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Ios swift – How to get all matches for regular expression when ending character of pattern matches more than once?

Here is my input string:

let input = "Hello, {un:23456} is under investigation {un:654, Bartek} along with Jason."

and simple extension:

extension String {
    func matchesFor(regularExpression: NSRegularExpression) -> [String] {
        let nsstring = self as NSString
        let matches = regularExpression.matches(in: self, range: NSRange(location: 0, length: nsstring.length))
        return matches.map { nsstring.substring(with: $0.range) }
    }
}

and the output is:

["{un:23456} is under investigation {un:654, Bartek}"]

instead of:

["{un:23456}", "{un:654, Bartek}"]

Why?

My regex is "\{un:(:?[0-9]{1,9}).*\}"

Tags:

2

Answers


  1. 0

    Change .* -> .*?. The ? stop * being greedy, match the string that ends with } and no more.

    Or, get all between { }:

    "\{(.*?)\}"

    As you said in your comment, you also need to remove these curly braces:

    let output = input.matchesFor(regularExpression: try .init(pattern: "\{(.*?)\}"))
    .map { $0.replacingOccurrences(of: "{", with: "").replacingOccurrences(of: "}", with: "") }

print(output)

//["un:23456", "un:654, Bartek"]

    Updated as @Larme suggested:

    func matchesFor(regularExpression: NSRegularExpression) -> [String] {
    let nsstring = self as NSString
    let matches = regularExpression.matches(in: self, range: NSRange(location: 0, length: nsstring.length))
    return matches.map { nsstring.substring(with: $0.range(at: 1)) } //<- here
}

let output = input.matchesFor(regularExpression: try .init(pattern: "\{(.*?)\}"))

print(output)

//["un:23456", "un:654, Bartek"]
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  2. 0

    I would use a character class to terminate the match at the first }. [^}] means "match anything except }"

    /{([^}]*)}/

    https://regex101.com/r/t9eLmP/1

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