unwrap Range($0.range, in: text)! is safe because
the NSRange refers to a substring of the given string text.
However, if you want to avoid it then use
return results.flatMap {
Range($0.range, in: text).map { String(text[$0]) }
}
instead.
(Older answer for Swift 3 and earlier:)
So you should convert the given Swift string to an NSString and then extract the
ranges. The result will be converted to a Swift string array automatically.
(The code for Swift 1.2 can be found in the edit history.)
Swift 2 (Xcode 7.3.1) :
func matchesForRegexInText(regex: String, text: String) -> [String] {
do {
let regex = try NSRegularExpression(pattern: regex, options: [])
let nsString = text as NSString
let results = regex.matchesInString(text,
options: [], range: NSMakeRange(0, nsString.length))
return results.map { nsString.substringWithRange($0.range)}
} catch let error as NSError {
print("invalid regex: \(error.localizedDescription)")
return []
}
}
Example:
let string = "🇩🇪€4€9"
let matches = matchesForRegexInText("[0-9]", text: string)
print(matches)
// ["4", "9"]
Swift 3 (Xcode 8)
func matches(for regex: String, in text: String) -> [String] {
do {
let regex = try NSRegularExpression(pattern: regex)
let nsString = text as NSString
let results = regex.matches(in: text, range: NSRange(location: 0, length: nsString.length))
return results.map { nsString.substring(with: $0.range)}
} catch let error {
print("invalid regex: \(error.localizedDescription)")
return []
}
}
Example:
let string = "🇩🇪€4€9"
let matched = matches(for: "[0-9]", in: string)
print(matched)
// ["4", "9"]
My answer builds on top of given answers but makes regex matching more robust by adding additional support:
Returns not only matches but returns also all capturing groups for each match (see examples below)
Instead of returning an empty array, this solution supports optional matches
Avoids do/catch by not printing to the console and makes use of the guard construct
Adds matchingStrings as an extension to String
Swift 4.2
//: Playground - noun: a place where people can play
import Foundation
extension String {
func matchingStrings(regex: String) -> [[String]] {
guard let regex = try? NSRegularExpression(pattern: regex, options: []) else { return [] }
let nsString = self as NSString
let results = regex.matches(in: self, options: [], range: NSMakeRange(0, nsString.length))
return results.map { result in
(0..<result.numberOfRanges).map {
result.range(at: $0).location != NSNotFound
? nsString.substring(with: result.range(at: $0))
: ""
}
}
}
}
"prefix12 aaa3 prefix45".matchingStrings(regex: "fix([0-9])([0-9])")
// Prints: [["fix12", "1", "2"], ["fix45", "4", "5"]]
"prefix12".matchingStrings(regex: "(?:prefix)?([0-9]+)")
// Prints: [["prefix12", "12"]]
"12".matchingStrings(regex: "(?:prefix)?([0-9]+)")
// Prints: [["12", "12"]], other answers return an empty array here
// Safely accessing the capture of the first match (if any):
let number = "prefix12suffix".matchingStrings(regex: "fix([0-9]+)su").first?[1]
// Prints: Optional("12")
Swift 3
//: Playground - noun: a place where people can play
import Foundation
extension String {
func matchingStrings(regex: String) -> [[String]] {
guard let regex = try? NSRegularExpression(pattern: regex, options: []) else { return [] }
let nsString = self as NSString
let results = regex.matches(in: self, options: [], range: NSMakeRange(0, nsString.length))
return results.map { result in
(0..<result.numberOfRanges).map {
result.rangeAt($0).location != NSNotFound
? nsString.substring(with: result.rangeAt($0))
: ""
}
}
}
}
"prefix12 aaa3 prefix45".matchingStrings(regex: "fix([0-9])([0-9])")
// Prints: [["fix12", "1", "2"], ["fix45", "4", "5"]]
"prefix12".matchingStrings(regex: "(?:prefix)?([0-9]+)")
// Prints: [["prefix12", "12"]]
"12".matchingStrings(regex: "(?:prefix)?([0-9]+)")
// Prints: [["12", "12"]], other answers return an empty array here
// Safely accessing the capture of the first match (if any):
let number = "prefix12suffix".matchingStrings(regex: "fix([0-9]+)su").first?[1]
// Prints: Optional("12")
Swift 2
extension String {
func matchingStrings(regex: String) -> [[String]] {
guard let regex = try? NSRegularExpression(pattern: regex, options: []) else { return [] }
let nsString = self as NSString
let results = regex.matchesInString(self, options: [], range: NSMakeRange(0, nsString.length))
return results.map { result in
(0..<result.numberOfRanges).map {
result.rangeAtIndex($0).location != NSNotFound
? nsString.substringWithRange(result.rangeAtIndex($0))
: ""
}
}
}
}
The fastest way to return all matches and capture groups in Swift 5
extension String {
func match(_ regex: String) -> [[String]] {
let nsString = self as NSString
return (try? NSRegularExpression(pattern: regex, options: []))?.matches(in: self, options: [], range: NSMakeRange(0, nsString.length)).map { match in
(0..<match.numberOfRanges).map { match.range(at: $0).location == NSNotFound ? "" : nsString.substring(with: match.range(at: $0)) }
} ?? []
}
}
Returns a 2-dimentional array of strings:
"prefix12suffix fix1su".match("fix([0-9]+)su")
returns...
