How to write regular expressions in Objective C (NSRegularExpression)?

Question

I have this regex working when I test it in PHP but it doesn\'t work in Objective C:

(?:www\\.)?((?!-)[a-zA-Z0-9-]{2,63}(?

        

Answer 1

A NSTextCheckingResult has multiple items obtained by indexing into it.

[match rangeAtIndex:0]; is the full match.
[match rangeAtIndex:1]; (if it exists) is the first capture group match.
etc.

You can use something like this:

NSString *searchedString = @"domain-name.tld.tld2";
NSRange   searchedRange = NSMakeRange(0, [searchedString length]);
NSString *pattern = @"(?:www\\.)?((?!-)[a-zA-Z0-9-]{2,63}(?<!-))\\.?((?:[a-zA-Z0-9]{2,})?(?:\\.[a-zA-Z0-9]{2,})?)";
NSError  *error = nil;

NSRegularExpression* regex = [NSRegularExpression regularExpressionWithPattern: pattern options:0 error:&error];
NSArray* matches = [regex matchesInString:searchedString options:0 range: searchedRange];
for (NSTextCheckingResult* match in matches) {
    NSString* matchText = [searchedString substringWithRange:[match range]];
    NSLog(@"match: %@", matchText);
    NSRange group1 = [match rangeAtIndex:1];
    NSRange group2 = [match rangeAtIndex:2];
    NSLog(@"group1: %@", [searchedString substringWithRange:group1]);
    NSLog(@"group2: %@", [searchedString substringWithRange:group2]);
}

NSLog output:

match: domain-name.tld.tld2
domain-name
tld.tld2

Do test that the match ranges are valid.

More simply in this case:

NSString *searchedString = @"domain-name.tld.tld2";
NSRange   searchedRange = NSMakeRange(0, [searchedString length]);
NSString *pattern = @"(?:www\\.)?((?!-)[a-zA-Z0-9-]{2,63}(?<!-))\\.?((?:[a-zA-Z0-9]{2,})?(?:\\.[a-zA-Z0-9]{2,})?)";
NSError  *error = nil;

NSRegularExpression* regex = [NSRegularExpression regularExpressionWithPattern:pattern options:0 error:&error];
NSTextCheckingResult *match = [regex firstMatchInString:searchedString options:0 range: searchedRange];
NSLog(@"group1: %@", [searchedString substringWithRange:[match rangeAtIndex:1]]);
NSLog(@"group2: %@", [searchedString substringWithRange:[match rangeAtIndex:2]]);

Swift 3.0:

let searchedString = "domain-name.tld.tld2"
let nsSearchedString = searchedString as NSString
let searchedRange = NSMakeRange(0, searchedString.characters.count)
let pattern = "(?:www\\.)?((?!-)[a-zA-Z0-9-]{2,63}(?<!-))\\.?((?:[a-zA-Z0-9]{2,})?(?:\\.[a-zA-Z0-9]{2,})?)"

do {
    let regex = try NSRegularExpression(pattern:pattern, options: [])
    let matches = regex.matches(in:searchedString, options:[], range:searchedRange)
    for match in matches {
        let matchText = nsSearchedString.substring(with:match.range);
        print("match: \(matchText)");

        let group1 : NSRange = match.rangeAt(1)
        let matchText1 = nsSearchedString.substring(with: group1)
        print("matchText1: \(matchText1)")

        let group2 = match.rangeAt(2)
        let matchText2 = nsSearchedString.substring(with: group2)
        print("matchText2: \(matchText2)")
    }
} catch let error as NSError {
    print(error.localizedDescription)
}

print output:

match: domain-name.tld.tld2
matchText1: domain-name
matchText2: tld.tld2

More simply in this case:

do {
    let regex = try NSRegularExpression(pattern:pattern, options: [])
    let match = regex.firstMatch(in:searchedString, options:[], range:searchedRange)

    let matchText1 = nsSearchedString.substring(with: match!.rangeAt(1))
    print("matchText1: \(matchText1)")

    let matchText2 = nsSearchedString.substring(with: match!.rangeAt(2))
    print("matchText2: \(matchText2)")

} catch let error as NSError {
    print(error.localizedDescription)
}

print output:

matchText1: domain-name
matchText2: tld.tld2

Answer 2

According to Apple's documentation, these characters must be quoted (using \) to be treated as literals:

* ? + [ ( ) { } ^ $ | \ . /

It would also help if you could explain what you are trying to achieve. Do you have any test fixtures?