I want to write a function that could be used like this:
let 👩👩👧👦 = "👩👩👧👧".replacingFirstOccurrence(of: "👧", with: "👦")
Given how odd both this string and Swift's String library are, is this possible in Swift?
Based on the insights gained at Why are emoji characters like 👩👩👧👦 treated so strangely in Swift strings?, a sensible approach might be to replace Unicode scalars:
extension String {
func replacingFirstOccurrence(of target: UnicodeScalar, with replacement: UnicodeScalar) -> String {
let uc = self.unicodeScalars
guard let idx = uc.index(of: target) else { return self }
let prefix = uc[uc.startIndex..<idx]
let suffix = uc[uc.index(after: idx) ..< uc.endIndex]
return "\(prefix)\(replacement)\(suffix)"
}
}
Example:
let family1 = "👩👩👧👦"
print(family1.characters.map { Array(String($0).unicodeScalars) })
// [["\u{0001F469}", "\u{200D}"], ["\u{0001F469}", "\u{200D}"], ["\u{0001F467}", "\u{200D}"], ["\u{0001F466}"]]
let family2 = family1.replacingFirstOccurrence(of: "👧", with: "👦")
print(family2) // 👩👩👦👦
print(family2.characters.map { Array(String($0).unicodeScalars) })
// [["\u{0001F469}", "\u{200D}"], ["\u{0001F469}", "\u{200D}"], ["\u{0001F466}", "\u{200D}"], ["\u{0001F466}"]]
And here is a possible version which locates and replaces the Unicode scalars of an arbitrary string:
extension String {
func replacingFirstOccurrence(of target: String, with replacement: String) -> String {
let uc = self.unicodeScalars
let tuc = target.unicodeScalars
// Target empty or too long:
if tuc.count == 0 || tuc.count > uc.count {
return self
}
// Current search position:
var pos = uc.startIndex
// Last possible position of `tuc` within `uc`:
let end = uc.index(uc.endIndex, offsetBy: tuc.count - 1)
// Locate first Unicode scalar
while let from = uc[pos..<end].index(of: tuc.first!) {
// Compare all Unicode scalars:
let to = uc.index(from, offsetBy: tuc.count)
if !zip(uc[from..<to], tuc).contains(where: { $0 != $1 }) {
let prefix = uc[uc.startIndex..<from]
let suffix = uc[to ..< uc.endIndex]
return "\(prefix)\(replacement)\(suffix)"
}
// Next search position:
uc.formIndex(after: &pos)
}
// Target not found.
return self
}
}
Using the range(of:options:range:locale:) the solution became quite concise:
extension String {
func replaceFirstOccurrence(of searchString: String, with replacementString: String) -> String {
guard let range = self.range(of: searchString, options: .literal) else { return self }
return self.replacingCharacters(in: range, with: replacementString)
}
}
This works by first finding the range of searchString within the instance, and if a range is found the range is replaced with replacementString. Otherwise the instance just returns itself. And, since the range(of:) method returns as soon as it finds a match, the returned range is guaranteed to be the first occurrence.
"221".replaceFirstOccurrence(of: "2", with: "3") // 321
"👩👩👧👦".replaceFirstOccurrence(of: "\u{1f469}", with: "\u{1f468}") // 👨👩👧👦
*To clarify, the last test case converts woman-woman-girl-boy to man-woman-girl-boy.
I have the need to parse some unknown data which should just be a numeric value, but may contain whitespace or other non-alphanumeric characters.
Is there a new way of doing this in Swift? All I can find online seems to be the old C way of doing things.
I am looking at stringByTrimmingCharactersInSet - as I am sure my inputs will only have whitespace/special characters at the start or end of the string. Are there any built in character sets I can use for this? Or do I need to create my own?
I was hoping there would be something like stringFromCharactersInSet() which would allow me to specify only valid characters to keep
I was hoping there would be something like stringFromCharactersInSet() which would allow me to specify only valid characters to keep.
You can either use trimmingCharacters with the inverted character set to remove characters from the start or the end of the string. In Swift 3 and later:
let result = string.trimmingCharacters(in: CharacterSet(charactersIn: "0123456789.").inverted)
Or, if you want to remove non-numeric characters anywhere in the string (not just the start or end), you can filter the characters, e.g. in Swift 4.2.1:
let result = string.filter("0123456789.".contains)
Or, if you want to remove characters from a CharacterSet from anywhere in the string, use:
let result = String(string.unicodeScalars.filter(CharacterSet.whitespaces.inverted.contains))
Or, if you want to only match valid strings of a certain format (e.g. ####.##), you could use regular expression. For example:
if let range = string.range(of: #"\d+(\.\d*)?"#, options: .regularExpression) {
let result = string[range] // or `String(string[range])` if you need `String`
}
The behavior of these different approaches differ slightly so it just depends on precisely what you're trying to do. Include or exclude the decimal point if you want decimal numbers, or just integers. There are lots of ways to accomplish this.
