This question already has answers here:
Swift find all occurrences of a substring
(7 answers)
Closed 5 years ago.
Swift 4 apparently has introduced a lot of new changes to String. I'm wondering if there is now a built-in method for finding all instances of a substring within a String.
Here's the kind of thing I'm looking for:
let searchSentence = "hello world, hello"
let wordToMatch = "hello"
let matchingIndexArray = searchSentence.indices(of: "wordToMatch")
'matchingIndexArray' would then be [0, 13]
import Foundation
let searchSentence = "hello world, hello"
var searchRange = searchSentence.startIndex..<searchSentence.endIndex
var ranges: [Range<String.Index>] = []
let searchTerm = "hello"
while let range = searchSentence.range(of: searchTerm, range: searchRange) {
ranges.append(range)
searchRange = range.upperBound..<searchRange.upperBound
}
print(ranges.map { "(\(searchSentence.distance(from: searchSentence.startIndex, to: $0.lowerBound)), \(searchSentence.distance(from: searchSentence.startIndex, to: $0.upperBound)))" })
outputs:
["(0, 5)", "(13, 18)"]
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Hello there fellow Swift devs!
I am a junior dev, and I'm trying to figure out a best way to tokenize / parse Swift String as an exercise.
What I have is a string which looks like this:
let string = "This is a {B}string{/B} and this is a substring."
What I would like to do is, tokenize the string, and change the "strings / tokens" inside the tags you see.
I can see using NSRegularExpression and it's matches, but it feels too generic. I would like to have only say 2 of these tags, that change the text. What would be the best approach in Swift 5.2^?
if let regex = try? NSRegularExpression(pattern: "\\{[a-z0-9]+\}", options: .caseInsensitive) {
let string = self as NSString
return regex.matches(in: self, options: [], range: NSRange(location: 0, length: string.length)).map {
// now $0 is the result? but it won't work for enclosing the tags :/
}
}
If the option of using html tags instead of {B}{/B} is acceptable, then you can use the StringEx library that I wrote for this purpose.
You can select a substring inside the html tag and replace it with another string like this:
let string = "This is a <b>string</b> and this is a substring."
let ex = string.ex
ex[.tag("b")].replace(with: "some value")
print(ex.rawString) // This is a <b>some value</b> and this is a substring.
print(ex.string) // This is a some value and this is a substring.
if necessary, you can also style the selected substrings and get NSAttributedString:
ex[.tag("b")].style([
.font(.boldSystemFont(ofSize: 16)),
.color(.black)
])
myLabel.attributedText = ex.attributedString
Not sure if you have solved it with NLTokenizer or not, but you can certainly solve it with Regx here is how (I have implemented it as generic, in future if you have to handle different kinds of tags and substite different string for them small tweak to the logic should do the job )
override func viewDidLoad() {
super.viewDidLoad()
let regexStr = "(\\{B\\}(\\s*\\w+\\s*)*\\{\\/B\\})"
let regex = try! NSRegularExpression(pattern: regexStr)
var string = "Sandeep {B}Bhandaari{/B} is here{B}Sandeep{/B}"
var foundRanges = [NSRange]()
regex.enumerateMatches(in: string, options: [], range: NSMakeRange(0, string.count)) { (match, flag, stop) in
if let matchRange = match?.range(at: 1) {
foundRanges.append(matchRange)
}
}
let substituteString = "abcd"
var replacedString = string as NSString
let foundRangesCount = foundRanges.count
var currentRange = 0
while foundRangesCount > currentRange {
let range = foundRanges[currentRange]
replacedString = replacedString.replacingCharacters(in: range, with: substituteString) as NSString
reEvaluateAllRanges(ranges: &foundRanges, byOffset: range.length - substituteString.count)
currentRange += 1
}
debugPrint(replacedString)
}
func reEvaluateAllRanges(ranges: inout [NSRange], byOffset: Int) {
var newFoundRange = [NSRange]()
for range in ranges {
newFoundRange.append(NSMakeRange(range.location - byOffset, range.length))
}
ranges = newFoundRange
}
Input: "Sandeep {B}Bhandaari{/B} is here"
Output: Sandeep abcd is here
Input: "Sandeep {B}Bhandaari{/B} is here{B}Sandeep{/B}"
Output: Sandeep abcd is hereabcd
Look at the edge case handling Longer strings replaced by smaller substitute strings and vice versa also detection of string enclosed in tag with / without space
EDIT 1:
Regx (\\{B\\}(\\s*\\w+\\s*)*\\{\\/B\\}) should be self explanatory, incase you need help with understanding it use cheat sheet
