Revisions to How to get last 4 characters of a string? [duplicate]

Added a "Swift 2" heading.

edited Sep 15, 2022 at 17:35

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Swift 2:

A solution is substringFromIndex

let a = "StackOverFlow" let last4 = a.substringFromIndex(a.endIndex.advancedBy(-4))

or suffix on characters

let last4 = String(a.characters.suffix(4))

code is Swift 2

Swift 3:

In Swift 3 the syntax for the first solution has been changed to

let last4 = a.substring(from:a.index(a.endIndex, offsetBy: -4))

Swift 4+:

In Swift 4 it becomes more convenient:

let last4 = a.suffix(4)

The type of the result is a new type Substring which behaves as a String in many cases. However if the substring is supposed to leave the scope where it's created in you have to create a new String instance.

let last4 = String(a.suffix(4))

A solution is substringFromIndex

let a = "StackOverFlow" let last4 = a.substringFromIndex(a.endIndex.advancedBy(-4))

or suffix on characters

let last4 = String(a.characters.suffix(4))

code is Swift 2

Swift 3:

In Swift 3 the syntax for the first solution has been changed to

let last4 = a.substring(from:a.index(a.endIndex, offsetBy: -4))

Swift 4+:

In Swift 4 it becomes more convenient:

let last4 = a.suffix(4)

The type of the result is a new type Substring which behaves as a String in many cases. However if the substring is supposed to leave the scope where it's created in you have to create a new String instance.

let last4 = String(a.suffix(4))

Swift 2:

A solution is substringFromIndex

let a = "StackOverFlow" let last4 = a.substringFromIndex(a.endIndex.advancedBy(-4))

or suffix on characters

let last4 = String(a.characters.suffix(4))

Swift 3:

In Swift 3 the syntax for the first solution has been changed to

let last4 = a.substring(from:a.index(a.endIndex, offsetBy: -4))

Swift 4+:

In Swift 4 it becomes more convenient:

let last4 = a.suffix(4)

The type of the result is a new type Substring which behaves as a String in many cases. However if the substring is supposed to leave the scope where it's created in you have to create a new String instance.

let last4 = String(a.suffix(4))

added 1 character in body

Source Link

edited Jul 9, 2020 at 10:51

vadian

286.2k
32
378
379

A solution is substringFromIndex

let a = "StackOverFlow" let last4 = a.substringFromIndex(a.endIndex.advancedBy(-4))

or suffix on characters

let last4 = String(a.characters.suffix(4))

code is Swift 2

Swift 3:

In Swift 3 the syntax for the first solution has been changed to

let last4 = a.substring(from:a.index(a.endIndex, offsetBy: -4))

Swift 44+:

In Swift 4 it becomes more convenient:

let last4 = a.suffix(4)

The type of the result is a new type Substring which behaves as a String in many cases. However if the substring is supposed to leave the scope where it's created in you have to create a new String instance.

let last4 = String(a.suffix(4))

A solution is substringFromIndex

let a = "StackOverFlow" let last4 = a.substringFromIndex(a.endIndex.advancedBy(-4))

or suffix on characters

let last4 = String(a.characters.suffix(4))

code is Swift 2

Swift 3:

In Swift 3 the syntax for the first solution has been changed to

let last4 = a.substring(from:a.index(a.endIndex, offsetBy: -4))

Swift 4:

In Swift 4 it becomes more convenient:

let last4 = a.suffix(4)

The type of the result is a new type Substring which behaves as a String in many cases. However if the substring is supposed to leave the scope where it's created in you have to create a new String instance.

let last4 = String(a.suffix(4))

A solution is substringFromIndex

let a = "StackOverFlow" let last4 = a.substringFromIndex(a.endIndex.advancedBy(-4))

or suffix on characters

let last4 = String(a.characters.suffix(4))

code is Swift 2

Swift 3:

In Swift 3 the syntax for the first solution has been changed to

let last4 = a.substring(from:a.index(a.endIndex, offsetBy: -4))

Swift 4+:

In Swift 4 it becomes more convenient:

let last4 = a.suffix(4)

The type of the result is a new type Substring which behaves as a String in many cases. However if the substring is supposed to leave the scope where it's created in you have to create a new String instance.

let last4 = String(a.suffix(4))

deleted 12 characters in body

Source Link

edited Nov 18, 2017 at 7:18

vadian

286.2k
32
378
379

A solution is substringFromIndex

let a = "StackOverFlow" let last4 = a.substringFromIndex(a.endIndex.advancedBy(-4))

or suffix on characters

let last4 = String(a.characters.suffix(4))

code is Swift 2

Swift 3:

In Swift 3 the syntax for the first solution has been changed to

let last4 = a.substring(from:a.index(a.endIndex, offsetBy: -4))

Swift 4:

In Swift 4 it becomes more convenient:

let last4 = a.suffix(4)

The type of the result is a new type Substring which behaves as a String in many cases. However if the substring is supposed to leave the scope where it's created in Apple recommendsyou have to create a regularnew String instance.

let last4 = String(a.suffix(4))

A solution is substringFromIndex

let a = "StackOverFlow" let last4 = a.substringFromIndex(a.endIndex.advancedBy(-4))

or suffix on characters

let last4 = String(a.characters.suffix(4))

code is Swift 2

Swift 3:

In Swift 3 the syntax for the first solution has been changed to

let last4 = a.substring(from:a.index(a.endIndex, offsetBy: -4))

Swift 4:

In Swift 4 it becomes more convenient:

let last4 = a.suffix(4)

The type of the result is a new type Substring which behaves as a String in many cases. However if the substring is supposed to leave the scope where it's created in Apple recommends to create a regular String instance.

let last4 = String(a.suffix(4))

A solution is substringFromIndex

let a = "StackOverFlow" let last4 = a.substringFromIndex(a.endIndex.advancedBy(-4))

or suffix on characters

let last4 = String(a.characters.suffix(4))

code is Swift 2

Swift 3:

In Swift 3 the syntax for the first solution has been changed to

let last4 = a.substring(from:a.index(a.endIndex, offsetBy: -4))

Swift 4:

In Swift 4 it becomes more convenient:

let last4 = a.suffix(4)

The type of the result is a new type Substring which behaves as a String in many cases. However if the substring is supposed to leave the scope where it's created in you have to create a new String instance.