For older, Swift 2 syntax, see previous revision of this answer.
let result = string.stringByReplacingOccurrencesOfString("[^0-9]", withString: "", options: NSStringCompareOptions.RegularExpressionSearch, range:nil).stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceCharacterSet())
Swift 3
let result = string.replacingOccurrences( of:"[^0-9]", with: "", options: .regularExpression)
You can upvote this answer.
I prefer this solution, because I like extensions, and it seems a bit cleaner to me. Solution reproduced here:
extension String {
var digits: String {
return components(separatedBy: CharacterSet.decimalDigits.inverted)
.joined()
}
}
You can filter the UnicodeScalarView of the string using the pattern matching operator for ranges, pass a UnicodeScalar ClosedRange from 0 to 9 and initialise a new String with the resulting UnicodeScalarView:
extension String {
private static var digits = UnicodeScalar("0")..."9"
var digits: String {
return String(unicodeScalars.filter(String.digits.contains))
}
}
"abc12345".digits // "12345"
edit/update:
Swift 4.2
extension RangeReplaceableCollection where Self: StringProtocol {
var digits: Self {
return filter(("0"..."9").contains)
}
}
or as a mutating method
extension RangeReplaceableCollection where Self: StringProtocol {
mutating func removeAllNonNumeric() {
removeAll { !("0"..."9" ~= $0) }
}
}
Swift 5.2 • Xcode 11.4 or later
In Swift5 we can use a new Character property called isWholeNumber:
extension RangeReplaceableCollection where Self: StringProtocol {
var digits: Self { filter(\.isWholeNumber) }
}
extension RangeReplaceableCollection where Self: StringProtocol {
mutating func removeAllNonNumeric() {
removeAll { !$0.isWholeNumber }
}
}
To allow a period as well we can extend Character and create a computed property:
extension Character {
var isDecimalOrPeriod: Bool { "0"..."9" ~= self || self == "." }
}
extension RangeReplaceableCollection where Self: StringProtocol {
var digitsAndPeriods: Self { filter(\.isDecimalOrPeriod) }
}
Playground testing:
"abc12345".digits // "12345"
var str = "123abc0"
str.removeAllNonNumeric()
print(str) //"1230"
"Testing0123456789.".digitsAndPeriods // "0123456789."
Swift 4
I found a decent way to get only alpha numeric characters set from a string.
For instance:-
func getAlphaNumericValue() {
var yourString = "123456789!##$%^&*()AnyThingYouWant"
let unsafeChars = CharacterSet.alphanumerics.inverted // Remove the .inverted to get the opposite result.
let cleanChars = yourString.components(separatedBy: unsafeChars).joined(separator: "")
print(cleanChars) // 123456789AnyThingYouWant
}
A solution using the filter function and rangeOfCharacterFromSet
let string = "sld [f]34é7*˜µ"
let alphaNumericCharacterSet = NSCharacterSet.alphanumericCharacterSet()
let filteredCharacters = string.characters.filter {
return String($0).rangeOfCharacterFromSet(alphaNumericCharacterSet) != nil
}
let filteredString = String(filteredCharacters) // -> sldf34é7µ
To filter for only numeric characters use
let string = "sld [f]34é7*˜µ"
let numericSet = "0123456789"
let filteredCharacters = string.characters.filter {
return numericSet.containsString(String($0))
}
let filteredString = String(filteredCharacters) // -> 347
or
let numericSet : [Character] = ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]
let filteredCharacters = string.characters.filter {
return numericSet.contains($0)
}
let filteredString = String(filteredCharacters) // -> 347
Swift 4
But without extensions or componentsSeparatedByCharactersInSet which doesn't read as well.
let allowedCharSet = NSCharacterSet.letters.union(.whitespaces)
let filteredText = String(sourceText.unicodeScalars.filter(allowedCharSet.contains))
let string = "+1*(234) fds567#-8/90-"
let onlyNumbers = string.components(separatedBy: CharacterSet.decimalDigits.inverted).joined()
print(onlyNumbers) // "1234567890"
or
extension String {
func removeNonNumeric() -> String {
return self.components(separatedBy: CharacterSet.decimalDigits.inverted).joined()
}
}
let onlyNumbers = "+1*(234) fds567#-8/90-".removeNonNumeric()
print(onlyNumbers)// "1234567890"
Swift 3, filters all except numbers
let myString = "dasdf3453453fsdf23455sf.2234"
let result = String(myString.characters.filter { String($0).rangeOfCharacter(from: CharacterSet(charactersIn: "0123456789")) != nil })
print(result)
Swift 4.2
let numericString = string.filter { (char) -> Bool in
return char.isNumber
}
You can do something like this...
let string = "[,myString1. \"" // string : [,myString1. "
let characterSet = NSCharacterSet(charactersInString: "[,. \"")
let finalString = (string.componentsSeparatedByCharactersInSet(characterSet) as NSArray).componentsJoinedByString("")
print(finalString)
//finalString will be "myString1"
The issue with Rob's first solution is stringByTrimmingCharactersInSet only filters the ends of the string rather than throughout, as stated in Apple's documentation:
Returns a new string made by removing from both ends of the receiver characters contained in a given character set.
