Range<String.Index> Versus String.Index - swift

I am having an issue understanding the difference between Range-String.Index- and String.Index
For example:
func returnHtmlContent() -> String {
let filePath = URL(string:"xxx.htm")
filePath?.startAccessingSecurityScopedResource();
let htmlFile = Bundle.main.path(forResource: "getData", ofType: "htm");
let html = try! String(contentsOfFile: htmlFile!, encoding: String.Encoding.utf8);
filePath?.stopAccessingSecurityScopedResource();
return html;
};
func refactorHtml(content: String) -> String {
let StartingString = "<div id=\"1\">";
let EndingString = "</div></form>";
func selectString() -> String {
var htmlContent = returnHtmlContent();
let charLocationStart = htmlContent.range(of: StartingString);
let charLocationEnd = htmlContent.range(of: EndingString);
htmlContent.remove(at: charLocationStart);
return htmlContent;
};
let formattedBody = selectString();
return formattedBody;
};
refactorHtml(content: returnHtmlContent());
The idea in pseudocode
Generated HTMLBody
Pass To Function that formats
Remove all characters before StartingString
Remove all Characters After EndingString
Send NewString to Variable
Now - when I try to find the index position I cant seem to get the right value type, this is the error I am getting
Cannot convert value of type 'Range<String.Index>?' to expected argument type 'String.Index'
This is running in a playground

String indices aren't integers. They're opaque objects (of type String.Index) which can be used to subscript into a String to obtain a character.
Ranges aren't limited to only Range<Int>. If you look at the declaration of Range, you can see it's generic over any Bound, so long as the Bound is Comparable (which String.Index is).
So a Range<String.Index> is just that. It's a range of string indices, and just like any other range, it has a lowerBound, and an upperBound.

Related

Split String or Substring with Regex pattern in Swift

First let me point out... I want to split a String or Substring with any character that is not an alphabet, a number, # or #. That means, I want to split with whitespaces(spaces & line breaks) and special characters or symbols excluding # and #
In Android Java, I am able to achieve this with:
String[] textArr = text.split("[^\\w_##]");
Now, I want to do the same in Swift. I added an extension to String and Substring classes
extension String {}
extension Substring {}
In both extensions, I added a method that returns an array of Substring
func splitWithRegex(by regexStr: String) -> [Substring] {
//let string = self (for String extension) | String(self) (for Substring extension)
let regex = try! NSRegularExpression(pattern: regexStr)
let range = NSRange(string.startIndex..., in: string)
return regex.matches(in: string, options: .anchored, range: range)
.map { match -> Substring in
let range = Range(match.range(at: 1), in: string)!
return string[range]
}
}
And when I tried to use it, (Only tested with a Substring, but I also think String will give me the same result)
let textArray = substring.splitWithRegex(by: "[^\\w_##]")
print("substring: \(substring)")
print("textArray: \(textArray)")
This is the out put:
substring: This,is a #random #text written for debugging
textArray: []
Please can Someone help me. I don't know if the problem if from my regex [^\\w_##] or from splitWithRegex method
The main reason why the code doesn't work is range(at: 1) which returns the content of the first captured group, but the pattern does not capture anything.
With just range the regex returns the ranges of the found matches, but I suppose you want the characters between.
To accomplish that you need a dynamic index starting at the first character. In the map closure return the string from the current index to the lowerBound of the found range and set the index to its upperBound. Finally you have to add manually the string from the upperBound of the last match to the end.
The Substring type is a helper type for slicing strings. It should not be used beyond a temporary scope.
extension String {
func splitWithRegex(by regexStr: String) -> [String] {
guard let regex = try? NSRegularExpression(pattern: regexStr) else { return [] }
let range = NSRange(startIndex..., in: self)
var index = startIndex
var array = regex.matches(in: self, range: range)
.map { match -> String in
let range = Range(match.range, in: self)!
let result = self[index..<range.lowerBound]
index = range.upperBound
return String(result)
}
array.append(String(self[index...]))
return array
}
}
let text = "This,is a #random #text written for debugging"
let textArray = text.splitWithRegex(by: "[^\\w_##]")
print(textArray) // ["This", "is", "a", "#random", "#text", "written", "for", "debugging"]
However in macOS 13 and iOS 16 there is a new API quite similar to the java API
let text = "This,is a #random #text written for debugging"
let textArray = Array(text.split(separator: /[^\w_##]/))
print(textArray)
The forward slashes indicate a regex literal

How to get the range of the first line in a string?

