This question already has answers here:
regex to get string between two % characters
(2 answers)
Closed 2 years ago.
Hi am trying to get values of a string which is imputed in the form "40|LQ,FP,MD,GR \"Dinner out\"". The string value I am trying to extract is Dinner out the text could be different though like ride in but it still follows the same pattern. how can I extract this string value from the rest of the character using regex of any alternative.
You can try
var text = "40|LQ,FP,MD,GR \"Dinner out\""
var regex = try? NSRegularExpression(pattern: "\\\".+?\\\"", options: [])
var matches = regex?.matches(in: text, options: []
, range: NSMakeRange(0, text.count)) ?? []
for match in matches {
let r = (text as NSString).substring(with: match.range)
print("found= \(r)")
}
I would recommend you to have a look at this question and its answers.
Swift Get string between 2 strings in a string
Quoting answer here just for your ease.
extension String {
func slice(from: String, to: String) -> String? {
return (range(of: from)?.upperBound).flatMap { substringFrom in
(range(of: to, range: substringFrom..<endIndex)?.lowerBound).map { substringTo in
String(self[substringFrom..<substringTo])
}
}
}
}
This question already has answers here:
How should I remove all the leading spaces from a string? - swift
(31 answers)
Closed 4 years ago.
i'm getting my contacts number locally from my mobile. There are some number in which there are white spaces between numbers. I'm trying to remove the white spaces from the number but it isn't working,this is how i'm removing the white spaces,
let number = contact.phoneNumbers.first?.value.stringValue
let formattedString = number?.replacingOccurrences(of: " ", with: "")
print(formattedString)
But when i print this is what i got in the console,
+92 324 4544783
The white sapces are still coming how can i remove that?
Here you go: source
For trimming white spaces from both ends, you can use:
let number = contact.phoneNumbers.first?.value.stringValue
let formattedString = number.trimmingCharacters(in: .whitespacesAndNewlines)
print(formattedString)
For removing whitespaces that might be inside the string, use:
let x = "+92 300 7681277"
let result = x.replacingOccurrences(of: " ", with: "")
You should get:
result = +923007681277
EDIT: I updated my answer.
let number = contact.phoneNumbers.first?.value.stringValue
let number_without_space = number.components(separatedBy: .whitespaces).joined()
print(number_without_space) //use this variable wherever you want to use
joined() is a function it will join your string after removing spaces like this
let str = "String Name"
str.components(separatedBy: .whitespaces).joined()
extension to remove spaces
extension String
{
func removeSpaces() -> String {
return components(separatedBy: .whitespaces).joined()
}
}
This question already has answers here:
How to split a string into substrings of equal length
(13 answers)
Closed 4 years ago.
Like the question says, if I have:
XQQ230IJFEKJLDSAIOUOIDSAUIFOPDSFE28
How can I split this string at every 8th character to get:
XQQ230IJ FEKJLDSA IOUOIDSA UIFOPDSA
Implement this function
extension String {
func inserting(separator: String, every n: Int) -> String {
var result: String = ""
let characters = Array(self.characters)
stride(from: 0, to: characters.count, by: n).forEach {
result += String(characters[$0..<min($0+n, characters.count)])
if $0+n < characters.count {
result += separator
}
}
return result
}
}
call it this way,
let str = "XQQ230IJFEKJLDSAIOUOIDSAUIFOPDSFE28"
let final = str.inserting(separator: " ", every: 8)
print(final)
Output will be like this,
XQQ230IJ FEKJLDSA IOUOIDSA UIFOPDSF E28
This will be generic solution if you want to add any character instead of space, it will work.
This question already has answers here:
How to capitalize each word in a string using Swift iOS
(8 answers)
Closed 5 years ago.
I'm getting strings from a server that is all caps like:
"HELLO WORLD"
but I'm trying to make is so each word is caps on its own like:
"Hello World"
I've tried this:
extension String {
func capitalizingFirstLetter() -> String {
let first = String(characters.prefix(1)).capitalized
let other = String(characters.dropFirst()).lowercased()
return first + other
}
mutating func capitalizeFirstLetter() {
self = self.capitalizingFirstLetter()
}
}
but the result is
"Hello world"
any idea on how to do this?
Apple already did that for you:
print("HELLO WORLD".capitalized)
Documentation: https://developer.apple.com/documentation/foundation/nsstring/1416784-capitalized
Every example of trimming strings in Swift remove both leading and trailing whitespace, but how can only trailing whitespace be removed?
For example, if I have a string:
" example "
How can I end up with:
" example"
Every solution I've found shows trimmingCharacters(in: CharacterSet.whitespaces), but I want to retain the leading whitespace.