[["fix12su", "12"], ["fix1su", "1"]]
// First element of sub-array is the match
// All subsequent elements are the capture groups
If you want to extract substrings from a String, not just the position, (but the actual String including emojis). Then, the following maybe a simpler solution.
extension String {
func regex (pattern: String) -> [String] {
do {
let regex = try NSRegularExpression(pattern: pattern, options: NSRegularExpressionOptions(rawValue: 0))
let nsstr = self as NSString
let all = NSRange(location: 0, length: nsstr.length)
var matches : [String] = [String]()
regex.enumerateMatchesInString(self, options: NSMatchingOptions(rawValue: 0), range: all) {
(result : NSTextCheckingResult?, _, _) in
if let r = result {
let result = nsstr.substringWithRange(r.range) as String
matches.append(result)
}
}
return matches
} catch {
return [String]()
}
}
}
Example Usage:
"someText 👿🏅👿⚽️ pig".regex("👿⚽️")
Will return the following:
["👿⚽️"]
Note using "\w+" may produce an unexpected ""
"someText 👿🏅👿⚽️ pig".regex("\\w+")
Will return this String array
["someText", "️", "pig"]
I found that the accepted answer's solution unfortunately does not compile on Swift 3 for Linux. Here's a modified version, then, that does:
import Foundation
func matches(for regex: String, in text: String) -> [String] {
do {
let regex = try RegularExpression(pattern: regex, options: [])
let nsString = NSString(string: text)
let results = regex.matches(in: text, options: [], range: NSRange(location: 0, length: nsString.length))
return results.map { nsString.substring(with: $0.range) }
} catch let error {
print("invalid regex: \(error.localizedDescription)")
return []
}
}
The main differences are:
Swift on Linux seems to require dropping the NS prefix on Foundation objects for which there is no Swift-native equivalent. (See Swift evolution proposal #86.)
Swift on Linux also requires specifying the options arguments for both the RegularExpression initialization and the matches method.
For some reason, coercing a String into an NSString doesn't work in Swift on Linux but initializing a new NSString with a String as the source does work.
This version also works with Swift 3 on macOS / Xcode with the sole exception that you must use the name NSRegularExpression instead of RegularExpression.
Swift 4 without NSString.
extension String {
func matches(regex: String) -> [String] {
guard let regex = try? NSRegularExpression(pattern: regex, options: [.caseInsensitive]) else { return [] }
let matches = regex.matches(in: self, options: [], range: NSMakeRange(0, self.count))
return matches.map { match in
return String(self[Range(match.range, in: self)!])