regex.enumerateMatches(in: string, options: [], range: NSMakeRange(0, string.count)) { (match, flag, stop) in
if let matchRange = match?.range(at: 1) {
foundRanges.append(matchRange)
}
}
I could have modified substring here itself, but if you have more than one match and if you mutate string evaluated ranges will be corrupted hence am saving all found ranges into an array and apply replace on each one of them later
let substituteString = "abcd"
var replacedString = string as NSString
let foundRangesCount = foundRanges.count
var currentRange = 0
while foundRangesCount > currentRange {
let range = foundRanges[currentRange]
replacedString = replacedString.replacingCharacters(in: range, with: substituteString) as NSString
reEvaluateAllRanges(ranges: &foundRanges, byOffset: range.length - substituteString.count)
currentRange += 1
}
Here am iterating through all found match ranges and replace character in range with substitute string, you can always have a switch / if else ladder inside while loop to look for different types of tags and pass different substitute strings for each tags
func reEvaluateAllRanges(ranges: inout [NSRange], byOffset: Int) {
var newFoundRange = [NSRange]()
for range in ranges {
newFoundRange.append(NSMakeRange(range.location - byOffset, range.length))
}
ranges = newFoundRange
}
This function modifies all the ranges in array using the offset, remember you need to only modify range's location, length remains same
One bit of optimisation you can do is probably get rid of ranges from array for which you have already applied substitute strings
This question already has an answer here:
Understanding the removeRange(_:) documentation
(1 answer)
Closed 2 years ago.
how can i remove character from string with rang. for example «banana» i want to remove only a from index (1..<3), i don’t want to remove the first and last character if they where «a»
i want from banana to bnna only removed the two midle.
the only thing i can do now is to remove the all “a”.
var charr = "a"
var somfruit = "banana"
var newString = ""
for i in somfruit{
if charr.contains(i) {
continue
}
newString.append(i)
}
print(newString)
In SWIFT 5 try:
var charr = "a"
var somfruit = "banana"
var newString = ""
let lower = somfruit.firstIndex(of: charr) + 1
let upper = somfruit.lastIndex(of: charr) - 1
newString = somfruit.replacingOccurrences(of: charr, with: '', option: nil, range: Range(lower, upper)
print(newString)
This is simplified. firstIndex and lastIndex returns Int? so you have to check they exist and they are not equals.
This is using the example code from the official Swift4 doc
let greeting = "Hi there! It's nice to meet you! 👋"
let endOfSentence = greeting.index(of: "!")!
let firstSentence = greeting[...endOfSentence]
// firstSentence == "Hi there!"
But lets say let greeting = "Hello there world!"
and I want to retrieve only the second word (substring) in this sentence? So I only want the word "there".
I've tried using "world!" as an argument like
let endOfSentence = greeting.index(of: "world!")! but Swift 4 Playground doesn't like that. It's expecting 'Character' and my argument is a string.
So how can I get a substring of a very precise subrange? Or get nth word in a sentence for greater use in the future?
You can search for substrings using range(of:).
import Foundation
let greeting = "Hello there world!"
if let endIndex = greeting.range(of: "world!")?.lowerBound {
print(greeting[..<endIndex])
}
outputs:
Hello there
EDIT:
If you want to separate out the words, there's a quick-and-dirty way and a good way. The quick-and-dirty way:
import Foundation
let greeting = "Hello there world!"
let words = greeting.split(separator: " ")
print(words[1])
And here's the thorough way, which will enumerate all the words in the string no matter how they're separated:
import Foundation
let greeting = "Hello there world!"
var words: [String] = []
greeting.enumerateSubstrings(in: greeting.startIndex..<greeting.endIndex, options: .byWords) { substring, _, _, _ in
if let substring = substring {
words.append(substring)
}
}
print(words[1])
EDIT 2: And if you're just trying to get the 7th through the 11th character, you can do this:
import Foundation
let greeting = "Hello there world!"
let startIndex = greeting.index(greeting.startIndex, offsetBy: 6)
let endIndex = greeting.index(startIndex, offsetBy: 5)
print(greeting[startIndex..<endIndex])
For swift4,
let string = "substring test"
let start = String.Index(encodedOffset: 0)
let end = String.Index(encodedOffset: 10)
let substring = String(string[start..<end])
In Swift 5 encodedOffset (swift 4 func) is deprecated.
