Swift 3 Get Start Index (As Int) of Substring

swift 3 get start index (as int) of substring

The distance(from:to:) method of String computes the difference
between two String.Index values:

let mystring = "hi this is my name"
if let range = mystring.range(of: "this") {
    let startPos = mystring.distance(from: mystring.startIndex, to: range.lowerBound)
    let endPos = mystring.distance(from: mystring.startIndex, to: range.upperBound)
    print(startPos, endPos) // 3 7
}

Actually it just forwards the call to the string's CharacterView, so
the above gives the same result as

let mystring = "hi this is my name"
if let range = mystring.range(of: "this") {
    let startPos = mystring.characters.distance(from: mystring.characters.startIndex, to: range.lowerBound)
    let endPos = mystring.characters.distance(from: mystring.characters.startIndex, to: range.upperBound)
    print(startPos, endPos) // 3 7
}

---

If you need all occurrences of the string:

let mystring = "this is this and that is that"
var searchPosition = mystring.startIndex
while let range = mystring.range(of: "this", range: searchPosition..<mystring.endIndex) {
    let startPos = mystring.distance(from: mystring.startIndex, to: range.lowerBound)
    let endPos = mystring.distance(from: mystring.startIndex, to: range.upperBound)
    print(startPos, endPos)

    searchPosition = range.upperBound
}

How does String substring work in Swift

All of the following examples use

var str = "Hello, playground"

Swift 4

Strings got a pretty big overhaul in Swift 4. When you get some substring from a String now, you get a Substring type back rather than a String. Why is this? Strings are value types in Swift. That means if you use one String to make a new one, then it has to be copied over. This is good for stability (no one else is going to change it without your knowledge) but bad for efficiency.

A Substring, on the other hand, is a reference back to the original String from which it came. Here is an image from the documentation illustrating that.

No copying is needed so it is much more efficient to use. However, imagine you got a ten character Substring from a million character String. Because the Substring is referencing the String, the system would have to hold on to the entire String for as long as the Substring is around. Thus, whenever you are done manipulating your Substring, convert it to a String.

let myString = String(mySubstring)

This will copy just the substring over and the memory holding old String can be reclaimed. Substrings (as a type) are meant to be short lived.

Another big improvement in Swift 4 is that Strings are Collections (again). That means that whatever you can do to a Collection, you can do to a String (use subscripts, iterate over the characters, filter, etc).

The following examples show how to get a substring in Swift.

Getting substrings

You can get a substring from a string by using subscripts or a number of other methods (for example, prefix, suffix, split). You still need to use String.Index and not an Int index for the range, though. (See my other answer if you need help with that.)

Beginning of a string

You can use a subscript (note the Swift 4 one-sided range):

let index = str.index(str.startIndex, offsetBy: 5)
let mySubstring = str[..<index] // Hello

or prefix:

let index = str.index(str.startIndex, offsetBy: 5)
let mySubstring = str.prefix(upTo: index) // Hello

or even easier:

let mySubstring = str.prefix(5) // Hello

End of a string

Using subscripts:

let index = str.index(str.endIndex, offsetBy: -10)
let mySubstring = str[index...] // playground

or suffix:

let index = str.index(str.endIndex, offsetBy: -10)
let mySubstring = str.suffix(from: index) // playground

or even easier:

let mySubstring = str.suffix(10) // playground

Note that when using the suffix(from: index) I had to count back from the end by using -10. That is not necessary when just using suffix(x), which just takes the last x characters of a String.

Range in a string

Again we simply use subscripts here.

let start = str.index(str.startIndex, offsetBy: 7)
let end = str.index(str.endIndex, offsetBy: -6)
let range = start..<end

let mySubstring = str[range]  // play

Converting Substring to String

Don't forget, when you are ready to save your substring, you should convert it to a String so that the old string's memory can be cleaned up.

let myString = String(mySubstring)

Using an Int index extension?

I'm hesitant to use an Int based index extension after reading the article Strings in Swift 3 by Airspeed Velocity and Ole Begemann. Although in Swift 4, Strings are collections, the Swift team purposely hasn't used Int indexes. It is still String.Index. This has to do with Swift Characters being composed of varying numbers of Unicode codepoints. The actual index has to be uniquely calculated for every string.

I have to say, I hope the Swift team finds a way to abstract away String.Index in the future. But until then, I am choosing to use their API. It helps me to remember that String manipulations are not just simple Int index lookups.

