How to remove '\u{ef}' character from String Swift - ios

let's say I have a string
var a = "#bb #cccc #ddddd\u{ef}"
and i am setting it to textview like this
let text = a.trimmingCharacters(in: .whitespacesAndNewlines)
let textRemoved = text?.replacingOccurrences(of: "\u{ef}", with: "", options: NSString.CompareOptions.literal, range:nil)
textView.text = textRemove
I am trying to remove the \u{ef} character here. But in textRemoved it is not happening. Please help me how to do it.
I am using Xcode 10. Looks like below Xcode version than 10 is working
fine. is it a bug of Xcode 10?

This is a late answer but I struggled to replace "\u{ef}" in string as well. During debugging when hovered over string it showed presence of \u{ef} but when print in description it only showed space.
let str = "\u{ef} Some Title"
print(str) //" Some Title"
I tried replacingOccurrences(of: "\u{ef}", with: "", options: NSString.CompareOptions.literal, range: nil).trimmingCharacters(in: .whitespaces) but it failed as well.
So I used below snippet and it worked like wonder.
let modifiedStr = str.replacingOccurrences(of: "\u{fffc}", with: "", options: NSString.CompareOptions.literal, range: nil).trimmingCharacters(in: .whitespaces)
print(modifiedStr) //"Some Title"
Hope this helps someone!!

i also faced same issue for "\u{e2}". i have searched a lot but unable to find any answer. then i have tried below code , which works for me.
var newString = ""
for char in strMainString.unicodeScalars{
if char.isASCII{
newString += String(char)
}
}
Hope that will also work for you too.

In Xcode 10 Playground, string replaces for \u{00EF} is working.
var a = "#bb #cccc #ddddd\u{ef}"
a = a.replacingOccurrences(of: "\u{00EF}", with: "")
I hope that will work for you.

I tried the following and it worked like a charm:
replacingOccurrences(of: "ïżœ", with: " ", options: NSString.CompareOptions.literal, range: nil)

e.g. 1
let text = "\u{ef}\u{ef}\u{ef}\u{ef}😇擩擩擩"
let text1 = text.replacingOccurrences(of: "\u{fffc}", with: "", options: String.CompareOptions.literal, range: nil)
let text2 = text.replacingOccurrences(of: "\u{ef}", with: "", options: String.CompareOptions.literal, range: nil).trimmingCharacters(in: .whitespaces)
runnable
<img src="https://i.stack.imgur.com/styVo.png"/>
e.g. 2
let strBefore = textDocumentProxy.documentContextBeforeInput
let strAfter = textDocumentProxy.documentContextAfterInput
var textInput = strBefore + strAfter
let textInput2 = textInput.replacingOccurrences(of: "\u{ef}", with: "", options: String.CompareOptions.literal, range: nil)
let textInput1 = textInput.replacingOccurrences(of: "\u{fffc}", with: "", options: String.CompareOptions.literal, range: nil).trimmingCharacters(in: .whitespaces)
runnable
<img src="https://i.stack.imgur.com/xGHtW.png"/>

Similar to question but with \u{e2} symbol (fix is the same):
\u{e2} is not a character rather subset of UTF8 plane which starts with 0xE2 byte.
So look here, E2 are general punctuation symbols.
There many symbols actually which started with \u{e2} but not limited to it and full char can be represented f.e. with e2 80 a8 bytes (line separator).
That explains why shown in Xcode \u{e2} can't be replaced with replacingOccurrences... function. In order to filter out correct symbol you have to know what exact symbol it is, f.e. by using the snippet below:
"\u{2028}&đŸ˜Č".forEach { (char) in
print(Data(char.utf8).map { String(format: "%02x", $0) }.joined(separator: " "))
}
it prints to console:
e2 80 a8
26
f0 9f 98 b2
which are byte representation for each symbol.
Next step is to filter your string, go here and search in 3d column your bytes and unicode code point value is what you need (first column) and write it in swift code like "\u{2028}\u{206A}..." (depending on your sorting).
The final function may look like:
func removingE2Symbols() -> String {
let specialChars = "\u{202A}\u{202C}"
return filter { !specialChars.contains($0) }
}

Try this
extension String {
var asciiString: String {
return String(self.unicodeScalars.filter{ $0.isASCII })
}
}

It,s working Please check again:
let a = "#bb #cccc #ddddd\u{ef}"
let text = a.trimmingCharacters(in: .whitespacesAndNewlines)
let textRemoved = text.replacingOccurrences(of: "\u{ef}", with: "", options: NSString.CompareOptions.literal, range:nil)
print(textRemoved)

Related

Swift: Getting range of text that includes emojis [duplicate]

