What would the regex be (to be used in IOS, "NSRegularExpression") to get the "words" from a string delimited by a space(s), i.e. could be " ", or " ", or " " etc as the delimited.
So therefore:
"26:43:33 S 153:02:51 E"
Would give:
1-"26:43:33"
2-"S"
3-"153:02:51"
4-"E"
So therefore:
"26:43:33 S 153:02:51 E"
Would give:
1-"26:43:33"
2-"S"
3-"153:02:51"
4-"E"
So if you're going to use a regex for this, you want to look for all contiguous stretches of not-space. Like this:
let s = "26:43:33 S 153:02:51 E" as NSString
let pattern = "[^ ]+"
let reg = try! NSRegularExpression(pattern: pattern, options: [])
let matches = reg.matchesInString(s as String, options: [], range: NSMakeRange(0, s.length))
let result = matches.map {s.substringWithRange($0.range)}
// result is: ["26:43:33", "S", "153:02:51", "E"]
As an alternative to regex, I would suggest using the split method on your string.
let string = "26:43:33 S 153:02:51 E"
let words = string.characters.split { $0 == " " }.map { String($0) }
Because calling split on the characters property will return an array of Character types, we need to use the map method to convert them back to strings. map will perform a closure on each element of a collection. In this case we just use it to cast each element to a String
Related
Need to remove part of the string based on brackets.
For ex:
let str = "My Account (_1234)"
I wanted to remove inside brackets (_1234) and the result should be string My Account
Expected Output:
My Account
How can we achieve this with regex format or else using swift default class,need support on this.
Tried with separatedBy but it splits string into array but that is not the expected output.
str.components(separatedBy: "(")
Use replacingOccurrences(…)
let result = str.replacingOccurrences(of: #"\(.*\)"#,
with: "",
options: .regularExpression)
.trimmingCharacters(in: .whitespaces))
Regex ist not necessary. Just get the range of ( and extract the substring up to the lower bound of the range
let str = "My Account (_1234)"
if let range = str.range(of: " (") {
let account = String(str[..<range.lowerBound])
print(account)
}
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.
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 ï 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
}
I have a string: "Hey #username that's funny". For a given string, how can I search the string to return all ranges of string with first character # and last character to get the username?
I suppose I can get all indexes of # and for each, get the substringToIndex of the next space character, but wondering if there's an easier way.
If your username can contain only letters and numbers, you can use regular expression for that:
let s = "Hey #username123 that's funny"
if let r = s.rangeOfString("#\\w+", options: NSStringCompareOptions.RegularExpressionSearch) {
let name = s.substringWithRange(r) // #username123"
}
#Vladimir's answer is correct, but if you're trying to find multiple occurrences of "username", this should also work:
let s = "Hey #username123 that's funny"
let ranges: [NSRange]
do {
// Create the regular expression.
let regex = try NSRegularExpression(pattern: "#\\w+", options: [])
// Use the regular expression to get an array of NSTextCheckingResult.
// Use map to extract the range from each result.
ranges = regex.matchesInString(s, options: [], range: NSMakeRange(0, s.characters.count)).map {$0.range}
}
catch {
// There was a problem creating the regular expression
ranges = []
}
for range in ranges {
print((s as NSString).substringWithRange(range))
}