}
}
}
#p4bloch if you want to capture results from a series of capture parentheses, then you need to use the rangeAtIndex(index) method of NSTextCheckingResult, instead of range. Here's #MartinR 's method for Swift2 from above, adapted for capture parentheses. In the array that is returned, the first result [0] is the entire capture, and then individual capture groups begin from [1]. I commented out the map operation (so it's easier to see what I changed) and replaced it with nested loops.
func matches(for regex: String!, in text: String!) -> [String] {
do {
let regex = try NSRegularExpression(pattern: regex, options: [])
let nsString = text as NSString
let results = regex.matchesInString(text, options: [], range: NSMakeRange(0, nsString.length))
var match = [String]()
for result in results {
for i in 0..<result.numberOfRanges {
match.append(nsString.substringWithRange( result.rangeAtIndex(i) ))
}
}
return match
//return results.map { nsString.substringWithRange( $0.range )} //rangeAtIndex(0)
} catch let error as NSError {
print("invalid regex: \(error.localizedDescription)")
return []
}
}
An example use case might be, say you want to split a string of title year eg "Finding Dory 2016" you could do this:
print ( matches(for: "^(.+)\\s(\\d{4})" , in: "Finding Dory 2016"))
// ["Finding Dory 2016", "Finding Dory", "2016"]
Most of the solutions above only give the full match as a result ignoring the capture groups e.g.: ^\d+\s+(\d+)
To get the capture group matches as expected you need something like (Swift4) :
public extension String {
public func capturedGroups(withRegex pattern: String) -> [String] {
var results = [String]()
var regex: NSRegularExpression
do {
regex = try NSRegularExpression(pattern: pattern, options: [])
} catch {
return results
}
let matches = regex.matches(in: self, options: [], range: NSRange(location:0, length: self.count))
guard let match = matches.first else { return results }
let lastRangeIndex = match.numberOfRanges - 1
guard lastRangeIndex >= 1 else { return results }
for i in 1...lastRangeIndex {
let capturedGroupIndex = match.range(at: i)
let matchedString = (self as NSString).substring(with: capturedGroupIndex)
results.append(matchedString)
}
return results
}
}
Update for iOS 16: Regex, RegexBuilder 👷♀️
Xcode previously supported Regex with the Find and Search tab. Many found Apple's NSRegularExpressions Swift API verbose and unwieldy, so Apple released Regex literal support and RegexBuilder this year.
The API has been simplified going forward to tidy up complex String range-based parsing logic in iOS 16 / macOS 13 as well as improve performance.
RegEx literals in Swift 5.7
func parseLine(_ line: Substring) throws -> MailmapEntry {
let regex = /\h*([^<#]+?)??\h*<([^>#]+)>\h*(?:#|\Z)/
guard let match = line.prefixMatch(of: regex) else {
throw MailmapError.badLine
}
return MailmapEntry(name: match.1, email: match.2)
}
At the moment, we are able to match using prefixMatch or wholeMatch to find a single match, but the API may improve in the future for multiple matches.
RegexBuilder in Swift 5.7
RegexBuilder is a new API released by Apple aimed at making RegEx code easier to write in Swift. We can translate the Regex literal /\h*([^<#]+?)??\h*<([^>#]+)>\h*(?:#|\Z)/ from above into a more declarative form using RegexBuilder if we want more readability.
Do note that we can use raw strings in a RegexBuilder and also interleave Regex Literals in the builder if we want to balance readability with conciseness.
import RegexBuilder
let regex = Regex {
ZeroOrMore(.horizontalWhitespace)
Optionally {
Capture(OneOrMore(.noneOf("<#")))
}
.repetitionBehavior(.reluctant)
ZeroOrMore(.horizontalWhitespace)
"<"
Capture(OneOrMore(.noneOf(">#")))
">"
ZeroOrMore(.horizontalWhitespace)
/#|\Z/
}
The RegEx literal /£|\Z/ is equivalent to:
ChoiceOf {
"#"
Anchor.endOfSubjectBeforeNewline
}
Composable RegexComponent
RegexBuilder syntax is similar to SwiftUI also in terms of composability because we can reuse RegexComponents within other RegexComponents:
struct MailmapLine: RegexComponent {
#RegexComponentBuilder
var regex: Regex<(Substring, Substring?, Substring)> {
ZeroOrMore(.horizontalWhitespace)
Optionally {
Capture(OneOrMore(.noneOf("<#")))
}
.repetitionBehavior(.reluctant)
ZeroOrMore(.horizontalWhitespace)
"<"
Capture(OneOrMore(.noneOf(">#")))
">"
ZeroOrMore(.horizontalWhitespace)
ChoiceOf {
"#"
Anchor.endOfSubjectBeforeNewline
}
}
}
This is how I did it, I hope it brings a new perspective how this works on Swift.