This question already has an answer here:
Swift Regex doesn't work
(1 answer)
Closed 5 years ago.
I'm trying to parse out "#mentions" from a user provided string. The regular expression itself seems to find them, but the range it provides is incorrect when emoji are present.
let text = "😂😘🙂 #joe "
let tagExpr = try? NSRegularExpression(pattern: "#\\S+")
tagExpr?.enumerateMatches(in: text, range: NSRange(location: 0, length: text.characters.count)) { tag, flags, pointer in
guard let tag = tag?.range else { return }
if let newRange = Range(tag, in: text) {
let replaced = text.replacingCharacters(in: newRange, with: "[email]")
print(replaced)
}
}
When running this
tag = (location: 7, length: 2)
And prints out
😂😘🙂 [email]oe
The expected result is
😂😘🙂 [email]
NSRegularExpression (and anything involving NSRange) operates on UTF16 counts / indexes. For that matter, NSString.count is the UTF16 count as well.
But in your code, you're telling NSRegularExpression to use a length of text.characters.count. This is the number of composed characters, not the UTF16 count. Your string "😂😘🙂 #joe " has 9 composed characters, but 12 UTF16 code units. So you're actually telling NSRegularExpression to only look at the first 9 UTF16 code units, which means it's ignoring the trailing "oe ".
The fix is to pass length: text.utf16.count.
let text = "😂😘🙂 #joe "
let tagExpr = try? NSRegularExpression(pattern: "#\\S+")
tagExpr?.enumerateMatches(in: text, range: NSRange(location: 0, length: text.utf16.count)) { tag, flags, pointer in
guard let tag = tag?.range else { return }
if let newRange = Range(tag, in: text) {
let replaced = text.replacingCharacters(in: newRange, with: "[email]")
print(replaced)
}
}

How to use NSRegular Expression and filter string array correctly in Swift

Here is my question, I have a string array that contains a bunch of
countries:
let myCountryStart = ["Africa/ABC", "America/BBC", "Asia/CBC", "Pacific/CBA", "Europe/CBB", "Indian/CAB"]
Is there have any solution to remove the specific words like
'Africa', 'America', 'Asia'...etc,.
Let the output result looks like the below followings:
let myCountryEnd = ["ABC", "BBC", "CBC", "CBA", "CBB", "CAB"]
Here are my code for now...
let 1stReplace = myCountryStart.replacingOccurrences(of: "/", with: "", options: .literal)
let 2ndReplace = 1stReplace.replacingOccurrences(of: "Africa", with: "", options: .literal)
let 3rdReplace = 2ndReplace.replacingOccurrences(of: "Asia", with: "", options: .literal)
I know this is a stupid solution. Hence, I prefer to use NSRegular Expression. But I encountered a problem
about String & String Array issue.
let target = myCountryStart
let regex = "/"
let RE = try? NSRegularExpression(pattern: "regex", options: .caseInsensitive)
let modified = RE?.stringByReplacingMatches(in: target, options: .reportProgress, range: nil, withTemplate: "") {
return modified
}
let myCountryEnd = modified
Therefore, I got a warning about I cannot use this method on String
array. What should I do to fix it?
Any suggestions or help will be greatly appreciated. Thanks from a Swift rookie.
You may use .map and .replacingOccurrences using a regex like .*/ or ^[^/]*/:
let myCountryStart = ["Africa/ABC", "America/BBC", "Asia/CBC", "Pacific/CBA", "Europe/CBB", "Indian/CAB"]
let myCountryEnd = myCountryStart.map{ $0.replacingOccurrences(of: ".*/", with: "", options: [.caseInsensitive,.regularExpression]) }
print(myCountryEnd)
// => ["ABC", "BBC", "CBC", "CBA", "CBB", "CAB"]
The .*/ pattern will match any 0 or more characters other than line break chars, as many as possible, up to the last /.
The ^[^/]*/ pattern will match any chars other than / from the start of the string till the first /.
Note you do not need the .caseInsensitive option, I kept it to show how you may combine several options in the options argument.