Instead use componentsSeparatedByCharactersInSet to first isolate all non-occurrences of the character set into arrays and subsequently join them with an empty string separator:
"$$1234%^56()78*9££".componentsSeparatedByCharactersInSet(NSCharacterSet(charactersInString: "0123456789").invertedSet)).joinWithSeparator("")
Which returns 123456789
Swift 3
extension String {
var keepNumericsOnly: String {
return self.components(separatedBy: CharacterSet(charactersIn: "0123456789").inverted).joined(separator: "")
}
}
Swift 4.0 version
extension String {
var numbers: String {
return String(describing: filter { String($0).rangeOfCharacter(from: CharacterSet(charactersIn: "0123456789")) != nil })
}
}
Swift 4
String.swift
import Foundation
extension String {
func removeCharacters(from forbiddenChars: CharacterSet) -> String {
let passed = self.unicodeScalars.filter { !forbiddenChars.contains($0) }
return String(String.UnicodeScalarView(passed))
}
func removeCharacters(from: String) -> String {
return removeCharacters(from: CharacterSet(charactersIn: from))
}
}
ViewController.swift
let character = "1Vi234s56a78l9"
let alphaNumericSet = character.removeCharacters(from: CharacterSet.decimalDigits.inverted)
print(alphaNumericSet) // will print: 123456789
let alphaNumericCharacterSet = character.removeCharacters(from: "0123456789")
print("no digits",alphaNumericCharacterSet) // will print: Vishal
Swift 4.2
let digitChars = yourString.components(separatedBy:
CharacterSet.decimalDigits.inverted).joined(separator: "")
Swift 3 Version
extension String
{
func trimmingCharactersNot(in charSet: CharacterSet) -> String
{
var s:String = ""
for unicodeScalar in self.unicodeScalars
{
if charSet.contains(unicodeScalar)
{
s.append(String(unicodeScalar))
}
}
return s
}
}
I want to convert the index of a letter contained within a string to an integer value. Attempted to read the header files but I cannot find the type for Index, although it appears to conform to protocol ForwardIndexType with methods (e.g. distanceTo).
var letters = "abcdefg"
let index = letters.characters.indexOf("c")!
// ERROR: Cannot invoke initializer for type 'Int' with an argument list of type '(String.CharacterView.Index)'
let intValue = Int(index) // I want the integer value of the index (e.g. 2)
Any help is appreciated.
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")
}
Works for Xcode 13 and Swift 5
let myString = "Hello World"
if let i = myString.firstIndex(of: "o") {
let index: Int = myString.distance(from: myString.startIndex, to: i)
print(index) // Prints 4
}
The function func distance(from start: String.Index, to end: String.Index) -> String.IndexDistance returns an IndexDistance which is just a typealias for Int
Swift 4
var str = "abcdefg"
let index = str.index(of: "c")?.encodedOffset // Result: 2
Note: If String contains same multiple characters, it will just get the nearest one from left
var str = "abcdefgc"
let index = str.index(of: "c")?.encodedOffset // Result: 2
encodedOffset has deprecated from Swift 4.2.
Deprecation message:
encodedOffset has been deprecated as most common usage is incorrect. Use utf16Offset(in:) to achieve the same behavior.
So we can use utf16Offset(in:) like this:
var str = "abcdefgc"
let index = str.index(of: "c")?.utf16Offset(in: str) // Result: 2
When searching for index like this
⛔️ guard let index = (positions.firstIndex { position <= $0 }) else {
it is treated as Array.Index. You have to give compiler a clue you want an integer
✅ guard let index: Int = (positions.firstIndex { position <= $0 }) else {
Swift 5
You can do convert to array of characters and then use advanced(by:) to convert to integer.
let myString = "Hello World"
if let i = Array(myString).firstIndex(of: "o") {
let index: Int = i.advanced(by: 0)
print(index) // Prints 4
}
To perform string operation based on index , you can not do it with traditional index numeric approach. because swift.index is retrieved by the indices function and it is not in the Int type. Even though String is an array of characters, still we can't read element by index.
This is frustrating.
