My app supports 5 languages. I have a string which has some double quotes in it. This string is translated into 5 languages in the localizable.strings files.
Example:
title_identifier = "Hi \"how\", are \"you\"";
I would like to bold out "how" and "you" in this string by finding the range of these words. So I am trying to fetch these quoted words out of the string and the result would be an array containing "how" and "you" or their range.
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 []
}
}
matches(for: "(?<=\")[^\"]*(?=\")", in: str)
The result is: ["how", ", are ", "you"] rather than ["how","you"]. I think this regex needs some addition to allow it to search for next quote once two quotes are found, so to avoid the words in between quotes.
Your problem is in the use of lookarounds that do not consume text but check if their patterns match and return either true or false. See your regex in action, the , are matches because the last " in the previous match was not consumed, the regex index remained right after w, so the next match could start with ". You need to use a consuming pattern here, "([^"]*)".
However, your code will only return full matches. You can just trim the first and last "s here with .map {$0.trimmingCharacters(in: ["\""])}, as the regex only matches one quote at the start and end:
matches(for: "\"[^\"]*\"", in: str).map {$0.trimmingCharacters(in: ["\""])}
Here is the regex demo.
Alternatively, access Group 1 value by appending (at: 1) after $0.range:
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(at: 1), in: text)!])
}
} catch let error {
print("invalid regex: \(error.localizedDescription)")
return []
}
}
let str = "Hi \"how\", are \"you\""
print(matches(for: "\"([^\"]*)\"", in: str))
// => ["how", "you"]
I'm bit confused by NSRegularExpression in swift, can any one help me?
task:1 given ("name","john","name of john")
then I should get ["name","john","name of john"]. Here I should avoid the brackets.
task:2 given ("name"," john","name of john")
then I should get ["name","john","name of john"]. Here I should avoid the brackets and extra spaces and finally get array of strings.
task:3 given key = value // comment
then I should get ["key","value","comment"]. Here I should get only strings in the line by avoiding = and //
I have tried below code for task 1 but not passed.
let string = "(name,john,string for user name)"
let pattern = "(?:\\w.*)"
do {
let regex = try NSRegularExpression(pattern: pattern, options: .caseInsensitive)
let matches = regex.matches(in: string, options: [], range: NSRange(location: 0, length: string.utf16.count))
for match in matches {
if let range = Range(match.range, in: string) {
let name = string[range]
print(name)
}
}
} catch {
print("Regex was bad!")
}
Thanks in advance.
RegEx in Swift
These posts might help you to explore regular expressions in swift:
Does a string match a pattern?
Swift extract regex matches
How can I use String slicing subscripts in Swift 4?
How to use regex with Swift?
Swift 3 - How do I extract captured groups in regular expressions?
How to group search regular expressions using swift?
Task 1 & 2
This expression might help you to match your desired outputs for both Task 1 and 2:
"(\s+)?([a-z\s]+?)(\s+)?"
Based on Rob's advice, you could much reduce the boundaries, such as the char list [a-z\s]. For example, here, we can also use:
"(\s+)?(.*?)(\s+)?"
or
"(\s+)?(.+?)(\s+)?"
to simply pass everything in between two " and/or space.
RegEx
If this wasn't your desired expression, you can modify/change your expressions in regex101.com.
RegEx Circuit
You can also visualize your expressions in jex.im:
JavaScript Demo
const regex = /"(\s+)?([a-z\s]+?)(\s+)?"/gm;
const str = `"name","john","name of john"
"name"," john","name of john"
" name "," john","name of john "
" name "," john"," name of john "`;
const subst = `\n$2`;
// The substituted value will be contained in the result variable
const result = str.replace(regex, subst);
console.log('Substitution result: ', result);
Task 3
This expression might help you to design an expression for the third task:
(.*?)([a-z\s]+)(.*?)
