In my case, I need to Implement Regex for my UITextField. Here, my textfield should allow only uppercase with number mixed values.
For Example:
AI1234
ER3456
I used below one, but not working
^[A-Z0-9]{3}?$
This regex matches the pattern above
2 Uppercase characters followed by 4 numbers
^[A-Z]{2}\\d{4}
You can test it on https://regexr.com/
Edit:
let str = """
AI1234
ER3456
"""
let pattern = try? NSRegularExpression(pattern: "[A-Z]{2}\\d{4}", options: [])
let range = NSRange(location: 0, length: str.utf16.count)
let matches = pattern?.matches(in: str, options: [], range: range)
print(matches)
Related
I have some Arabic text which has some decimals as well.
for example this text
"بِسۡمِ اللّٰہِ الرَّحۡمٰنِ الرَّحِیۡمِ ﴿۱﴾"
"وَاِذَا قِیۡلَ لَہُمۡ اٰمِنُوۡا کَمَاۤ اٰمَنَ النَّاسُ قَالُوۡۤا اَنُؤۡمِنُ کَمَاۤ اٰمَنَ السُّفَہَآءُ ؕ اَلَاۤ اِنَّہُمۡ ہُمُ السُّفَہَآءُ وَلٰکِنۡ لَّا یَعۡلَمُوۡنَ ﴿۱۴﴾"
This text has verse numbers as Arabic digits in the end.
I wanted to find out all the matches for the verse numbers in these verses.
In swift I am tring to use the regular expression but somehow i am not coming up with the correct regex.
Here is my code:
func getRegex() {
// unicode for the arabic digits
let regexStr = "[\u{0660}-\u{0669}]+"
//let regexStr = "[\\p{N}]+"
//let regexStr = "[۹۸۷۶۵۴۳۲۱۰]+"
do {
let regex = try NSRegularExpression(pattern: regexStr, options: .caseInsensitive)
let matches = regex.matches(in: self.arabicText, options: .anchored, range: NSRange(location: 0, length: self.arabicText.count))
print("Matches count : \(matches.count)")
} catch {
print(error)
}
}
Can somebody guide me on how I can get the matches for the Arabic digits in the example Arabic text?
The .anchored argument makes the pattern only match at the start of string, so you need to remove it.
Also, as your string is not ASCII, you need to use self.arabicText.utf16.count string property rather than accessing the self.arabicText.count directly.
So, you can use
let regexStr = "[۹۸۷۶۵۴۳۲۱۰]+"
and then
let matches = regex.matches(in: self.arabicText, options: [], range: NSRange(location: 0, length: self.arabicText.utf16.count))
I'm parsing an XML doc (using XMLParser) and some of the values have php-like placeholders, e.g. %1$s, and I would like to convert those to {x-1}.
Examples:
%1$s ---> {0}
%2$s ---> {1}
I'm doing this in a seemingly hacky way, using regex:
But there must be a better implementation of this regex.
Consider a string:
let str = "lala fawesfgeksgjesk 3rf3f %1$s rk32mrk3mfa %2$s fafafczcxz %3$s czcz $#$##%## %4$s qqq %5$s"
Now we're going to extract the integer strings between strings % and $s:
let regex = try! NSRegularExpression(pattern: "(?<=%)[^$s]+")
let range = NSRange(location: 0, length: str.utf16.count)
let matches = regex.matches(in: str, options: [], range: range)
matches.map {
print(String(str[Range($0.range, in: str)!]))
}
Works quite fine. The issue is that the "4" value got mixed up because of the preceding random strings before the %4$s.
Prints:
1
2
3
## %4
5
Is there any better way to do this?
This might not be a very efficient (or swifty :)) way but it gets the job done. What it does is that it searches for a given reg ex and uses the matched substring to extract the numeric value and decrease it and then perform a simple replace between the substring and a newly constructed placeholder value. This is executed in a loop until no more matches are found.
let pattern = #"%(\d*)\$s"#
while let range = str.range(of: pattern, options: .regularExpression) {
let placeholder = str[range]
let number = placeholder.trimmingCharacters(in: CharacterSet(charactersIn: "0123456789.").inverted)
if let value = Int(number) {
str = str.replacingOccurrences(of: placeholder, with: "{\(value - 1)}")
}
}
Im trying to replace matched strings using regex in swift, my requirement is as below
originalString = "It is live now at Germany(DE)"
i want the string within the (" ") i.eDE to be separated by space i.e. "D E"
so replacedString should be "It is live now at Germany(D E)"
i tried below code
var value: NSMutableString = "It is live now at Germany(DE)"
let pattern = "(\\([A-Za-z ]+\\))"
let regex = try? NSRegularExpression(pattern: pattern)
regex?.replaceMatches(in: value, options: .reportProgress, range:
NSRange(location: 0,length: value.length), withTemplate: " $1 ")
print(value)
output is It is live now at Germany (DE), i know it's not what is required.
here it is based on the template where we cannot modify based on matched string value. Is there any way to achieve this ?
