method to capture a swift regex

Question

Here I try to find a way to get the second word of my array and created a new list up with the words retrieve
Here are the array

[Orange, Orange Ekstraklasa, Orange (entreprise), Orange (Vaucluse), Orange mécanique, Orange-Nassau, Orange (fruit), Orange Cinémax, Orange TV, Orange Stinger]

Here is an example of the word that I want to recover

• Ekstraklasa
• entreprise
• Vaucluse
• mécanique
• fruit

etc...

I think I have created a regular expression and a space for parthentèse recovers the second word.

I find this method to match my expression and I ask if there was another one that creates a new array with the word select

here is the method

    let myStringToBeMatched = "ThisIsMyString"
    let myRegex = "^This"

    if let match = myStringToBeMatched.rangeOfString(myRegex, options: .RegularExpressionSearch){
        println("\(myStringToBeMatched) is matching!")
    }

I came across a new problem :

I tried the solution this solution works well outside the alamofire method:

let list = ["Orange", "Orange Ekstraklasa", "Orange (entreprise)", "Orange (Vaucluse)", "Orange mécanique", "Orange-Nassau", "Orange (fruit)", "Orange Cinémax", "Orange TV", "Orange Stinger"]

let results = list.map { split($0, {$0 == " " || $0 == "(" || $0 == ")"}) }
              .filter { $0.count > 1 }
              .map { $0[1] }

So I adapted to functioning in alamofire this

Alamofire.request(.GET, "http://fr.wikipedia.org/w/api.php", parameters: parameters)
        .responseJSON { (_, _, data, _) in

            let json = JSON(object: data!)
            var list = json[1].arrayValue!

            let results = list.map { split($0, {$0 == " " || $0 == "(" || $0 == ")"}) }
                .filter { $0.count > 1 }
                .map { $0[1] }

    }

at my variable results it myself but as error: Expression was too complex to be solved in reasonable time; consider breaking up the expression into distinct sub-expressions

Thanks

Answer 1

In addition to regex, you could also do something like:

let list = ["Orange", "Orange Ekstraklasa", "Orange (entreprise)", "Orange (Vaucluse)", "Orange mécanique", "Orange-Nassau", "Orange (fruit)", "Orange Cinémax", "Orange TV", "Orange Stinger"]

let results = list.map { split($0, {$0 == " " || $0 == "(" || $0 == ")"}) }
                  .filter { $0.count > 1 }
                  .map { $0[1] }

That builds an array of words for each item in the original list, filters out those with less than 2 items, and then builds new array of the second items.

A little less elegant, but probably more efficient, would be

let results = list
    .map {
        (string) -> String! in

        let words = split(string, {$0 == " " || $0 == "(" || $0 == ")"})
        if words.count > 1 {
            return words[1]
        } else {
            return nil
        }
    }
    .filter { $0 != nil }

That builds array of the second words (or nil if there was only one word) and then filters out the nil values.

The regex equivalent would be:

var error: NSError?
let regex = NSRegularExpression(pattern: "[\\w-]+", options: nil, error: &error)

let results = list
    .map {
        (string) -> String! in

        let matches = regex?.matchesInString(string, options: nil, range: NSMakeRange(0, countElements(string))) as [NSTextCheckingResult]?
        if matches?.count > 1 {
            return (string as NSString).substringWithRange(matches![1].range)
        } else {
            return nil
        }
    }
    .filter { $0 != nil }

If you'd prefer traditional for loops:

var results = [String]()
for string in list {
    let words = split(string, {$0 == " " || $0 == "(" || $0 == ")"})
    if words.count > 1 {
        results.append(words[1])
    }
}

or, its regex equivalent:

var error: NSError?
let regex = NSRegularExpression(pattern: "[\\w-]+", options: nil, error: &error)

var results = [String]()
for string in list {
    let matches = regex?.matchesInString(string, options: nil, range: NSMakeRange(0, countElements(string))) as [NSTextCheckingResult]?
    if matches?.count > 1 {
        results.append((string as NSString).substringWithRange(matches![1].range))
    }
}

Answer 2

do not understand why he told me this line that I get an error like:
Could not find member 'stringValue'

the error is at the println ()

if let match = json[1][index].stringValue?.rangeOfString(myRegex, options: .RegularExpressionSearch) {
    println(json[1][match].stringValue?)
}

Here is the rest of the code

let json = JSON(object: data!)
let list: Array<JSON> = json[1].arrayValue!
var index:Int

for(index = 0 ; index < list.count ; index++) {
    let myRegex = "\\((.*?)\\)"
    if let match = json[1][index].stringValue?.rangeOfString(myRegex, options: .RegularExpressionSearch){
        println(json[1][match].stringValue?)
    }
}

Answer 3

The rangeOfString returns a range, you can use it to create a new string instance.

let myStringToBeMatched = "ThisIsMyString"
let myRegex = "^This"

if let match = myStringToBeMatched.rangeOfString(myRegex, options: .RegularExpressionSearch){
    let result = myStringToBeMatched[match]
    println("\(result) is matching!")
}