So ,to create new substring of every even character of string , check below code.
let mystr = "abcdefghijklmnopqrstuvwxyz"
let mystrArray = Array(mystr)
let strLength = mystrArray.count
var resultStrArray : [Character] = []
var i = 0
while i < strLength {
if i % 2 == 0 {
resultStrArray.append(mystrArray[i])
}
i += 1
}
let resultString = String(resultStrArray)
print(resultString)
Output : acegikmoqsuwy
Thanks In advance
Here is an extension that will let you access the bounds of a substring as Ints instead of String.Index values:
import Foundation
/// This extension is available at
/// https://gist.github.com/zackdotcomputer/9d83f4d48af7127cd0bea427b4d6d61b
extension StringProtocol {
/// Access the range of the search string as integer indices
/// in the rendered string.
/// - NOTE: This is "unsafe" because it may not return what you expect if
/// your string contains single symbols formed from multiple scalars.
/// - Returns: A `CountableRange<Int>` that will align with the Swift String.Index
/// from the result of the standard function range(of:).
func countableRange<SearchType: StringProtocol>(
of search: SearchType,
options: String.CompareOptions = [],
range: Range<String.Index>? = nil,
locale: Locale? = nil
) -> CountableRange<Int>? {
guard let trueRange = self.range(of: search, options: options, range: range, locale: locale) else {
return nil
}
let intStart = self.distance(from: startIndex, to: trueRange.lowerBound)
let intEnd = self.distance(from: trueRange.lowerBound, to: trueRange.upperBound) + intStart
return Range(uncheckedBounds: (lower: intStart, upper: intEnd))
}
}
Just be aware that this can lead to weirdness, which is why Apple has chosen to make it hard. (Though that's a debatable design decision - hiding a dangerous thing by just making it hard...)
You can read more in the String documentation from Apple, but the tldr is that it stems from the fact that these "indices" are actually implementation-specific. They represent the indices into the string after it has been rendered by the OS, and so can shift from OS-to-OS depending on what version of the Unicode spec is being used. This means that accessing values by index is no longer a constant-time operation, because the UTF spec has to be run over the data to determine the right place in the string. These indices will also not line up with the values generated by NSString, if you bridge to it, or with the indices into the underlying UTF scalars. Caveat developer.
In case you got an "index is out of bounds" error. You may try this approach. Working in Swift 5
extension String{
func countIndex(_ char:Character) -> Int{
var count = 0
var temp = self
for c in self{
if c == char {
//temp.remove(at: temp.index(temp.startIndex,offsetBy:count))
//temp.insert(".", at: temp.index(temp.startIndex,offsetBy: count))
return count
}
count += 1
}
return -1
}
}
I have not yet been able to figure out how to get a substring of a String in Swift:
var str = “Hello, playground”
func test(str: String) -> String {
return str.substringWithRange( /* What goes here? */ )
}
test (str)
I'm not able to create a Range in Swift. Autocomplete in the Playground isn’t super helpful - this is what it suggests:
return str.substringWithRange(aRange: Range<String.Index>)
I haven't found anything in the Swift Standard Reference Library that helps. Here was another wild guess:
return str.substringWithRange(Range(0, 1))
And this:
let r:Range<String.Index> = Range<String.Index>(start: 0, end: 2)
return str.substringWithRange(r)
I've seen other answers (Finding index of character in Swift String) that seem to suggest that since String is a bridge type for NSString, the "old" methods should work, but it's not clear how - e.g., this doesn't work either (doesn't appear to be valid syntax):
let x = str.substringWithRange(NSMakeRange(0, 3))
Thoughts?
You can use the substringWithRange method. It takes a start and end String.Index.
var str = "Hello, playground"
str.substringWithRange(Range<String.Index>(start: str.startIndex, end: str.endIndex)) //"Hello, playground"
To change the start and end index, use advancedBy(n).
var str = "Hello, playground"
str.substringWithRange(Range<String.Index>(start: str.startIndex.advancedBy(2), end: str.endIndex.advancedBy(-1))) //"llo, playgroun"
You can also still use the NSString method with NSRange, but you have to make sure you are using an NSString like this:
let myNSString = str as NSString
myNSString.substringWithRange(NSRange(location: 0, length: 3))
Note: as JanX2 mentioned, this second method is not safe with unicode strings.
Swift 2
Simple
let str = "My String"
let subStr = str[str.startIndex.advancedBy(3)...str.startIndex.advancedBy(7)]
//"Strin"
Swift 3
let startIndex = str.index(str.startIndex, offsetBy: 3)
let endIndex = str.index(str.startIndex, offsetBy: 7)
str[startIndex...endIndex] // "Strin"
str.substring(to: startIndex) // "My "
str.substring(from: startIndex) // "String"
Swift 4
substring(to:) and substring(from:) are deprecated in Swift 4.