Convert String.Index? to Int

Try this out:

s.indexOf("World").encodedOffset ?? -1

Index of a substring in a string with Swift

edit/update:

Xcode 11.4 • Swift 5.2 or later

import Foundation

extension StringProtocol {
    func index<S: StringProtocol>(of string: S, options: String.CompareOptions = []) -> Index? {
        range(of: string, options: options)?.lowerBound
    }
    func endIndex<S: StringProtocol>(of string: S, options: String.CompareOptions = []) -> Index? {
        range(of: string, options: options)?.upperBound
    }
    func indices<S: StringProtocol>(of string: S, options: String.CompareOptions = []) -> [Index] {
        ranges(of: string, options: options).map(\.lowerBound)
    }
    func ranges<S: StringProtocol>(of string: S, options: String.CompareOptions = []) -> [Range<Index>] {
        var result: [Range<Index>] = []
        var startIndex = self.startIndex
        while startIndex < endIndex,
            let range = self[startIndex...]
                .range(of: string, options: options) {
                result.append(range)
                startIndex = 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]   // ab
    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.indices(of: "play")    // [7, 19, 31]
str.ranges(of: "play")     // [{lowerBound 7, upperBound 11}, {lowerBound 19, upperBound 23}, {lowerBound 31, upperBound 35}]

---

case insensitive sample

let query = "Play"
let ranges = str.ranges(of: query, options: .caseInsensitive)
let matches = ranges.map { str[$0] }   //
print(matches)  // ["play", "play", "play"]

---

regular expression sample

let query = "play"
let escapedQuery = NSRegularExpression.escapedPattern(for: query)
let pattern = "\\b\(escapedQuery)\\w+"  // matches any word that starts with "play" prefix

let ranges = str.ranges(of: pattern, options: .regularExpression)
let matches = ranges.map { str[$0] }

print(matches) //  ["playground", "playground", "playground"]

In swift 3, how do you advance an index?

Looks pretty clear as the first item in the Swift 3 migration guide:

The most visible change is that indexes no longer have successor(), predecessor(), advancedBy(_:), advancedBy(_:limit:), or distanceTo(_:) methods. Instead, those operations are moved to the collection, which is now responsible for incrementing and decrementing its indices.

myIndex.successor()  =>  myCollection.index(after: myIndex)
myIndex.predecessor()  =>  myCollection.index(before: myIndex)
myIndex.advance(by: …) => myCollection.index(myIndex, offsetBy: …)

So it looks like you want something like:

let greeting = "hello"
let secondCharIndex = greeting.index(after: greeting.startIndex)
let enryTheEighthGreeting = greeting.substring(from: secondCharIndex) // -> "ello"

(Note also that if you want Collection functionality—like index management—on a String, it sometimes helps to use its characters view. String methods like startIndex and index(after:) are just conveniences that forward to the characters view. That part isn't new, though... Strings stopped being Collections themselves in Swift 2 IIRC.)

There's more about the collection indexes change in SE-0065 - A New Model for Collections and Indices.

How to convert Index to type Int in Swift?

edit/update:

Xcode 11 • Swift 5.1 or later

extension StringProtocol {
    func distance(of element: Element) -> Int? { firstIndex(of: element)?.distance(in: self) }
    func distance<S: StringProtocol>(of string: S) -> Int? { range(of: string)?.lowerBound.distance(in: self) }
}

---

extension Collection {
    func distance(to index: Index) -> Int { distance(from: startIndex, to: index) }
}

---

extension String.Index {
    func distance<S: StringProtocol>(in string: S) -> Int { string.distance(to: self) }
}

---

Playground testing

let letters = "abcdefg"

let char: Character = "c"
if let distance = letters.distance(of: char) {
    print("character \(char) was found at position #\(distance)")   // "character c was found at position #2\n"
} else {
    print("character \(char) was not found")
}

---

let string = "cde"
if let distance = letters.distance(of: string) {
    print("string \(string) was found at position #\(distance)")   // "string cde was found at position #2\n"
} else {
    print("string \(string) was not found")
}

Swift: get index of start of a substring in a string

You can reach using the rangeOfString function that finds and returns the range of the first occurrence of a given string within a given range like in the following way:

var text = "Hello Victor Sigler"

if let indexOf = text.rangeOfString("Victor")?.startIndex {
   println(indexOf) //6
}

I hope this help you.

String may not be indexed with 'Int' Swift3

Try this:

    if let contact = contact
    {
        var firstLetter = !contact.firstName.isEmpty ? contact?.firstName.characters.first?.description.uppercased() : contact?.lastName.characters.first?.description.uppercased()
    }

I will be able to help you more if you tell the type/definition of "contact".

String.Substring in Swift 3 when start = end

You want to drop the last two characters.

There are multiple solutions. One is

let string = "0123"
let minutes = Int(String(string.characters.dropLast(2)))!

or

let range = string.index(string.endIndex, offsetBy: -2)..<string.endIndex
let minutes = Int(string.substring(to: range.lowerBound))!

And since you want Int anyway, why not simply

let minutes = Int(string)! / 100

Take character from a specific position to last (swift 3)

Use map(_:) with array and then simply use substring(from:) with shorthand argument.

let strArray = ["Md. Monir-Uz-Zaman Monir", "Md. Monir-Uz-Zaman Monir01", "Md. Monir-Uz-Zaman Monir876"]
let nameArray = strArray.map { $0.substring(from: $0.index($0.startIndex, offsetBy: 19)) }
print(nameArray) //["Monir", "Monir01", "Monir876"]

---

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