In this example below I will get the any string between []
var sample = "this is an [hello] amazing [world]"
var regex = NSRegularExpression(pattern: "\\[.+?\\]"
, options: NSRegularExpressionOptions.CaseInsensitive
, error: nil)
var matches = regex?.matchesInString(sample, options: nil
, range: NSMakeRange(0, countElements(sample))) as Array<NSTextCheckingResult>
for match in matches {
let r = (sample as NSString).substringWithRange(match.range)//cast to NSString is required to match range format.
println("found= \(r)")
}
This is a very simple solution that returns an array of string with the matches
Swift 3.
internal func stringsMatching(regularExpressionPattern: String, options: NSRegularExpression.Options = []) -> [String] {
guard let regex = try? NSRegularExpression(pattern: regularExpressionPattern, options: options) else {
return []
}
let nsString = self as NSString
let results = regex.matches(in: self, options: [], range: NSMakeRange(0, nsString.length))
return results.map {
nsString.substring(with: $0.range)
}
}
update #Mike Chirico's to Swift 5
extension String{
func regex(pattern: String) -> [String]?{
do {
let regex = try NSRegularExpression(pattern: pattern, options: NSRegularExpression.Options(rawValue: 0))
let all = NSRange(location: 0, length: count)
var matches = [String]()
regex.enumerateMatches(in: self, options: NSRegularExpression.MatchingOptions(rawValue: 0), range: all) {
(result : NSTextCheckingResult?, _, _) in
if let r = result {
let nsstr = self as NSString
let result = nsstr.substring(with: r.range) as String
matches.append(result)
}
}
return matches
} catch {
return nil
}
}
}
basic phone number matching
let phoneNumbers = ["+79990001101", "+7 (800) 000-11-02", "+34 507 574 147 ", "+1-202-555-0118"]
let match: (String) -> String = {
$0.replacingOccurrences(of: #"[^\d+]"#, with: "", options: .regularExpression)
}
print(phoneNumbers.map(match))
// ["+79990001101", "+78000001102", "+34507574147", "+12025550118"]
Big thanks to Lars Blumberg his answer for capturing groups and full matches with Swift 4, which helped me out a lot. I also made an addition to it for the people who do want an error.localizedDescription response when their regex is invalid:
extension String {
func matchingStrings(regex: String) -> [[String]] {
do {
let regex = try NSRegularExpression(pattern: regex)
let nsString = self as NSString
let results = regex.matches(in: self, options: [], range: NSMakeRange(0, nsString.length))
return results.map { result in
(0..<result.numberOfRanges).map {
result.range(at: $0).location != NSNotFound
? nsString.substring(with: result.range(at: $0))
: ""
}
}
} catch let error {
print("invalid regex: \(error.localizedDescription)")
return []
}
}
}
For me having the localizedDescription as error helped understand what went wrong with escaping, since it's displays which final regex swift tries to implement.
You can use matching(regex:) on the string like:
let array = try "Your String To Search".matching(regex: ".")
using this simple extension:
public extension String {
func matching(regex: String) throws -> [String] {
let regex = try NSRegularExpression(pattern: regex)
let results = regex.matches(in: self, range: NSRange(startIndex..., in: self))
return results.map { String(self[Range($0.range, in: self)!]) }
}
}
I'm trying to create an internal function for the String class to get only AlphaNumeric characters and return a string. I'm running into a few errors with how to convert the matches back into a string using Regex. Can someone tell me how to fix the code or if there's an easier way?
I want something like this
let testString = "_<$abc$>_"
let alphaNumericString = testString.alphaNumeric() //abc
So far I have:
extension String {
internal func alphaNumeric() -> String {
let regex = try? NSRegularExpression(pattern: "[^a-z0-9]", options: .caseInsensitive)
let string = self as NSString
let results = regex?.matches(in: self, options: [], range: NSRange(location: 0, length: string.length))
let matches = results.map {
String(self[Range($0.range, in: self)!])