Replace just bold word in string, Swift

I have a UILabel with text "hello world, hello". There are 2 hello words.
And I want to replace the only 'bold hello' to 'thanks' without bold.
I use this code:
uiLabel1.text = "hello world, hello"
let target = "hello"
let replace = "thanks"
uiLabel1.text.replacingOccurrences(of: target, with: replace, options:
NSString.CompareOptions.literal, range: nil)
And the result is: "thanks world, thanks"
The result I want: "hello world, thanks"
Okay, so there's probably an easier way to do this...
So, I went through the API (like super quick) and looked for something like lastIndexOf, which lead me on a little trail to String#range(of:options), which allows you to search backwards, hmmm, interesting.
This returns a Range<String.Index> ... okay, so how can I use that?! Hmm, maybe String#replacingOccurrences(of:with:options:range:) đŸ€”
So, crack open a play ground and...
var str = "hello world, hello"
let lastIndexOf = str.range(of: "hello", options: .backwards)
str = str.replacingOccurrences(of: "hello", with: "thanks", options: .caseInsensitive, range: lastIndexOf)
str now equals "hello world, thanks"
Hi #MadProgrammer, your code is to replace the last hello word to thanks, right? But my question is to replace hello with the bold attribute, it may in the first, middle or at the end of a string.
Okay, so clearly we're missing some context...
Assuming, now, you're using a NSAttributedString, it becomes slightly more complicated
Building the string itself is not hard, figuring out how to find string components by attribute, a little more difficult.
Lucky for us, we have the Internet. So, the following is based on ideas I got from:
NSAttributedString by example
Detect whether a font is bold/italic on iOS?
One of the important things to remember when trying to solve an issue, you'll be lucky to find a single answer which does it all, instead, you need to break your issue down and focus on solving individual elements, and be prepared to go back to the start 😉
So, once again, unto the play ground...
import UIKit
var str = "hello world, "
//let lastIndexOf = str.range(of: "hello", options: .backwards)
//str = str.replacingOccurrences(of: "hello", with: "thanks", options: .caseInsensitive, range: lastIndexOf)
extension UIFont {
var isBold: Bool {
return fontDescriptor.symbolicTraits.contains(.traitBold)
}
var isItalic: Bool {
return fontDescriptor.symbolicTraits.contains(.traitItalic)
}
}
// Just so I can see that the style ;)
let fontSize = CGFloat(24.0)
let boldAttrs = [
NSAttributedString.Key.font: UIFont.boldSystemFont(ofSize: fontSize),
NSAttributedString.Key.foregroundColor: UIColor.white // Playground
]
// Playground only
let plainAttrs = [
NSAttributedString.Key.foregroundColor: UIColor.white // Playground
]
let boldText = NSMutableAttributedString(string: "hello", attributes: boldAttrs)
let styledText = NSMutableAttributedString(string: str, attributes: plainAttrs)
let someMoreBoldText = NSMutableAttributedString(string: "not to be replaced", attributes: boldAttrs)
// Attributes can be combined with their appear together ;)
styledText.append(boldText)
styledText.append(NSMutableAttributedString(string: " ", attributes: plainAttrs))
styledText.append(someMoreBoldText)
styledText.append(NSMutableAttributedString(string: " ", attributes: plainAttrs))
styledText.append(boldText)
styledText.enumerateAttribute(NSAttributedString.Key.font, in: NSRange(0..<styledText.length)) { (value, range, stop) in
guard let font = value as? UIFont, font.isBold else {
return;
}
let subText = styledText.attributedSubstring(from: range)
guard subText.string == "hello" else {
return
}
styledText.replaceCharacters(in: range, with: "thanks")
}
styledText
Which outputs...
The important things for me are:
The style has not be changed
Only the individual "hello" values, which are bolded, have been changed
Here is the code. But actually this is hardcoded. If the target enclosed in between <b></b>, it will work.
var text = "hello world, <b>hello</b>"
let target = "hello"
let replace = "thanks"
text = text.replacingOccurrences(of: "<b>\(target)</b>", with: replace, options: .literal, range: nil) //hello world, thanks

Remove U\0000fffc unicode scalar from string

I receive an NSAttributedString that contains a NSTextAttachment. I want to remove that attachment, and it looks like it is represented as "\u{ef}" in the string. Printing the unicode scalars of such string, it also seems that unicode scalar for the "\u{ef}" is U\0000fffc.
I tried to do this:
noAttachmentsText = text.replacingOccurrences(of: "\u{ef}", with: "")
with no success, so I'm trying by comparing unicode scalars:
var scalars = Array(text.unicodeScalars)
for scalar in scalars {
// compare `scalar` to `U\0000fffc`
}
but I'm not able either to succeed in the comparison.
How could I do this?
But this code works for me from How do I remove "\U0000fffc" from a string in Swift?
let original = "First part \u{ef} Last part"
let originalRange = Range<String.Index>(start: original.startIndex, end: original.endIndex)
let target = original.stringByReplacingOccurrencesOfString("\u{ef}", withString: "", options: NSStringCompareOptions.LiteralSearch, range: originalRange)
print(target)
Output :
"First part ĂŻ Last part"
to
First part Last part
U can use similar code for swift 3 just replace unicode using replacingOccurrences option for exapmle :
func stringTocleanup(str: String) -> String {
var result = str
result = result.replacingOccurrences(of: "\"", with: "\"")
.replacingOccurrences(of: "\u{10}", with: "")
return result
}