I would like to change the formatting of the first line of text in an NSTextView (give it a different font size and weight to make it look like a headline). Therefore, I need the range of the first line. One way to go is this:
guard let firstLineString = textView.string.components(separatedBy: .newlines).first else {
return
}
let range = NSRange(location: 0, length: firstLineString.count)
However, I might be working with quite long texts so it appears to be inefficient to first split the entire string into line components when all I need is the first line component. Thus, it seems to make sense to use the firstIndex(where:) method:
let firstNewLineIndex = textView.string.firstIndex { character -> Bool in
return CharacterSet.newlines.contains(character)
}
// Then: Create an NSRange from 0 up to firstNewLineIndex.
This doesn't work and I get an error:
Cannot convert value of type '(Unicode.Scalar) -> Bool' to expected argument type 'Character'
because the contains method accepts not a Character but a Unicode.Scalar as a parameter (which doesn't really make sense to me because then it should be called a UnicodeScalarSet and not a CharacterSet, but nevermind...).
My question is:
How can I implement this in an efficient way, without first slicing the whole string?
(It doesn't necessarily have to use the firstIndex(where:) method, but appears to be the way to go.)
A String.Index range for the first line in string can be obtained with
let range = string.lineRange(for: ..<string.startIndex)
If you need that as an NSRange then
let nsRange = NSRange(range, in: string)
does the trick.
You can use rangeOfCharacter, which returns the Range<String.Index> of the first character from a set in your string:
extension StringProtocol where Index == String.Index {
var partialRangeOfFirstLine: PartialRangeUpTo<String.Index> {
return ..<(rangeOfCharacter(from: .newlines)?.lowerBound ?? endIndex)
}
var rangeOfFirstLine: Range<Index> {
return startIndex..<partialRangeOfFirstLine.upperBound
}
var firstLine: SubSequence {
return self[partialRangeOfFirstLine]
}
}
You can use it like so:
var str = """
some string
with new lines
"""
var attributedString = NSMutableAttributedString(string: str)
let firstLine = NSAttributedString(string: String(str.firstLine))
// change firstLine as you wish
let range = NSRange(str.rangeOfFirstLine, in: str)
attributedString.replaceCharacters(in: range, with: firstLine)

How to remove certain characters in a string?

The string value varies sometimes it's
93.93% - 94.13, 85.34, %74.90, 88.21%
I just need to extract the double value like this.
93.93, 85.34, 74.90, 88.21
You can use regex to extract numbers from your string like this:
let sourceString = "93.93% - 94.13, 85.34, %74.90, 88.21%"
func getNumbers(from string : String) -> [String] {
let pattern = "((\\+|-)?([0-9]+)(\\.[0-9]+)?)|((\\+|-)?\\.?[0-9]+)" // Change this according to your requirement
let regex = try! NSRegularExpression(pattern: pattern)
let matches = regex.matches(in: string, range: NSRange(string.startIndex..., in: string))
let result = matches.map { (match) -> String in
let range = Range(match.range, in: string)!
return String(string[range])
}
return result
}
let numberArray = getNumbers(from: sourceString)
print(numberArray)
Result:
["93.93", "94.13", "85.34", "74.90", "88.21"]
you should try using a regex like this for example :
[0-9]{2}.[0-9]{2}
This regex find all string that match two numbers, then a dot and two numbers again.
for each value such as var str='%74.90'; use this line -
var double=str.match(/[+-]?\d+(\.\d+)?/g).map(function(v) { return parseFloat(v); })[0];
Use Scanner to scan the values. Scanner is highly configurable and designed for scanning string and numeric values from loosely demarcated strings. Below is the example:
let characterSet = CharacterSet.init(charactersIn: "0123456789.").inverted
let scanner = Scanner(string: "93.93% - 94.13, 85.34, %74.90, 88.21%")
scanner.charactersToBeSkipped = characterSet
var numStr: NSString?
while scanner.scanUpToCharacters(from: characterSet, into: &numStr) {
print(numStr ?? "")
}
Output:
93.93
94.13
85.34
74.90
88.21
It is easier to understand comparatively regex.

