Swift: How to get substring from start to last index of character
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.
For example, convert "www.stackoverflow.com" to "www.stackoverflow". What code snippet would do that, and being the most swift-like? (I hope this doesn't bring a debate, but I can't find good lesson on how to handle substrings in Swift.
Just accessing backward
The best way is to use substringToIndex
combined to the endIndex
property and the advance
global function.
var string1 = "www.stackoverflow.com"
var index1 = advance(string1.endIndex, -4)
var substring1 = string1.substringToIndex(index1)
Looking for a string starting from the back
Use rangeOfString
and set options
to .BackwardsSearch
var string2 = "www.stackoverflow.com"
var index2 = string2.rangeOfString(".", options: .BackwardsSearch)?.startIndex
var substring2 = string2.substringToIndex(index2!)
No extensions, pure idiomatic Swift
Swift 2.0
advance
is now a part of Index
and is called advancedBy
. You do it like:
var string1 = "www.stackoverflow.com"
var index1 = string1.endIndex.advancedBy(-4)
var substring1 = string1.substringToIndex(index1)
Swift 3.0
You can't call advancedBy
on a String
because it has variable size elements. You have to use index(_, offsetBy:)
.
var string1 = "www.stackoverflow.com"
var index1 = string1.index(string1.endIndex, offsetBy: -4)
var substring1 = string1.substring(to: index1)
A lot of things have been renamed. The cases are written in camelCase, startIndex
became lowerBound
.
var string2 = "www.stackoverflow.com"
var index2 = string2.range(of: ".", options: .backwards)?.lowerBound
var substring2 = string2.substring(to: index2!)
Also, I wouldn't recommend force unwrapping index2
. You can use optional binding or map
. Personally, I prefer using map
:
var substring3 = index2.map(string2.substring(to:))
Swift 4
The Swift 3 version is still valid but now you can now use subscripts with indexes ranges:
let string1 = "www.stackoverflow.com"
let index1 = string1.index(string1.endIndex, offsetBy: -4)
let substring1 = string1[..<index1]
The second approach remains unchanged:
let string2 = "www.stackoverflow.com"
let index2 = string2.range(of: ".", options: .backwards)?.lowerBound
let substring3 = index2.map(string2.substring(to:))
Swift 3, XCode 8
func lastIndexOfCharacter(_ c: Character) -> Int? {
return range(of: String(c), options: .backwards)?.lowerBound.encodedOffset
}
Since advancedBy(Int)
is gone since Swift 3 use String
's method index(String.Index, Int)
. Check out this String
extension with substring and friends:
public extension String {
//right is the first encountered string after left
func between(_ left: String, _ right: String) -> String? {
guard let leftRange = range(of: left), let rightRange = range(of: right, options: .backwards)
, leftRange.upperBound <= rightRange.lowerBound
else { return nil }
let sub = self.substring(from: leftRange.upperBound)
let closestToLeftRange = sub.range(of: right)!
return sub.substring(to: closestToLeftRange.lowerBound)
}
var length: Int {
get {
return self.characters.count
}
}
func substring(to : Int) -> String {
let toIndex = self.index(self.startIndex, offsetBy: to)
return self.substring(to: toIndex)
}
func substring(from : Int) -> String {
let fromIndex = self.index(self.startIndex, offsetBy: from)
return self.substring(from: fromIndex)
}
func substring(_ r: Range<Int>) -> String {
let fromIndex = self.index(self.startIndex, offsetBy: r.lowerBound)
let toIndex = self.index(self.startIndex, offsetBy: r.upperBound)
return self.substring(with: Range<String.Index>(uncheckedBounds: (lower: fromIndex, upper: toIndex)))
}
func character(_ at: Int) -> Character {
return self[self.index(self.startIndex, offsetBy: at)]
}
func lastIndexOfCharacter(_ c: Character) -> Int? {
guard let index = range(of: String(c), options: .backwards)?.lowerBound else
{ return nil }
return distance(from: startIndex, to: index)
}
}
UPDATED extension for Swift 5
public extension String {
//right is the first encountered string after left
func between(_ left: String, _ right: String) -> String? {
guard
let leftRange = range(of: left), let rightRange = range(of: right, options: .backwards)
, leftRange.upperBound <= rightRange.lowerBound
else { return nil }
let sub = self[leftRange.upperBound...]
let closestToLeftRange = sub.range(of: right)!
return String(sub[..<closestToLeftRange.lowerBound])
}
var length: Int {
get {
return self.count
}
}
func substring(to : Int) -> String {
let toIndex = self.index(self.startIndex, offsetBy: to)
return String(self[...toIndex])
}
func substring(from : Int) -> String {
let fromIndex = self.index(self.startIndex, offsetBy: from)
return String(self[fromIndex...])
}
func substring(_ r: Range<Int>) -> String {
let fromIndex = self.index(self.startIndex, offsetBy: r.lowerBound)
let toIndex = self.index(self.startIndex, offsetBy: r.upperBound)
let indexRange = Range<String.Index>(uncheckedBounds: (lower: fromIndex, upper: toIndex))
return String(self[indexRange])
}
func character(_ at: Int) -> Character {
return self[self.index(self.startIndex, offsetBy: at)]
}
func lastIndexOfCharacter(_ c: Character) -> Int? {
guard let index = range(of: String(c), options: .backwards)?.lowerBound else
{ return nil }
return distance(from: startIndex, to: index)
}
}
Usage:
let text = "www.stackoverflow.com"
let at = text.character(3) // .
let range = text.substring(0..<3) // www
let from = text.substring(from: 4) // stackoverflow.com
let to = text.substring(to: 16) // www.stackoverflow
let between = text.between(".", ".") // stackoverflow
let substringToLastIndexOfChar = text.lastIndexOfCharacter(".") // 17
P.S. It's really odd that developers forced to deal with String.Index
instead of plain Int
. Why should we bother about internal String
mechanics and not just have simple substring()
methods?
I would do it using a subscript (s[start..<end]
):
Swift 3, 4, 5
let s = "www.stackoverflow.com"
let start = s.startIndex
let end = s.index(s.endIndex, offsetBy: -4)
let substring = s[start..<end] // www.stackoverflow