Swift - Remove white spaces from string doesn't work

i'm trying to remove white spaces and some characters from a string, please check my code below
// giving phoneString = +39 333 3333333
var phoneString = ABMultiValueCopyValueAtIndex(phone, indexPhone).takeRetainedValue() as! String
// Remove spaces from string
phoneString = phoneString.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceCharacterSet())
// Remove +39 if exist
if phoneString.rangeOfString("+39") != nil{
phoneString = phoneString.stringByReplacingOccurrencesOfString("\0", withString: "+39", options: NSStringCompareOptions.LiteralSearch, range: nil)
}
print(phoneString) // output +39 333 3333333
it seems like all the changes has no effect over my string, why this happen?
EDIT #V S
EDIT 2:
I tried to convert my string in utf 8, check the result:
43 51 57 194 160 51 51 51 194 160 51 51 51 51 51 51 51
where:
43 = +
51 = 3
57 = 9
160 = space
194 = wtf?!? is this?
what do you try to do is
// your input string
let str = "+39 333 3333333"
let arr = str.characters.split(" ").map(String.init) // ["+39", "333", "3333333"]
// remove country code and reconstruct the rest as one string without whitespaces
let str2 = arr.dropFirst().joinWithSeparator("") // "3333333333"
to filter out country code, only if exists (as Eendje asks)
let str = "+39 123 456789"
let arr = str.characters.split(" ").map(String.init)
let str3 = arr.filter { !$0.hasPrefix("+") }.joinWithSeparator("") // "123456789"
UPDATE, based on your update.
160 represents no-breakable space. just modify next line in my code
let arr = str.characters.split{" \u{00A0}".characters.contains($0)}.map(String.init)
there is " \u{00A0}".characters.contains($0) expression where you can extend the string to as much whitespace characters, as you need. 160 is \u{00A0} see details here.
Update for Swift 4
String.characters is deprecated. So the correct answer would now be
// your input string
let str = "+39 333 3333333"
let arr = str.components(separatedBy: .whitespaces) // ["+39", "333", "3333333"]
// remove country code and reconstruct the rest as one string without whitespaces
let str2 = arr.dropFirst().joined() // "3333333333"
Firstly, stringByTrimmingCharactersInSet only trims the string - i.e. removes leading & trailing spaces - you need to use stringByReplacingOccurrencesOfString replacing " " with "".
Secondly, your parameters on stringByReplacingOccurrencesOfString for the country code are the wrong way round.
Thirdly, "\0" is not what you want- that's ASCII null, not zero.
Swift 3 / Swift 4
let withoutSpaces = phoneNumber.replacingOccurrences(of: "\\s", with: "", options: .regularExpression)
Swift 5
//MARK:- 3 ways to resolve it
var tempphone = "0345 55500 93"
//MARK:- No 1
tempphone = tempphone.replacingOccurrences(of: " ", with: "")
//MARK:- No 2
tempphone = tempphone.replacingOccurrences(of: "\\s", with: "", options: .regularExpression)
//MARK:- No 3
tempphone = tempphone.trimmingCharacters(in: .whitespaces)
phoneString = phoneString.stringByReplacingOccurrencesOfString("+39", withString: "", options: NSStringCompareOptions.LiteralSearch, range: nil)
phoneString = phoneString.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceCharacterSet())
Try this. This has worked for me:
if phoneString.rangeOfString("+39") != nil{
freshString = phoneString.stringByReplacingOccurrencesOfString("\0", withString: "+39", options: NSStringCompareOptions.LiteralSearch, range: nil)
}
var strings = freshString.componentsSeparatedByString(" ") as NSArray
var finalString = strings.componentsJoinedByString("")
//outputs +393333333333
You can use this replace the whitespace
phoneNumber.replacingOccurrences(of: "\u{00A0}", with: "")
let trimmedPhoneString = String(phoneString).stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceCharacterSet())
To Remove +39 if exist, you can use stringByReplacingOccurrencesOfString instead
var phoneString = "+39 333 3333333"
phoneString = phoneString.stringByReplacingOccurrencesOfString(" ", withString:"")
if phoneString.rangeOfString("+39") != nil
{
phoneString = phoneString.stringByReplacingOccurrencesOfString("+39", withString: "", options: NSStringCompareOptions.LiteralSearch, range: nil)
}
print(phoneString) // output 3333333333

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