Swift: How to get substring from start to last index of character

Question

I want to learn the best/simplest way to turn a string into another string but with only a subset, starting at the beginning and going to the last index of a character.

Answer 1

Here's how I do it. You could do it the same way, or use this code for ideas.

let s = "www.stackoverflow.com"
s.substringWithRange(0..<s.lastIndexOf("."))

Here are the extensions I use:

import Foundation
extension String {

  var length: Int {
    get {
      return countElements(self)
    }
  }

  func indexOf(target: String) -> Int {
    var range = self.rangeOfString(target)
    if let range = range {
      return distance(self.startIndex, range.startIndex)
    } else {
      return -1
    }
  }

  func indexOf(target: String, startIndex: Int) -> Int {
    var startRange = advance(self.startIndex, startIndex)        
    var range = self.rangeOfString(target, options: NSStringCompareOptions.LiteralSearch, range: Range<String.Index>(start: startRange, end: self.endIndex))
    if let range = range {
      return distance(self.startIndex, range.startIndex)
    } else {
      return -1
    }
  }

  func lastIndexOf(target: String) -> Int {
    var index = -1
    var stepIndex = self.indexOf(target)
    while stepIndex > -1 {
      index = stepIndex
      if stepIndex + target.length < self.length {
        stepIndex = indexOf(target, startIndex: stepIndex + target.length)
      } else {
        stepIndex = -1
      }
    }
    return index
  } 

  func substringWithRange(range:Range<Int>) -> String {
    let start = advance(self.startIndex, range.startIndex)
    let end = advance(self.startIndex, range.endIndex)
    return self.substringWithRange(start..<end)
  }

}

Credit albertbori / Common Swift String Extensions

Generally I am a strong proponent of extensions, especially for needs like string manipulation, searching, and slicing.

Answer 2

edit/update:

In Swift 4 or later (Xcode 10.0+) you can use the new BidirectionalCollection method lastIndex(of:)

func lastIndex(of element: Element) -> Int?

---

let string = "www.stackoverflow.com"
if let lastIndex = string.lastIndex(of: ".") {
    let subString = string[..<lastIndex]  // "www.stackoverflow"
}

Answer 3

Swift 3

let string = "www.stackoverflow.com"
let first3Characters = String(string.characters.prefix(3)) // www
let lastCharacters = string.characters.dropFirst(4) // stackoverflow.com (it would be a collection)

//or by index 
let indexOfFouthCharacter = olNumber.index(olNumber.startIndex, offsetBy: 4)
let first3Characters = olNumber.substring(to: indexOfFouthCharacter) // www
let lastCharacters = olNumber.substring(from: indexOfFouthCharacter) // .stackoverflow.com

Good article for understanding, why do we need this

Answer 4

Swift 4:

extension String {

    /// the length of the string
    var length: Int {
        return self.characters.count
    }

    /// Get substring, e.g. "ABCDE".substring(index: 2, length: 3) -> "CDE"
    ///
    /// - parameter index:  the start index
    /// - parameter length: the length of the substring
    ///
    /// - returns: the substring
    public func substring(index: Int, length: Int) -> String {
        if self.length <= index {
            return ""
        }
        let leftIndex = self.index(self.startIndex, offsetBy: index)
        if self.length <= index + length {
            return self.substring(from: leftIndex)
        }
        let rightIndex = self.index(self.endIndex, offsetBy: -(self.length - index - length))
        return self.substring(with: leftIndex..<rightIndex)
    }