const regex = /(.*?)([a-z\s]+)(.*?)/gm;
const str = `key = value // comment
key = value with some text // comment`;
const subst = `$2,`;
// The substituted value will be contained in the result variable
const result = str.replace(regex, subst);
console.log('Substitution result: ', result);
Separate the string by non alpha numeric characters except white spaces. Then trim the elements with white spaces.
extension String {
func words() -> [String] {
return self.components(separatedBy: CharacterSet.alphanumerics.inverted.subtracting(.whitespaces))
.filter({ !$0.isEmpty })
.map({ $0.trimmingCharacters(in: .whitespaces) })
}
}
let string1 = "(name,john,string for user name)"
let string2 = "(name, john,name of john)"
let string3 = "key = value // comment"
print(string1.words())//["name", "john", "string for user name"]
print(string2.words())//["name", "john", "name of john"]
print(string3.words())//["key", "value", "comment"]
Here I have done with after understanding all of above comments.
let text = """
Capturing and non-capturing groups are somewhat advanced topics. You’ll encounter examples of capturing and non-capturing groups later on in the tutorial
"""
extension String {
func rex (_ expr : String)->[String] {
return try! NSRegularExpression(pattern: expr, options: [.caseInsensitive])
.matches(in: self, options: [], range: NSRange(location: 0, length: self.count))
.map {
String(self[Range($0.range, in: self)!])
}
}
}
let r = text.rex("(?:\\w+-\\w+)") // pass any rex
A single pattern, works for test:1...3, in Swift.
let string =
//"(name,john,string for user name)" //test:1
//#"("name"," john","name of john")"# //test:2
"key = value // comment" //test:3
let pattern = #"(?:\w+)(?:\s+\w+)*"# //Swift 5+ only
//let pattern = "(?:\\w+)(?:\\s+\\w+)*"
do {
let regex = try NSRegularExpression(pattern: pattern)
let matches = regex.matches(in: string, range: NSRange(0..<string.utf16.count))
let matchingWords = matches.map {
String(string[Range($0.range, in: string)!])
}
print(matchingWords) //(test:3)->["key", "value", "comment"]
} catch {
print("Regex was bad!")
}
Let’s consider:
let string = "(name,José,name is José)"
I’d suggest a regex that looks for strings where:
It’s the substring either after the ( at the start of the full string or after a comma, i.e., look behind assertion of (?<=^\(|,);
It’s the substring that does not contain , within it, i.e., [^,]+?;
It’s the substring that is terminated by either a comma or ) at the end of the full string, i.e., look ahead assertion of (?=,|\)$), and
If you want to have it skip white space before and after the substrings, throw in the \s*+, too.
Thus:
let pattern = #"(?<=^\(|,)\s*+([^,]+?)\s*+(?=,|\)$)"#
let regex = try! NSRegularExpression(pattern: pattern)
regex.enumerateMatches(in: string, range: NSRange(string.startIndex..., in: string)) { match, _, _ in
if let nsRange = match?.range(at: 1), let range = Range(nsRange, in: string) {
let substring = String(string[range])
// do something with `substring` here
}
}
Note, I’m using the Swift 5 extended string delimiters (starting with #" and ending with "#) so that I don’t have to escape my backslashes within the string. If you’re using Swift 4 or earlier, you’ll want to escape those back slashes:
let pattern = "(?<=^\\(|,)\\s*+([^,]+?)\\s*+(?=,|\\)$)"
I'm trying to validate if in my first name field there exists a special character or number inputted by the user.
I've tried this code but when my name has a space like for example Mark John it is considered an error.
let characterset = CharacterSet(charactersIn: "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ")
if name.rangeOfCharacter(from: characterset.inverted) != nil {
return false
}else{
return true
}
You can add a space character to the string you use to initialize your CharacterSet.
Or you can use:
let characterSet = CharacterSet.letters.union(.whitespaces)
This allows letters from any alphabet.