String(str[..<startIndex]) // "My "
String(str[startIndex...]) // "String"
String(str[startIndex...endIndex]) // "Strin"
At the time I'm writing, no extension is perfectly Swift 4.2 compatible, so here is one that covers all the needs I could think of:
extension String {
func substring(from: Int?, to: Int?) -> String {
if let start = from {
guard start < self.count else {
return ""
}
}
if let end = to {
guard end >= 0 else {
return ""
}
}
if let start = from, let end = to {
guard end - start >= 0 else {
return ""
}
}
let startIndex: String.Index
if let start = from, start >= 0 {
startIndex = self.index(self.startIndex, offsetBy: start)
} else {
startIndex = self.startIndex
}
let endIndex: String.Index
if let end = to, end >= 0, end < self.count {
endIndex = self.index(self.startIndex, offsetBy: end + 1)
} else {
endIndex = self.endIndex
}
return String(self[startIndex ..< endIndex])
}
func substring(from: Int) -> String {
return self.substring(from: from, to: nil)
}
func substring(to: Int) -> String {
return self.substring(from: nil, to: to)
}
func substring(from: Int?, length: Int) -> String {
guard length > 0 else {
return ""
}
let end: Int
if let start = from, start > 0 {
end = start + length - 1
} else {
end = length - 1
}
return self.substring(from: from, to: end)
}
func substring(length: Int, to: Int?) -> String {
guard let end = to, end > 0, length > 0 else {
return ""
}
let start: Int
if let end = to, end - length > 0 {
start = end - length + 1
} else {
start = 0
}
return self.substring(from: start, to: to)
}
}
And then, you can use:
let string = "Hello,World!"
string.substring(from: 1, to: 7)gets you: ello,Wo
string.substring(to: 7)gets you: Hello,Wo
string.substring(from: 3)gets you: lo,World!
string.substring(from: 1, length: 4)gets you: ello
string.substring(length: 4, to: 7)gets you: o,Wo
Updated substring(from: Int?, length: Int) to support starting from zero.
NOTE: #airspeedswift makes some very insightful points on the trade-offs of this approach, particularly the hidden performance impacts. Strings are not simple beasts, and getting to a particular index may take O(n) time, which means a loop that uses a subscript can be O(n^2). You have been warned.
You just need to add a new subscript function that takes a range and uses advancedBy() to walk to where you want:
import Foundation
extension String {
subscript (r: Range<Int>) -> String {
get {
let startIndex = self.startIndex.advancedBy(r.startIndex)
let endIndex = startIndex.advancedBy(r.endIndex - r.startIndex)
return self[Range(start: startIndex, end: endIndex)]
}
}
}
var s = "Hello, playground"
println(s[0...5]) // ==> "Hello,"
println(s[0..<5]) // ==> "Hello"
(This should definitely be part of the language. Please dupe rdar://17158813)
For fun, you can also add a + operator onto the indexes:
func +<T: ForwardIndex>(var index: T, var count: Int) -> T {
for (; count > 0; --count) {
index = index.succ()
}
return index
}
s.substringWithRange(s.startIndex+2 .. s.startIndex+5)
(I don't know yet if this one should be part of the language or not.)
SWIFT 2.0
simple:
let myString = "full text container"
let substring = myString[myString.startIndex..<myString.startIndex.advancedBy(3)] // prints: ful
SWIFT 3.0
let substring = myString[myString.startIndex..<myString.index(myString.startIndex, offsetBy: 3)] // prints: ful
SWIFT 4.0
Substring operations return an instance of the Substring type, instead of String.
let substring = myString[myString.startIndex..<myString.index(myString.startIndex, offsetBy: 3)] // prints: ful
// Convert the result to a String for long-term storage.
let newString = String(substring)
It is much more simple than any of the answers here, once you find the right syntax.
I want to take away the [ and ]
let myString = "[ABCDEFGHI]"
let startIndex = advance(myString.startIndex, 1) //advance as much as you like
let endIndex = advance(myString.endIndex, -1)
let range = startIndex..<endIndex
let myNewString = myString.substringWithRange( range )
result will be "ABCDEFGHI"
the startIndex and endIndex could also be used in
let mySubString = myString.substringFromIndex(startIndex)
and so on!
PS: As indicated in the remarks, there are some syntax changes in swift 2 which comes with xcode 7 and iOS9!
Please look at this page
For example to find the first name (up to the first space) in my full name:
let name = "Joris Kluivers"
let start = name.startIndex
let end = find(name, " ")
if end {
let firstName = name[start..end!]
} else {
// no space found
}
start and end are of type String.Index here and are used to create a Range<String.Index> and used in the subscript accessor (if a space is found at all in the original string).
It's hard to create a String.Index directly from an integer position as used in the opening post. This is because in my name each character would be of equal size in bytes. But characters using special accents in other languages could have used several more bytes (depending on the encoding used). So what byte should the integer refer to?
It's possible to create a new String.Index from an existing one using the methods succ and pred which will make sure the correct number of bytes are skipped to get to the next code point in the encoding. However in this case it's easier to search for the index of the first space in the string to find the end index.
