Index of a substring in a string with Swift

Question

I'm used to do this in JavaScript:

var domains = "abcde".substring(0, "abcde".indexOf("cd")) // Returns "ab"

Swift doesn't have this function, how to do something similar?

Answer 1

Doing this in Swift is possible but it takes more lines, here is a function indexOf() doing what is expected:

func indexOf(source: String, substring: String) -> Int? {
    let maxIndex = source.characters.count - substring.characters.count
    for index in 0...maxIndex {
        let rangeSubstring = source.startIndex.advancedBy(index)..<source.startIndex.advancedBy(index + substring.characters.count)
        if source.substringWithRange(rangeSubstring) == substring {
            return index
        }
    }
    return nil
}

var str = "abcde"
if let indexOfCD = indexOf(str, substring: "cd") {
    let distance = str.startIndex.advancedBy(indexOfCD)
    print(str.substringToIndex(distance)) // Returns "ab"
}

This function is not optimized but it does the job for short strings.

Answer 2

In Swift 4 :

Getting Index of a character in a string :

let str = "abcdefghabcd"
if let index = str.index(of: "b") {
   print(index) // Index(_compoundOffset: 4, _cache: Swift.String.Index._Cache.character(1))
}

Creating SubString (prefix and suffix) from String using Swift 4:

let str : String = "ilike"
for i in 0...str.count {
    let index = str.index(str.startIndex, offsetBy: i) // String.Index
    let prefix = str[..<index] // String.SubSequence
    let suffix = str[index...] // String.SubSequence
    print("prefix \(prefix), suffix : \(suffix)")
}

Output

prefix , suffix : ilike
prefix i, suffix : like
prefix il, suffix : ike
prefix ili, suffix : ke
prefix ilik, suffix : e
prefix ilike, suffix : 

If you want to generate a substring between 2 indices , use :

let substring1 = string[startIndex...endIndex] // including endIndex
let subString2 = string[startIndex..<endIndex] // excluding endIndex

Answer 3

In the Swift version 3, String doesn't have functions like -

str.index(of: String)

If the index is required for a substring, one of the ways to is to get the range. We have the following functions in the string which returns range -

str.range(of: <String>)
str.rangeOfCharacter(from: <CharacterSet>)
str.range(of: <String>, options: <String.CompareOptions>, range: <Range<String.Index>?>, locale: <Locale?>)

For example to find the indexes of first occurrence of play in str

var str = "play play play"
var range = str.range(of: "play")
range?.lowerBound //Result : 0
range?.upperBound //Result : 4

Note : range is an optional. If it is not able to find the String it will make it nil. For example

var str = "play play play"
var range = str.range(of: "zoo") //Result : nil
range?.lowerBound //Result : nil
range?.upperBound //Result : nil

Answer 4

Tested for Swift 4.2/4.1/4.0/3.0

Using String[Range<String.Index>] subscript you can get the sub string. You need starting index and last index to create the range and you can do it as below

let str = "abcde"
if let range = str.range(of: "cd") {
  let substring = str[..<range.lowerBound] // or str[str.startIndex..<range.lowerBound]
  print(substring)  // Prints ab
}
else {
  print("String not present")
}

If you don't define the start index this operator ..< , it take the starting index. You can also use str[str.startIndex..<range.lowerBound] instead of str[..<range.lowerBound]

Answer 5

Xcode 9 • Swift 4 or later

extension StringProtocol where Index == String.Index {
    func index(of string: Self, options: String.CompareOptions = []) -> Index? {
        return range(of: string, options: options)?.lowerBound
    }
    func endIndex(of string: Self, options: String.CompareOptions = []) -> Index? {
        return range(of: string, options: options)?.upperBound
    }
    func indexes(of string: Self, options: String.CompareOptions = []) -> [Index] {
        var result: [Index] = []
        var start = startIndex
        while start < endIndex,
            let range = self[start..<endIndex].range(of: string, options: options) {
            result.append(range.lowerBound)
            start = range.lowerBound < range.upperBound ? range.upperBound :
                    index(range.lowerBound, offsetBy: 1, limitedBy: endIndex) ?? endIndex
        }
        return result
    }
    func ranges(of string: Self, options: String.CompareOptions = []) -> [Range<Index>] {
        var result: [Range<Index>] = []
        var start = startIndex
        while start < endIndex,
            let range = self[start..<endIndex].range(of: string, options: options) {
            result.append(range)
            start = range.lowerBound < range.upperBound ? range.upperBound :
                    index(range.lowerBound, offsetBy: 1, limitedBy: endIndex) ?? endIndex
        }
        return result
    }
}

---

usage:

let str = "abcde"
if let index = str.index(of: "cd") {
    let substring = str[..<index]
    let string = String(substring)
    print(string)  // "ab\n"
}

---

let str = "Hello, playground, playground, playground"
str.index(of: "play")      // 7
str.endIndex(of: "play")   // 11
str.indexes(of: "play")    // [7, 19, 31]
str.ranges(of: "play")     // [{lowerBound 7, upperBound 11}, {lowerBound 19, upperBound 23}, {lowerBound 31, upperBound 35}]

Answer 6

There are three closely connected issues here:

• All the substring-finding methods are over in the Cocoa NSString world (Foundation)

• Foundation NSRange has a mismatch with Swift Range; the former uses start and length, the latter uses endpoints

• In general, Swift characters are indexed using String.Index, not Int, but Foundation characters are indexed using Int, and there is no simple direct translation between them (because Foundation and Swift have different ideas of what constitutes a character)

Given all that, let's think about how to write:

func substring(of s: String, from:Int, toSubstring s2 : String) -> Substring? {
    // ?
}

The substring s2 must be sought in s using a String Foundation method. The resulting range comes back to us, not as an NSRange (even though this is a Foundation method), but as a Range of String.Index (wrapped in an Optional, in case we didn't find the substring at all). However, the other number, from, is an Int. Thus we cannot form any kind of range involving them both.

But we don't have to! All we have to do is slice off the end of our original string using a method that takes a String.Index, and slice off the start of our original string using a method that takes an Int. Fortunately, such methods exist! Like this:

func substring(of s: String, from:Int, toSubstring s2 : String) -> Substring? {
    guard let r = s.range(of:s2) else {return nil}
    var s = s.prefix(upTo:r.lowerBound)
    s = s.dropFirst(from)
    return s
}

Or, if you prefer to be able to apply this method directly to a string, like this...

let output = "abcde".substring(from:0, toSubstring:"cd")

...then make it an extension on String:

extension String {
    func substring(from:Int, toSubstring s2 : String) -> Substring? {
        guard let r = self.range(of:s2) else {return nil}
        var s = self.prefix(upTo:r.lowerBound)
        s = s.dropFirst(from)
        return s
    }
}