Thanks in advance
You may use
var value: NSMutableString = "It is live now at Germany(DE) or (SOFE)"
let pattern = "(?<=\\G(?<!\\A)|\\()[A-Za-z](?=[A-Za-z]+\\))"
let regex = try? NSRegularExpression(pattern: pattern)
regex?.replaceMatches(in: value, options: .reportProgress, range: NSRange(location: 0,length: value.length), withTemplate: "$0 ")
print(value)
Or just
let val = "It is live now at Germany(DE) or (SOFE)"
let pattern = "(?<=\\G(?<!\\A)|\\()[A-Za-z](?=[A-Za-z]+\\))"
print( val.replacingOccurrences(of: pattern, with: "$0 ", options: .regularExpression, range: nil) )
Output: It is live now at Germany(D E) or (S O F E)
Pattern details
(?<=\\G(?<!\\A)|\\() - a positive lookbehind that matches a location right after ( or at the end of the preceding successful match
[A-Za-z] - matches and consumes any ASCII letter
(?=[A-Za-z]+\\)) - a positive lookahead that matches a location that is immediately followed with 1+ ASCII letters and then a ) char.
The $0 in the replacement inserts the whole match value back into the resulting string.
Imagine the following sentence:
The **quick** brown **fox** ...
If I run the following regex
let boldPattern = "\\*{2}([\\w ]+)\\*{2}"
let boldRegex = try NSRegularExpression(pattern: boldPattern)
let str = "The **quick** brown **fox** ..."
let results = regex.matches(in: str, range: NSRange(str.startIndex..., in: str))
results.forEach {
print("$0")
}
I'm able to get all the words between the **.
I have read about negative regex, which returns every words except the ones we're trying to "avoid".
Given the above sentence, is there a way to use a negative regex to get words that doesn't match the boldPattern, so I would get The brown ..., avoiding the **words** ??
Edit
I'm looking for something around this pattern ((?!\\*{2}([\\w ]+)\\*{2}).*) , something that searches for words that doesn't start with ** word **
One way to do it is to use NSRegularExpression's stringByReplacingMatches method.
Example:
let boldPattern = "\\*{2}([\\w ]+)\\*{2}"
let boldRegex = try NSRegularExpression(pattern: boldPattern)
let str = "The **quick** brown **fox** ..."
let unmatchedString = boldRegex.stringByReplacingMatches(in: str, options: [], range: NSMakeRange(0, str.count), withTemplate: "")
print(unmatchedString)
// prints: The brown ...
I am trying to find the range of specific substrings of a string. Each substring begins with a hashtag and can have any character it likes within it (including emojis). Duplicate hashtags should be detected at distinct ranges. A kind user from here suggested this code:
var str = "The range of #hashtag should be different to this #hashtag"
let regex = try NSRegularExpression(pattern: "(#[A-Za-z0-9]*)", options: [])
let matches = regex.matchesInString(str, options:[], range:NSMakeRange(0, str.characters.count))
for match in matches {
print("match = \(match.range)")
}
However, this code does not work for emojis. What would be the regex expression to include emojis? Is there a way to detect a #, followed by any character up until a space/line break?
Similarly as in Swift extract regex matches,
you have to pass an NSRange to the match functions, and the
returned ranges are NSRanges as well. This can be achieved
by converting the given text to an NSString.
The #\S+ pattern matches a # followed by one or more
non-whitespace characters.
let text = "The 😀range of #hashtag🐶 should 👺 be 🇩🇪 different to this #hashtag🐮"
let nsText = text as NSString
let regex = try NSRegularExpression(pattern: "#\\S+", options: [])
for match in regex.matchesInString(text, options: [], range: NSRange(location: 0, length: nsText.length)) {
print(match.range)
print(nsText.substringWithRange(match.range))
}
Output:
(15,10)
#hashtag🐶
(62,10)
#hashtag🐮
You can also convert between NSRange and Range<String.Index>
using the methods from NSRange to Range<String.Index>.
Remark: As #WiktorStribiżew correctly noticed, the above pattern
will include trailing punctuation (commas, periods, etc). If
that is not desired then
let regex = try NSRegularExpression(pattern: "#[^[:punct:][:space:]]+", options: [])
would be an alternative.