ios - example - substring to string swift 4

Swift Obtener cadena entre 2 cadenas en una cadena

Question

Estoy obteniendo una cadena de análisis html que es;

string = "javascript:getInfo(1,''Info/99/something'', ''City Hall'',1, 99);"

mi código es algo como

var startIndex = text.rangeOfString("''")
var endIndex = text.rangeOfString("'',")
var range2 = startIndex2...endIndex
substr= string.substringWithRange(range)

no estoy seguro de si mi segunda cadena de división debería ser "''" o "'',"

quiero mi resultado como

substr = "Info/99/something"

Answer 1

Considere usar una expresión regular para unir todo entre comillas simples.

let string = "javascript:getInfo(1,''Info/99/something'', ''City Hall'',1, 99);"

let pattern = "''(.+?)''"
let regex = NSRegularExpression(pattern: pattern, options: nil, error: nil)
let results = regex!.matchesInString(string, options: nil, range: NSMakeRange(0, count(string)))  as! [NSTextCheckingResult]

let nsstring = string as NSString
let matches = results.map { result in return nsstring.substringWithRange(result.range)}

// First match
println(matches[0])

Answer 2

En Swift 4 o posterior, puede crear un método de extensión en StringProtocol para admitir también subcadenas. Puede devolver una Substring lugar de una nueva cadena:

edit / u [fecha: Swift 5

let string = "javascript:getInfo(1,''Info/99/something'', ''City Hall'',1, 99);"
let substr = string.substring(from: "''")                   // "Info/99/something'', ''City Hall'',1, 99);"
let subString = string.substring(from: "''", to: "'',")  // "Info/99/something"

let subStringCaseInsensitive = string.substring(from: "''info/", to: "/something", options: .caseInsensitive)  // "99"

    func find(inText text: String, pattern: String) -> [String]? {
        do {
            let regex = try NSRegularExpression(pattern: pattern, options: .caseInsensitive)
            let result = regex.matches(in: text, options: .init(rawValue: 0), range: NSRange(location: 0, length: text.count))

            let matches = result.map { result in
                return (text as NSString).substring(with: result.range)
            }

            return matches
        } catch {
            print(error)
        }
        return nil
    }

Answer 3

Esto funciona si siempre es la segunda división:

let subString = split(string, isSeparator: "''")[1]

Answer 4

Puede usar var arr = str.componentsSeparatedByString(",") como su segunda división que le devolverá su matriz

Answer 5

Reescribí una de las mejores respuestas de Swift 4 para comprender que estaba haciendo con el map . Prefiero una versión con guard , IMO

Sería fácil agregar opciones para Búsqueda hacia atrás, etc.

extension String { func slice(from: String, to: String) -> String? { guard let rangeFrom = range(of: from)?.upperBound else { return nil } guard let rangeTo = self[rangeFrom...].range(of: to)?.lowerBound else { return nil } return String(self[rangeFrom..<rangeTo]) } } let test1 = "a[b]c".slice(from: "[", to: "]") // "b" let test2 = "abc".slice(from: "[", to: "]") // nil let test3 = "a]b[c".slice(from: "[", to: "]") // nil let test4 = "[a[b]c]".slice(from: "[", to: "]") // "a[b"

Answer 6

Swift 4.2:

extension StringProtocol  {
    func substring(from start: Self, to end: Self? = nil, options: String.CompareOptions = []) -> SubSequence? {
        guard let lower = range(of: start, options: options)?.upperBound else { return nil }
        guard let end = end else { return self[lower...] }
        guard let upper = self[lower...].range(of: end, options: options)?.lowerBound else { return nil }
        return self[lower..<upper]
    }
}

Answer 7

Usaría una expresión regular para extraer subcadenas de entradas complejas como esta.

Swift 3.1:

let test = "javascript:getInfo(1,''Info/99/something'', ''City Hall'',1, 99);"

if let match = test.range(of: "(?<='')[^'']+", options: .regularExpression) {
    print(test.substring(with: match))
}

// Prints: Info/99/something

Swift 2.0:

let test = "javascript:getInfo(1,''Info/99/something'', ''City Hall'',1, 99);"

if let match = test.rangeOfString("(?<='')[^'']+", options: .RegularExpressionSearch) {
    print(test.substringWithRange(match))
}

// Prints: Info/99/something

Answer 8

Versión Swift 4 de @litso. Para encontrar todos los valores en el texto

func find(inText text: String, pattern: String) -> [String]? { do { let regex = try NSRegularExpression(pattern: pattern, options: .caseInsensitive) let result = regex.matches(in: text, options: .init(rawValue: 0), range: NSRange(location: 0, length: text.count)) let matches = result.map { result in return (text as NSString).substring(with: result.range) } return matches } catch { print(error) } return nil }

Answer 9

Swift 4

extension String {

    func slice(from: String, to: String) -> String? {

        return (range(of: from)?.upperBound).flatMap { substringFrom in
            (range(of: to, range: substringFrom..<endIndex)?.lowerBound).map { substringTo in
                String(self[substringFrom..<substringTo])
            }
        }
    }
}

Swift 3

extension String {

    func slice(from: String, to: String) -> String? {

        return (range(of: from)?.upperBound).flatMap { substringFrom in
            (range(of: to, range: substringFrom..<endIndex)?.lowerBound).map { substringTo in
                substring(with: substringFrom..<substringTo)
            }
        }
    }
}

Vieja respuesta:

import Foundation

extension String {
  func sliceFrom(start: String, to: String) -> String? {
    return (rangeOfString(start)?.endIndex).flatMap { sInd in
      (rangeOfString(to, range: sInd..<endIndex)?.startIndex).map { eInd in
        substringWithRange(sInd..<eInd)
      }
    }
  }
}

"javascript:getInfo(1,''Info/99/something'', ''City Hall'',1, 99);"
  .sliceFrom("''", to: "'',")