Regex catch word at the start and end of a UITextView - regex

Question

I'm trying to catch when a word is used in a UITextView. I've got it working for words in the interior of the view.
The problem is when the word is first or last in the view.
My code so far:
private func filteredTermFor(_ word: String) -> String {
let punctuationFilter = "([\\A|\\W|\\d|\\z| ])"
let wordInParens = "(\(word))"
return punctuationFilter + wordInParens + punctuationFilter
}
I checked and found I should use ^ for the start of input and $ for the end of input. When I add either of these, for example:
"([^|\\A|\\W|\\d|\\z| ])"
they don't seem to have any effect when the word in question is the first or last in the view.
*For the sake of being verbose with my question, the return value from the function above is being used as searchTerm in this:
func highlightedTextInString(with searchTerm: String, targetString: String) -> NSAttributedString? {
let attributedString = NSMutableAttributedString(string: targetString)
do {
let regex = try NSRegularExpression(pattern: searchTerm, options: .caseInsensitive)
let range = NSRange(location: 0, length: targetString.utf16.count)
for match in regex.matches(in: targetString, options: .withTransparentBounds, range: range) {
let fontColor = UIColor.red
attributedString.addAttribute(NSForegroundColorAttributeName, value: fontColor, range: match.range)
}
return attributedString
} catch _ {
print("Error creating regular expression")
return nil
}
}
** Edit **
Since this was marked as a duplicate
The question this was reported a duplicate of does not cover edge cases when the word is typed next to a punctuation mark or digit without spaces.
For example:
.word , word9 , ?word?

Answer 1

Note that ([^|\\A|\\W|\\d|\\z| ]) is a capturing group ((...)) containing a character class that matches a single char defined inside it. The ^ after [ makes the class a negated one, and it matches any char but the one(s) defined in the set. So, [^|\\A|\\W|\\d|\\z| ] matches a single char other than | (it is no longer an alternation operator inside a character class), A (the \ in front is not considered, is omitted), a non-word char, a digit, z and space. It effectively matches _ and any letters other than A and z.
You state that the words you need to match may occur within word boundaries or digits.
You may use
return "(?<![^\\W\\d])(\(word))(?![^\\W\\d])"
See the regex demo.
Here, "(?<![^\\W\\d])" is a negative lookbehind that matches a location that is NOT immediately preceded with a character other than a non-word and a digit char. This sounds cumbersome, but the main point here is that [^\W\d] matches the same texts as \w excluding digits (\w matches letters, digit, and _. So, "(?<![^\\W\\d])" makes sure there is a start of string or a non-letter and non-_ char right before the word. If you allow a word to match after _, just use (?<!\\p{L}) (where \p{L} matches any Unicode letter).
The "(?![^\\W\\d])" is a negative lookahead that makes sure there is an end of string or a non-letter and non-_ (there can be punctuation, symbols and digits) immediately to the right of the word. Again, if you want to match a word if it is followed with _, you may replace this lookahead with "(?!\\p{L})" (just no letter after the word is allowed).