Since String is a bridge type for NSString, the "old" methods should work, but it's not clear how - e.g., this doesn't work either (doesn't appear to be valid syntax):
let x = str.substringWithRange(NSMakeRange(0, 3))
To me, that is the really interesting part of your question. String is bridged to NSString, so most NSString methods do work directly on a String. You can use them freely and without thinking. So, for example, this works just as you expect:
// delete all spaces from Swift String stateName
stateName = stateName.stringByReplacingOccurrencesOfString(" ", withString:"")
But, as so often happens, "I got my mojo workin' but it just don't work on you." You just happened to pick one of the rare cases where a parallel identically named Swift method exists, and in a case like that, the Swift method overshadows the Objective-C method. Thus, when you say str.substringWithRange, Swift thinks you mean the Swift method rather than the NSString method — and then you are hosed, because the Swift method expects a Range<String.Index>, and you don't know how to make one of those.
The easy way out is to stop Swift from overshadowing like this, by casting explicitly:
let x = (str as NSString).substringWithRange(NSMakeRange(0, 3))
Note that no significant extra work is involved here. "Cast" does not mean "convert"; the String is effectively an NSString. We are just telling Swift how to look at this variable for purposes of this one line of code.
The really weird part of this whole thing is that the Swift method, which causes all this trouble, is undocumented. I have no idea where it is defined; it is not in the NSString header and it's not in the Swift header either.
The short answer is that this is really hard in Swift right now. My hunch is that there is still a bunch of work for Apple to do on convenience methods for things like this.
String.substringWithRange() is expecting a Range<String.Index> parameter, and as far as I can tell there isn't a generator method for the String.Index type. You can get String.Index values back from aString.startIndex and aString.endIndex and call .succ() or .pred() on them, but that's madness.
How about an extension on the String class that takes good old Ints?
extension String {
subscript (r: Range<Int>) -> String {
get {
let subStart = advance(self.startIndex, r.startIndex, self.endIndex)
let subEnd = advance(subStart, r.endIndex - r.startIndex, self.endIndex)
return self.substringWithRange(Range(start: subStart, end: subEnd))
}
}
func substring(from: Int) -> String {
let end = countElements(self)
return self[from..<end]
}
func substring(from: Int, length: Int) -> String {
let end = from + length
return self[from..<end]
}
}
let mobyDick = "Call me Ishmael."
println(mobyDick[8...14]) // Ishmael
let dogString = "This 🐶's name is Patch."
println(dogString[5..<6]) // 🐶
println(dogString[5...5]) // 🐶
println(dogString.substring(5)) // 🐶's name is Patch.
println(dogString.substring(5, length: 1)) // 🐶
Update: Swift beta 4 resolves the issues below!
As it stands [in beta 3 and earlier], even Swift-native strings have some issues with handling Unicode characters. The dog icon above worked, but the following doesn't:
let harderString = "1:1️⃣"
for character in harderString {
println(character)
}
Output:
1
:
1
️
⃣
In new Xcode 7.0 use
//: Playground - noun: a place where people can play
import UIKit
var name = "How do you use String.substringWithRange?"
let range = name.startIndex.advancedBy(0)..<name.startIndex.advancedBy(10)
name.substringWithRange(range)
//OUT:
You can use this extensions to improve substringWithRange
Swift 2.3
extension String
{
func substringWithRange(start: Int, end: Int) -> String
{
if (start < 0 || start > self.characters.count)
{
print("start index \(start) out of bounds")
return ""
}
else if end < 0 || end > self.characters.count
{
print("end index \(end) out of bounds")
return ""
}
let range = Range(start: self.startIndex.advancedBy(start), end: self.startIndex.advancedBy(end))
return self.substringWithRange(range)
}
func substringWithRange(start: Int, location: Int) -> String
{
if (start < 0 || start > self.characters.count)
{
print("start index \(start) out of bounds")
return ""
}
else if location < 0 || start + location > self.characters.count
{
print("end index \(start + location) out of bounds")
return ""
}
let range = Range(start: self.startIndex.advancedBy(start), end: self.startIndex.advancedBy(start + location))
return self.substringWithRange(range)
}
}
Swift 3
extension String
{
func substring(start: Int, end: Int) -> String
{
if (start < 0 || start > self.characters.count)
{
print("start index \(start) out of bounds")
return ""
}
else if end < 0 || end > self.characters.count
{
print("end index \(end) out of bounds")
return ""
}
let startIndex = self.characters.index(self.startIndex, offsetBy: start)
let endIndex = self.characters.index(self.startIndex, offsetBy: end)
let range = startIndex..<endIndex
return self.substring(with: range)
}
func substring(start: Int, location: Int) -> String
{
if (start < 0 || start > self.characters.count)
{
print("start index \(start) out of bounds")
return ""
}
else if location < 0 || start + location > self.characters.count
{
print("end index \(start + location) out of bounds")
return ""
}
let startIndex = self.characters.index(self.startIndex, offsetBy: start)
let endIndex = self.characters.index(self.startIndex, offsetBy: start + location)
let range = startIndex..<endIndex
return self.substring(with: range)
}
}
Usage:
let str = "Hello, playground"
let substring1 = str.substringWithRange(0, end: 5) //Hello
let substring2 = str.substringWithRange(7, location: 10) //playground
Sample Code for how to get substring in Swift 2.0
(i) Substring from starting index
Input:-
var str = "Swift is very powerful language!"