Why does swift substring with range require a special type of Range

Consider this function to build a string of random characters:
func makeToken(length: Int) -> String {
let chars: String = "abcdefghijklmnopqrstuvwxyz0123456789!?##$%ABCDEFGHIJKLMNOPQRSTUVWXYZ"
var result: String = ""
for _ in 0..<length {
let idx = Int(arc4random_uniform(UInt32(chars.characters.count)))
let idxEnd = idx + 1
let range: Range = idx..<idxEnd
let char = chars.substring(with: range)
result += char
}
return result
}
This throws an error on the substring method:
Cannot convert value of type 'Range<Int>' to expected argument
type 'Range<String.Index>' (aka 'Range<String.CharacterView.Index>')
I'm confused why I can't simply provide a Range with 2 integers, and why it's making me go the roundabout way of making a Range<String.Index>.
So I have to change the Range creation to this very over-complicated way:
let idx = Int(arc4random_uniform(UInt32(chars.characters.count)))
let start = chars.index(chars.startIndex, offsetBy: idx)
let end = chars.index(chars.startIndex, offsetBy: idx + 1)
let range: Range = start..<end
Why isn't it good enough for Swift for me to simply create a range with 2 integers and the half-open range operator? (..<)
Quite the contrast to "swift", in javascript I can simply do chars.substr(idx, 1)
I suggest converting your String to [Character] so that you can index it easily with Int:
func makeToken(length: Int) -> String {
let chars = Array("abcdefghijklmnopqrstuvwxyz0123456789!?##$%ABCDEFGHIJKLMNOPQRSTUVWXYZ".characters)
var result = ""
for _ in 0..<length {
let idx = Int(arc4random_uniform(UInt32(chars.count)))
result += String(chars[idx])
}
return result
}
Swift takes great care to provide a fully Unicode-compliant, type-safe, String abstraction.
Indexing a given Character, in an arbitrary Unicode string, is far from a trivial task. Each Character is a sequence of one or more Unicode scalars that (when combined) produce a single human-readable character. In particular, hiding all this complexity behind a simple Int based indexing scheme might result in the wrong performance mental model for programmers.
Having said that, you can always convert your string to a Array<Character> once for easy (and fast!) indexing. For instance:
let chars: String = "abcdefghijklmnop"
var charsArray = Array(chars.characters)
...
let resultingString = String(charsArray)

Conversion stringByAddingPercentEncodingWithAllowedCharacters(.URLHostAllowedCharacterSet()) from Swift2 to Swift3 [duplicate]

