ios - 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.

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.

Answer 1

Just accessing backward

The best way is to use substringToIndex combined to the endIndexproperty 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:))