print(str)
str = str.substringToIndex(str.startIndex.advancedBy(5))
print(str)
Output:-
Swift is very powerful language!
Swift
(ii) Substring from particular index
Input:-
var str = "Swift is very powerful language!"
print(str)
str = str.substringFromIndex(str.startIndex.advancedBy(6)).substringToIndex(str.startIndex.advancedBy(2))
print(str)
Output:-
Swift is very powerful language!
is
I hope it will help you!
Easy solution with little code.
Make an extension that includes basic subStringing that nearly all other languages have:
extension String {
func subString(start: Int, end: Int) -> String {
let startIndex = self.index(self.startIndex, offsetBy: start)
let endIndex = self.index(startIndex, offsetBy: end)
let finalString = self.substring(from: startIndex)
return finalString.substring(to: endIndex)
}
}
Simply call this with
someString.subString(start: 0, end: 6)
This works in my playground :)
String(seq: Array(str)[2...4])
Updated for Xcode 7. Adds String extension:
Use:
var chuck: String = "Hello Chuck Norris"
chuck[6...11] // => Chuck
Implementation:
extension String {
/**
Subscript to allow for quick String substrings ["Hello"][0...1] = "He"
*/
subscript (r: Range<Int>) -> String {
get {
let start = self.startIndex.advancedBy(r.startIndex)
let end = self.startIndex.advancedBy(r.endIndex - 1)
return self.substringWithRange(start..<end)
}
}
}
try this in playground
var str:String = "Hello, playground"
let range = Range(start:advance(str.startIndex,1), end: advance(str.startIndex,8))
it will give you "ello, p"
However where this gets really interesting is that if you make the last index bigger than the string in playground it will show any strings that you defined after str :o
Range() appears to be a generic function so that it needs to know the type it is dealing with.
You also have to give it the actual string your interested in playgrounds as it seems to hold all stings in a sequence one after another with their variable name afterwards.
So
var str:String = "Hello, playground"
var str2:String = "I'm the next string"
let range = Range(start:advance(str.startIndex,1), end: advance(str.startIndex,49))
gives "ello, playground�str���I'm the next string�str2�"
works even if str2 is defined with a let
:)
Rob Napier had already given a awesome answer using subscript. But i felt one drawback in that as there is no check for out of bound conditions. This can tend to crash. So i modified the extension and here it is
extension String {
subscript (r: Range<Int>) -> String? { //Optional String as return value
get {
let stringCount = self.characters.count as Int
//Check for out of boundary condition
if (stringCount < r.endIndex) || (stringCount < r.startIndex){
return nil
}
let startIndex = self.startIndex.advancedBy(r.startIndex)
let endIndex = self.startIndex.advancedBy(r.endIndex - r.startIndex)
return self[Range(start: startIndex, end: endIndex)]
}
}
}
Output below
var str2 = "Hello, World"
var str3 = str2[0...5]
//Hello,
var str4 = str2[0..<5]
//Hello
var str5 = str2[0..<15]
//nil
So i suggest always to check for the if let
if let string = str[0...5]
{
//Manipulate your string safely
}
In Swift3
For ex: a variable "Duke James Thomas", we need to get "James".
let name = "Duke James Thomas"
let range: Range<String.Index> = name.range(of:"James")!
let lastrange: Range<String.Index> = img.range(of:"Thomas")!
var middlename = name[range.lowerBound..<lstrange.lowerBound]
print (middlename)
Taking a page from Rob Napier, I developed these Common String Extensions, two of which are:
subscript (r: Range<Int>) -> String
{
get {
let startIndex = advance(self.startIndex, r.startIndex)
let endIndex = advance(self.startIndex, r.endIndex - 1)
return self[Range(start: startIndex, end: endIndex)]
}
}
func subString(startIndex: Int, length: Int) -> String
{
var start = advance(self.startIndex, startIndex)
var end = advance(self.startIndex, startIndex + length)
return self.substringWithRange(Range<String.Index>(start: start, end: end))
}
Usage:
"Awesome"[3...7] //"some"
"Awesome".subString(3, length: 4) //"some"
This is how you get a range from a string:
var str = "Hello, playground"
let startIndex = advance(str.startIndex, 1)
let endIndex = advance(startIndex, 8)
let range = startIndex..<endIndex
let substr = str[range] //"ello, pl"
The key point is that you are passing a range of values of type String.Index (this is what advance returns) instead of integers.