You will need to use utf160Offset
// Swift 5
let string = "Hi there! It's nice to meet you!"
let startIndex = 10 // random for this example
let endIndex = string.count
let start = String.Index(utf16Offset: startIndex, in: string)
let end = String.Index(utf16Offset: endIndex, in: string)
let substring = String(string[start..<end])
prints -> It's nice to meet you!
There is one mistake in the first answer.
Range<String.Index>.upperBound
The upperBound property should be the endIndex
For Example:
let text = "From Here Hello World"
if let result = text.range(of: "Hello World") {
let startIndex = result.upperBound
let endIndex = result.lowerBound
print(String(text[startIndex..<endIndex])) //"Hello World"
}
the simplest way I use is :
var str = "abcdefg"
String(Array(str)[2...4])
Old habits die hard. I did it the "Java" way and split the string up by spaces, then accessed the second word.
print(greeting.split(separator: " ")[1]) // "there /n"
This question already has answers here:
Reversing the order of a string value
(13 answers)
Closed 5 years ago.
Needed to reverse a swift string, managed to do so with this.
var str = "Hello, playground"
let reactedText = str.characters.reversed()
let nowBackwards = Array(reactedText)
let answer = String(nowBackwards)
And since I find nothing on SO in the subject I post it for some positive votes :) or indeed a better [shorter, as in different] solution.
Assuming you are using Swift 4 :
let string = "Hello, playground"
let reversedString = String(string.reversed())
Since in Swift 4, Strings are Character arrays again, you can call reversed on the String itself and the result will be of type [Character], from which you can initialize a String.
let stringToBeReversed = "Hello, playground"
let reversedString = String(stringToBeReversed.reversed()) //"dnuorgyalp ,olleH"
reversed method is available in String library
let str = "Hello, world!"
let reversed = String(str.reversed())
print(reversed)
Last night I had to convert my Swift 2.3 code to Swift 3.0 and my code is a mess after the conversion.
In Swift 2.3 I had the following code:
let maxChar = 40;
let val = "some long string";
var startRange = val.startIndex;
var endRange = val.startIndex.advancedBy(maxChar, limit: val.endIndex);
let index = val.rangeOfString(" ", options: NSStringCompareOptions.BackwardsSearch , range: startRange...endRange , locale: nil)?.startIndex;
Xcode converted my code to this which doesn't work:
let maxChar = 40;
let val = "some long string";
var startRange = val.startIndex;
var endRange = val.characters.index(val.startIndex, offsetBy: maxChar, limitedBy: val.endIndex);
let index = val.range(of: " ", options: NSString.CompareOptions.backwards , range: startRange...endRange , locale: nil)?.lowerBound
The error is in the parameter range in val.rage, saying No '...' candidates produce the expected contextual result type 'Range?'.
I tried using Range(startRange...endRange) as suggestd in the docs but I'm getting en error saying: connot invoke initiliazer for type ClosedRange<_> with an arguement list of type (ClosedRange). Seems like I'm missing something fundametnal.
Any help is appreciated.
Thanks!
Simple answer: the fundamental thing you are missing is that a closed range is now different from a range. So, change startRange...endRange to startRange..<endRange.
In more detail, here's an abbreviated version of your code (without the maxChar part):
let val = "some long string";
var startRange = val.startIndex;
var endRange = val.endIndex;
let index = val.range(
of: " ", options: .backwards, range: startRange..<endRange)?.lowerBound
// 9
Now you can use that as a basis to restore your actual desired functionality.
However, if all you want to do is split the string, then reinventing the wheel is kind of silly:
let arr = "hey ho ha".characters.split(separator:" ").map{String($0)}
arr // ["hey", "ho", "ha"]