Swift equivalent for Java indexOf and lastIndexOf of a string

Question

I come from Java. I studied the Swift documentation and understood most of the concepts.

What I now looking for is an equivalent to the Java indexOf and lastIndexOf methods to finding positions of substring in a string.

I already found a solution with rangeOfString and using the startIndex property. That looks helpful for me to define the indexOf method.

But I think that rangeOfString only starts searching from the beginning of a string. Is this correct? And if so, how can I search in reverse direction (from end to start of the string)?

What I mean is to have f.e. the string "hello world" and if I start searching for "l" then I want to find the letter at position 9 and not at position 2.

Have a look at http://stackoverflow.com/questions/28182441/swift-how-to-get-substring-from-start-to-last-index-of-character, it shows various methods, including the `.BackwardsSearch` option. — Martin R, Sep 30 '16 at 11:41
You can use "some string".characters.indexOf('someChar') but it only works for chars.. — Dan, Sep 30 '16 at 12:06

score 26 · Answer 1 · answered Apr 05 '17 at 20:04

26

In Swift 3

Java indexOf equivalent:

var index1 = string1.index(string1.endIndex, offsetBy: -4)

Java lastIndexOf equivalent:

 var index2 = string2.range(of: ".", options: .backwards)?.lowerBound

answered Apr 05 '17 at 20:04

Danyl

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Why do you have `offsetBy: -4` for the indexOf ? – John Difool Feb 25 '18 at 03:23
@Laurent indexOf has an optional fromIndex: https://beginnersbook.com/2013/12/java-string-indexof-method-example/ – Patrick Aug 08 '19 at 11:51
In Swift 5 the "lastIndexOf" code returns a `String.Index`, not an `Int` (like in Java). You could as well just do `mystring.lastIndex(of: ".")`, which is shorter and also returns a `String.Index`. The problem is that there doesn't seem to be an easy way to convert that into a proper `Int` you can work with. – Neph Mar 03 '21 at 10:15

score 8 · Answer 2 · answered Apr 14 '17 at 19:23

8

extension String {
    func indexOf(_ input: String,
                 options: String.CompareOptions = .literal) -> String.Index? {
        return self.range(of: input, options: options)?.lowerBound
    }

    func lastIndexOf(_ input: String) -> String.Index? {
        return indexOf(input, options: .backwards)
    }
}

"hello world".indexOf("l") // 2
"hello world".lastIndexOf("l") // 9

answered Apr 14 '17 at 19:23

Vladimir Kofman

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Helped me. Thank you – mcabe Feb 09 '18 at 21:46
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Printing the tests results in: 1. `Optional(Swift.String.Index(_rawBits: 131072))`, 2. `Optional(Swift.String.Index(_rawBits: 589824))` - how did you get the actual int from it? – Neph Mar 03 '21 at 14:16
Problem is using this in substring. – chitgoks Aug 28 '22 at 12:24

score 7 · Answer 3 · answered Jan 30 '18 at 15:48

7

If you want the returned value to be an Int:

extension String {

    func lastIndex(of string: String) -> Int? {
        guard let index = range(of: string, options: .backwards) else { return nil }
        return self.distance(from: self.startIndex, to: index.lowerBound)
    }
}

answered Jan 30 '18 at 15:48

Aviv Ben Shabat

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Thank you, this helped me so much! – Xcoder Mar 20 '18 at 15:35

score 0 · Answer 4 · answered Dec 16 '18 at 01:01

There is a built in swift function for finding the index of last occurrence of a character. The function signature is below

public func lastIndex(of element: Character) -> String.Index?{}

We can get the index for a character like below:

let url = "http://www.google.com/abc"
guard let lastIndexOfChar = url.lastIndex(of: "/") else { return nil }
let startIndex = url.index(lastIndexOfChar, offsetBy:1)
let substring = url[startIndex..<url.endIndex] // prints "abc"

Swift equivalent for Java indexOf and lastIndexOf of a string

4 Answers4