    /// Get substring, e.g. -> "ABCDE".substring(left: 0, right: 2) -> "ABC"
    ///
    /// - parameter left:  the start index
    /// - parameter right: the end index
    ///
    /// - returns: the substring
    public func substring(left: Int, right: Int) -> String {
        if length <= left {
            return ""
        }
        let leftIndex = self.index(self.startIndex, offsetBy: left)
        if length <= right {
            return self.substring(from: leftIndex)
        }
        else {
            let rightIndex = self.index(self.endIndex, offsetBy: -self.length + right + 1)
            return self.substring(with: leftIndex..<rightIndex)
        }
    }
}

you can test it as follows:

    print("test: " + String("ABCDE".substring(index: 2, length: 3) == "CDE"))
    print("test: " + String("ABCDE".substring(index: 0, length: 3) == "ABC"))
    print("test: " + String("ABCDE".substring(index: 2, length: 1000) == "CDE"))
    print("test: " + String("ABCDE".substring(left: 0, right: 2) == "ABC"))
    print("test: " + String("ABCDE".substring(left: 1, right: 3) == "BCD"))
    print("test: " + String("ABCDE".substring(left: 3, right: 1000) == "DE"))

Check https://gitlab.com/seriyvolk83/SwiftEx library. It contains these and other helpful methods.

Answer 5

var url = "www.stackoverflow.com"
let str = path.suffix(3)
print(str) //low

Answer 6

Do you want to get a substring of a string from start index to the last index of one of its characters? If so, you may choose one of the following Swift 2.0+ methods.

Methods that require Foundation

Get a substring that includes the last index of a character:

import Foundation

let string = "www.stackoverflow.com"
if let rangeOfIndex = string.rangeOfCharacterFromSet(NSCharacterSet(charactersInString: "."), options: .BackwardsSearch) {
    print(string.substringToIndex(rangeOfIndex.endIndex))
}

// prints "www.stackoverflow."

Get a substring that DOES NOT include the last index of a character:

import Foundation

let string = "www.stackoverflow.com"
if let rangeOfIndex = string.rangeOfCharacterFromSet(NSCharacterSet(charactersInString: "."), options: .BackwardsSearch) {
    print(string.substringToIndex(rangeOfIndex.startIndex))
}

// prints "www.stackoverflow"

---

If you need to repeat those operations, extending String can be a good solution:

import Foundation

extension String {
    func substringWithLastInstanceOf(character: Character) -> String? {
        if let rangeOfIndex = rangeOfCharacterFromSet(NSCharacterSet(charactersInString: String(character)), options: .BackwardsSearch) {
            return self.substringToIndex(rangeOfIndex.endIndex)
        }
        return nil
    }
    func substringWithoutLastInstanceOf(character: Character) -> String? {
        if let rangeOfIndex = rangeOfCharacterFromSet(NSCharacterSet(charactersInString: String(character)), options: .BackwardsSearch) {
            return self.substringToIndex(rangeOfIndex.startIndex)
        }
        return nil
    }
}

print("www.stackoverflow.com".substringWithLastInstanceOf("."))
print("www.stackoverflow.com".substringWithoutLastInstanceOf("."))

/*
prints:
Optional("www.stackoverflow.")
Optional("www.stackoverflow")
*/

---

Methods that DO NOT require Foundation

Get a substring that includes the last index of a character:

let string = "www.stackoverflow.com"
if let reverseIndex = string.characters.reverse().indexOf(".") {
    print(string[string.startIndex ..< reverseIndex.base])
}

// prints "www.stackoverflow."

Get a substring that DOES NOT include the last index of a character:

let string = "www.stackoverflow.com"
if let reverseIndex = string.characters.reverse().indexOf(".") {
    print(string[string.startIndex ..< reverseIndex.base.advancedBy(-1)])
}

// prints "www.stackoverflow"

---

If you need to repeat those operations, extending String can be a good solution:

extension String {
    func substringWithLastInstanceOf(character: Character) -> String? {
        if let reverseIndex = characters.reverse().indexOf(".") {
            return self[self.startIndex ..< reverseIndex.base]
        }
        return nil
    }
    func substringWithoutLastInstanceOf(character: Character) -> String? {
        if let reverseIndex = characters.reverse().indexOf(".") {
            return self[self.startIndex ..< reverseIndex.base.advancedBy(-1)]
        }
        return nil
    }
}

print("www.stackoverflow.com".substringWithLastInstanceOf("."))
print("www.stackoverflow.com".substringWithoutLastInstanceOf("."))

/*
prints:
Optional("www.stackoverflow.")
Optional("www.stackoverflow")
*/