Unrelated but you can replace that if/else with:
return name.rangeOfCharacter(from: characterset) != nil
You can use a regular expression like.
do {
let regex = try NSRegularExpression(pattern: ".*[^A-Za-z ].*", options: [])
if regex.firstMatch(in: nameValue, options: [], range: NSMakeRange(0, nameValue.characters.count)) != nil {
self.showAlert(message: "Must not contain Number in Name")
} else {
}
}
catch {
}
How to check whether a WHOLE string can be matches to regex? In Java is method String.matches(regex)
You need to use anchors, ^ (start of string anchor) and $ (end of string anchor), with range(of:options:range:locale:), passing the .regularExpression option:
import Foundation
let phoneNumber = "123-456-789"
let result = phoneNumber.range(of: "^\\d{3}-\\d{3}-\\d{3}$", options: .regularExpression) != nil
print(result)
Or, you may pass an array of options, [.regularExpression, .anchored], where .anchored will anchor the pattern at the start of the string only, and you will be able to omit ^, but still, $ will be required to anchor at the string end:
let result = phoneNumber.range(of: "\\d{3}-\\d{3}-\\d{3}$", options: [.regularExpression, .anchored]) != nil
See the online Swift demo
Also, using NSPredicate with MATCHES is an alternative here:
The left hand expression equals the right hand expression using a regex-style comparison according to ICU v3 (for more details see the ICU User Guide for Regular Expressions).
MATCHES actually anchors the regex match both at the start and end of the string (note this might not work in all Swift 3 builds):
let pattern = "\\d{3}-\\d{3}-\\d{3}"
let predicate = NSPredicate(format: "self MATCHES [c] %#", pattern)
let result = predicate.evaluate(with: "123-456-789")
What you are looking for is range(of:options:range:locale:) then you can then compare the result of range(of:option:) with whole range of comparing string..
Example:
let phoneNumber = "(999) 555-1111"
let wholeRange = phoneNumber.startIndex..<phoneNumber.endIndex
if let match = phoneNumber.range(of: "\\(?\\d{3}\\)?\\s\\d{3}-\\d{4}", options: .regularExpression), wholeRange == match {
print("Valid number")
}
else {
print("Invalid number")
}
//Valid number
Edit: You can also use NSPredicate and compare your string with evaluate(with:) method of its.
let pattern = "^\\(?\\d{3}\\)?\\s\\d{3}-\\d{4}$"
let predicate = NSPredicate(format: "self MATCHES [c] %#", pattern)
if predicate.evaluate(with: "(888) 555-1111") {
print("Valid")
}
else {
print("Invalid")
}
Swift extract regex matches
with little bit of edit
import Foundation
func matches(for regex: String, in text: String) -> Bool {
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.isEmpty
} catch let error {
print("invalid regex: \(error.localizedDescription)")
return false
}
}
Example usage from link above:
let string = "19320"
let matched = matches(for: "^[1-9]\\d*$", in: string)
print(matched) // will match
let string = "a19320"
let matched = matches(for: "^[1-9]\\d*$", in: string)
print(matched) // will not match
I try to include ' symbol to Regular Expressions
I use this function
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 {
text.substring(with: Range($0.range, in: text)!)
}
} catch let error {
print("invalid regex: \(error.localizedDescription)")
return []
}
}
and this Regular Expressions
let matched = matches(for: "^[‘]|[0-9]|[a-zA-Z]+$", in: string)
when I search I can find numbers and english letters
but not ' symbol
I guess that what you really want is this:
"['0-9a-zA-Z]+"
Note that I have removed the ^ (text start) and $ (text end) characters because then your whole text would have to match.
I have merged the groups because otherwise you would not match the text as a whole word. You would get separate apostrophe and then the word.
I have changed the ‘ character into the proper ' character. The automatic conversion from the simple apostrophe is caused by iOS 11 Smart Punctuation. You can turn it off on an input using:
input.smartQuotesType = .no
See https://developer.apple.com/documentation/uikit/uitextinputtraits/2865931-smartquotestype