I was using this, in Swift 1.2
let urlwithPercentEscapes = myurlstring.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding)
This now gives me a warning asking me to use
stringByAddingPercentEncodingWithAllowedCharacters
I need to use a NSCharacterSet as an argument, but there are so many and I cannot determine what one will give me the same outcome as the previously used method.
An example URL I want to use will be like this
http://www.mapquestapi.com/geocoding/v1/batch?key=YOUR_KEY_HERE&callback=renderBatch&location=Pottsville,PA&location=Red Lion&location=19036&location=1090 N Charlotte St, Lancaster, PA
The URL Character Set for encoding seems to contain sets the trim my
URL. i.e,
The path component of a URL is the component immediately following the
host component (if present). It ends wherever the query or fragment
component begins. For example, in the URL
http://www.example.com/index.php?key1=value1, the path component is
/index.php.
However I don't want to trim any aspect of it.
When I used my String, for example myurlstring it would fail.
But when used the following, then there were no issues. It encoded the string with some magic and I could get my URL data.
let urlwithPercentEscapes = myurlstring.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding)
As it
Returns a representation of the String using a given encoding to
determine the percent escapes necessary to convert the String into a
legal URL string
Thanks
For the given URL string the equivalent to
let urlwithPercentEscapes = myurlstring.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding)
is the character set URLQueryAllowedCharacterSet
let urlwithPercentEscapes = myurlstring.stringByAddingPercentEncodingWithAllowedCharacters( NSCharacterSet.URLQueryAllowedCharacterSet())
Swift 3:
let urlwithPercentEscapes = myurlstring.addingPercentEncoding( withAllowedCharacters: .urlQueryAllowed)
It encodes everything after the question mark in the URL string.
Since the method stringByAddingPercentEncodingWithAllowedCharacters can return nil, use optional bindings as suggested in the answer of Leo Dabus.
It will depend on your url. If your url is a path you can use the character set
urlPathAllowed
let myFileString = "My File.txt"
if let urlwithPercentEscapes = myFileString.addingPercentEncoding(withAllowedCharacters: .urlPathAllowed) {
print(urlwithPercentEscapes) // "My%20File.txt"
}
Creating a Character Set for URL Encoding
urlFragmentAllowed
urlHostAllowed
urlPasswordAllowed
urlQueryAllowed
urlUserAllowed
You can create also your own url character set:
let myUrlString = "http://www.mapquestapi.com/geocoding/v1/batch?key=YOUR_KEY_HERE&callback=renderBatch&location=Pottsville,PA&location=Red Lion&location=19036&location=1090 N Charlotte St, Lancaster, PA"
let urlSet = CharacterSet.urlFragmentAllowed
.union(.urlHostAllowed)
.union(.urlPasswordAllowed)
.union(.urlQueryAllowed)
.union(.urlUserAllowed)
extension CharacterSet {
static let urlAllowed = CharacterSet.urlFragmentAllowed
.union(.urlHostAllowed)
.union(.urlPasswordAllowed)
.union(.urlQueryAllowed)
.union(.urlUserAllowed)
}
if let urlwithPercentEscapes = myUrlString.addingPercentEncoding(withAllowedCharacters: .urlAllowed) {
print(urlwithPercentEscapes) // "http://www.mapquestapi.com/geocoding/v1/batch?key=YOUR_KEY_HERE&callback=renderBatch&location=Pottsville,PA&location=Red%20Lion&location=19036&location=1090%20N%20Charlotte%20St,%20Lancaster,%20PA"
}
Another option is to use URLComponents to properly create your url
Swift 3.0 (From grokswift)
Creating URLs from strings is a minefield for bugs. Just miss a single / or accidentally URL encode the ? in a query and your API call will fail and your app won’t have any data to display (or even crash if you didn’t anticipate that possibility). Since iOS 8 there’s a better way to build URLs using NSURLComponents and NSURLQueryItems.
func createURLWithComponents() -> URL? {
var urlComponents = URLComponents()
urlComponents.scheme = "http"
urlComponents.host = "www.mapquestapi.com"
urlComponents.path = "/geocoding/v1/batch"
let key = URLQueryItem(name: "key", value: "YOUR_KEY_HERE")
let callback = URLQueryItem(name: "callback", value: "renderBatch")
let locationA = URLQueryItem(name: "location", value: "Pottsville,PA")
let locationB = URLQueryItem(name: "location", value: "Red Lion")
let locationC = URLQueryItem(name: "location", value: "19036")
let locationD = URLQueryItem(name: "location", value: "1090 N Charlotte St, Lancaster, PA")
urlComponents.queryItems = [key, callback, locationA, locationB, locationC, locationD]
return urlComponents.url
}
Below is the code to access url using guard statement.
guard let url = createURLWithComponents() else {
print("invalid URL")
return nil
}
print(url)
Output:
http://www.mapquestapi.com/geocoding/v1/batch?key=YOUR_KEY_HERE&callback=renderBatch&location=Pottsville,PA&location=Red%20Lion&location=19036&location=1090%20N%20Charlotte%20St,%20Lancaster,%20PA
In Swift 3.1, I am using something like the following:
let query = "param1=value1&param2=" + valueToEncode.addingPercentEncoding(withAllowedCharacters: .alphanumeric)
It's safer than .urlQueryAllowed and the others, because it this will encode every characters other than A-Z, a-z and 0-9. This works better when the value you are encoding may use special characters like ?, &, =, + and spaces.
In my case where the last component was non latin characters I did the following in Swift 2.2:
extension String {
func encodeUTF8() -> String? {
//If I can create an NSURL out of the string nothing is wrong with it
if let _ = NSURL(string: self) {
return self
}
//Get the last component from the string this will return subSequence
let optionalLastComponent = self.characters.split { $0 == "/" }.last
if let lastComponent = optionalLastComponent {
//Get the string from the sub sequence by mapping the characters to [String] then reduce the array to String
let lastComponentAsString = lastComponent.map { String($0) }.reduce("", combine: +)
//Get the range of the last component
if let rangeOfLastComponent = self.rangeOfString(lastComponentAsString) {
//Get the string without its last component
let stringWithoutLastComponent = self.substringToIndex(rangeOfLastComponent.startIndex)
//Encode the last component
if let lastComponentEncoded = lastComponentAsString.stringByAddingPercentEncodingWithAllowedCharacters(NSCharacterSet.alphanumericCharacterSet()) {
//Finally append the original string (without its last component) to the encoded part (encoded last component)
let encodedString = stringWithoutLastComponent + lastComponentEncoded
//Return the string (original string/encoded string)
return encodedString
}
}
}
return nil;
}
}
Swift 4.0
let encodedData = myUrlString.addingPercentEncoding(withAllowedCharacters: CharacterSet.urlHostAllowed)