The reason why this is necessary, is that strings in Swift don't have random access (because of variable length of Unicode characters basically). You also can't do str[1]. String.Index is designed to work with their internal structure.
You can create an extension with a subscript though, that does this for you, so you can just pass a range of integers (see e.g. Rob Napier's answer).
I tried to come up with something Pythonic.
All the subscripts here are great, but the times I really need something simple is usually when I want to count from back, e.g. string.endIndex.advancedBy(-1)
It supports nil values, for both start and end index, where nil would mean index at 0 for start, string.characters.count for end.
extension String {
var subString: (Int?) -> (Int?) -> String {
return { (start) in
{ (end) in
let startIndex = start ?? 0 < 0 ? self.endIndex.advancedBy(start!) : self.startIndex.advancedBy(start ?? 0)
let endIndex = end ?? self.characters.count < 0 ? self.endIndex.advancedBy(end!) : self.startIndex.advancedBy(end ?? self.characters.count)
return startIndex > endIndex ? "" : self.substringWithRange(startIndex ..< endIndex)
}
}
}
}
let dog = "Dog‼🐶"
print(dog.subString(nil)(-1)) // Dog!!
EDIT
A more elegant solution:
public extension String {
struct Substring {
var start: Int?
var string: String
public subscript(end: Int?) -> String {
let startIndex = start ?? 0 < 0 ? string.endIndex.advancedBy(start!) : string.startIndex.advancedBy(start ?? 0)
let endIndex = end ?? string.characters.count < 0 ? string.endIndex.advancedBy(end!) : string.startIndex.advancedBy(end ?? string.characters.count)
return startIndex > endIndex ? "" : string.substringWithRange(startIndex ..< endIndex)
}
}
public subscript(start: Int?) -> Substring {
return Substring(start: start, string: self)
}
}
let dog = "Dog‼🐶"
print(dog[nil][-1]) // Dog!!
First create the range, then the substring. You can use fromIndex..<toIndex syntax like so:
let range = fullString.startIndex..<fullString.startIndex.advancedBy(15) // 15 first characters of the string
let substring = fullString.substringWithRange(range)
here is a example to get video-Id only .i.e (6oL687G0Iso) from the whole URL in swift
let str = "https://www.youtube.com/watch?v=6oL687G0Iso&list=PLKmzL8Ib1gsT-5LN3V2h2H14wyBZTyvVL&index=2"
var arrSaprate = str.componentsSeparatedByString("v=")
let start = arrSaprate[1]
let rangeOfID = Range(start: start.startIndex,end:start.startIndex.advancedBy(11))
let substring = start[rangeOfID]
print(substring)
let startIndex = text.startIndex
var range = startIndex.advancedBy(1) ..< text.endIndex.advancedBy(-4)
let substring = text.substringWithRange(range)
Full sample you can see here
http://www.learnswiftonline.com/reference-guides/string-reference-guide-for-swift/
shows that this works well:
var str = "abcd"
str = str.substringToIndex(1)
Well, I had the same issue and solved with the "bridgeToObjectiveC()" function:
var helloworld = "Hello World!"
var world = helloworld.bridgeToObjectiveC().substringWithRange(NSMakeRange(6,6))
println("\(world)") // should print World!
Please note that in the example, substringWithRange in conjunction with NSMakeRange take the part of the string starting at index 6 (character "W") and finishing at index 6 + 6 positions ahead (character "!")
Cheers.
You can use any of the substring methods in a Swift String extension I wrote https://bit.ly/JString.
var string = "hello"
var sub = string.substringFrom(3) // or string[3...5]
println(sub)// "lo"
If you have an NSRange, bridging to NSString works seamlessly. For example, I was doing some work with UITextFieldDelegate and I quickly wanted to compute the new string value when it asked if it should replace the range.
func textField(textField: UITextField, shouldChangeCharactersInRange range: NSRange, replacementString string: String) -> Bool {
let newString = (textField.text as NSString).stringByReplacingCharactersInRange(range, withString: string)
println("Got new string: ", newString)
}
Simple extension for String:
extension String {
func substringToIndex(index: Int) -> String {
return self[startIndex...startIndex.advancedBy(min(index, characters.count - 1))]
}
}
If you don't care about performance... this is probably the most concise solution in Swift 4
extension String {
subscript(range: CountableClosedRange<Int>) -> String {
return enumerated().filter{$0.offset >= range.first! && $0.offset < range.last!}
.reduce(""){$0 + String($1.element)}
}
}
It enables you to do something like this:
let myStr = "abcd"
myStr[0..<2] // produces "ab"