}
return matches.join()
}
}
You may directly use replacingOccurrences (that removes all non-overlapping matches from the input string) with [^A-Za-z0-9]+ pattern:
let str = "_<$abc$>_"
let pattern = "[^A-Za-z0-9]+"
let result = str.replacingOccurrences(of: pattern, with: "", options: [.regularExpression])
print(result) // => abc
The [^A-Za-z0-9]+ pattern is a negated character class that matches any char but the ones defined in the class, one or more occurrences (due to + quantifier).
See the regex demo.
Try below extension:
extension String {
var alphanumeric: String {
return self.components(separatedBy: CharacterSet.alphanumerics.inverted).joined().lowercased()
}
}
Usage: print("alphanumeric :", "_<$abc$>_".alphanumeric)
Output : abc
You can also use characterset for this like
extension String {
var alphaNumeric: String {
components(separatedBy: CharacterSet.alphanumerics.inverted).joined()
}
}
I want to write a function that could be used like this:
let 👩👩👧👦 = "👩👩👧👧".replacingFirstOccurrence(of: "👧", with: "👦")
Given how odd both this string and Swift's String library are, is this possible in Swift?
Based on the insights gained at Why are emoji characters like 👩👩👧👦 treated so strangely in Swift strings?, a sensible approach might be to replace Unicode scalars:
extension String {
func replacingFirstOccurrence(of target: UnicodeScalar, with replacement: UnicodeScalar) -> String {
let uc = self.unicodeScalars
guard let idx = uc.index(of: target) else { return self }
let prefix = uc[uc.startIndex..<idx]
let suffix = uc[uc.index(after: idx) ..< uc.endIndex]
return "\(prefix)\(replacement)\(suffix)"
}
}
Example:
let family1 = "👩👩👧👦"
print(family1.characters.map { Array(String($0).unicodeScalars) })
// [["\u{0001F469}", "\u{200D}"], ["\u{0001F469}", "\u{200D}"], ["\u{0001F467}", "\u{200D}"], ["\u{0001F466}"]]
let family2 = family1.replacingFirstOccurrence(of: "👧", with: "👦")
print(family2) // 👩👩👦👦
print(family2.characters.map { Array(String($0).unicodeScalars) })
// [["\u{0001F469}", "\u{200D}"], ["\u{0001F469}", "\u{200D}"], ["\u{0001F466}", "\u{200D}"], ["\u{0001F466}"]]
And here is a possible version which locates and replaces the Unicode scalars of an arbitrary string:
extension String {
func replacingFirstOccurrence(of target: String, with replacement: String) -> String {
let uc = self.unicodeScalars
let tuc = target.unicodeScalars
// Target empty or too long:
if tuc.count == 0 || tuc.count > uc.count {
return self
}
// Current search position:
var pos = uc.startIndex
// Last possible position of `tuc` within `uc`:
let end = uc.index(uc.endIndex, offsetBy: tuc.count - 1)
// Locate first Unicode scalar
while let from = uc[pos..<end].index(of: tuc.first!) {
// Compare all Unicode scalars:
let to = uc.index(from, offsetBy: tuc.count)
if !zip(uc[from..<to], tuc).contains(where: { $0 != $1 }) {
let prefix = uc[uc.startIndex..<from]
let suffix = uc[to ..< uc.endIndex]
return "\(prefix)\(replacement)\(suffix)"
}
// Next search position:
uc.formIndex(after: &pos)
}
// Target not found.
return self
}
}
Using the range(of:options:range:locale:) the solution became quite concise:
extension String {
func replaceFirstOccurrence(of searchString: String, with replacementString: String) -> String {
guard let range = self.range(of: searchString, options: .literal) else { return self }
return self.replacingCharacters(in: range, with: replacementString)
}
}
This works by first finding the range of searchString within the instance, and if a range is found the range is replaced with replacementString. Otherwise the instance just returns itself. And, since the range(of:) method returns as soon as it finds a match, the returned range is guaranteed to be the first occurrence.
"221".replaceFirstOccurrence(of: "2", with: "3") // 321
"👩👩👧👦".replaceFirstOccurrence(of: "\u{1f469}", with: "\u{1f468}") // 👨👩👧👦
*To clarify, the last test case converts woman-woman-girl-boy to man-woman-girl-boy.