Swift: How to get substring from start to last index of character
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?
edit/update:
In Swift 4 or later (Xcode 10.0+) you can use the new BidirectionalCollection method lastIndex(of:)
func lastIndex(of element: Self.Element) -> Self.Index?
let string = "www.stackoverflow.com"
if let lastIndex = string.lastIndex(of: ".") {
let subString = string[..<lastIndex] // "www.stackoverflow"
}
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:))
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