RegEx is a possibility, or a range can be derived to determine index of characters to remove, but I can't seem to find an elegant solution for this.
With regular expressions:
let string = " example "
let trimmed = string.replacingOccurrences(of: "\\s+$", with: "", options: .regularExpression)
print(">" + trimmed + "<")
// > example<
\s+ matches one or more whitespace characters, and $ matches
the end of the string.
In Swift 4 & Swift 5
This code will also remove trailing new lines.
It works based on a Character struct's method .isWhitespace
var trailingSpacesTrimmed: String {
var newString = self
while newString.last?.isWhitespace == true {
newString = String(newString.dropLast())
}
return newString
}
This short Swift 3 extension of string uses the .anchored and .backwards option of rangeOfCharacter and then calls itself recursively if it needs to loop. Because the compiler is expecting a CharacterSet as the parameter, you can just supply the static when calling, e.g. "1234 ".trailing(.whitespaces) will return "1234". (I've not done timings, but would expect faster than regex.)
extension String {
func trailingTrim(_ characterSet : CharacterSet) -> String {
if let range = rangeOfCharacter(from: characterSet, options: [.anchored, .backwards]) {
return self.substring(to: range.lowerBound).trailingTrim(characterSet)
}
return self
}
}
In Foundation you can get ranges of indices matching a regular expression. You can also replace subranges. Combining this, we get:
import Foundation
extension String {
func trimTrailingWhitespace() -> String {
if let trailingWs = self.range(of: "\\s+$", options: .regularExpression) {
return self.replacingCharacters(in: trailingWs, with: "")
} else {
return self
}
}
}
You can also have a mutating version of this:
import Foundation
extension String {
mutating func trimTrailingWhitespace() {
if let trailingWs = self.range(of: "\\s+$", options: .regularExpression) {
self.replaceSubrange(trailingWs, with: "")
}
}
}
If we match against \s* (as Martin R. did at first) we can skip the if let guard and force-unwrap the optional since there will always be a match. I think this is nicer since it's obviously safe, and remains safe if you change the regexp. I did not think about performance.
Handy String extension In Swift 4
extension String {
func trimmingTrailingSpaces() -> String {
var t = self
while t.hasSuffix(" ") {
t = "" + t.dropLast()
}
return t
}
mutating func trimmedTrailingSpaces() {
self = self.trimmingTrailingSpaces()
}
}
Swift 4
extension String {
var trimmingTrailingSpaces: String {
if let range = rangeOfCharacter(from: .whitespacesAndNewlines, options: [.anchored, .backwards]) {
return String(self[..<range.lowerBound]).trimmingTrailingSpaces
}
return self
}
}
Demosthese's answer is a useful solution to the problem, but it's not particularly efficient. This is an upgrade to their answer, extending StringProtocol instead, and utilizing Substring to remove the need for repeated copying.
extension StringProtocol {
#inline(__always)
var trailingSpacesTrimmed: Self.SubSequence {
var view = self[...]
while view.last?.isWhitespace == true {
view = view.dropLast()
}
return view
}
}
No need to create a new string when dropping from the end each time.
extension String {
func trimRight() -> String {
String(reversed().drop { $0.isWhitespace }.reversed())
}
}
This operates on the collection and only converts the result back into a string once.
It's a little bit hacky :D
let message = " example "
var trimmed = ("s" + message).trimmingCharacters(in: .whitespacesAndNewlines)
trimmed = trimmed.substring(from: trimmed.index(after: trimmed.startIndex))
Without regular expression there is not direct way to achieve that.Alternatively you can use the below function to achieve your required result :
func removeTrailingSpaces(with spaces : String) -> String{
var spaceCount = 0
for characters in spaces.characters{
if characters == " "{
print("Space Encountered")
spaceCount = spaceCount + 1
}else{
break;
}
}
var finalString = ""
let duplicateString = spaces.replacingOccurrences(of: " ", with: "")
while spaceCount != 0 {
finalString = finalString + " "
spaceCount = spaceCount - 1
}
return (finalString + duplicateString)
}
You can use this function by following way :-
let str = " Himanshu "
print(removeTrailingSpaces(with : str))
One line solution with Swift 4 & 5
As a beginner in Swift and iOS programming I really like #demosthese's solution above with the while loop as it's very easy to understand. However the example code seems longer than necessary. The following uses essentially the same logic but implements it as a single line while loop.
// Remove trailing spaces from myString
while myString.last == " " { myString = String(myString.dropLast()) }
This can also be written using the .isWhitespace property, as in #demosthese's solution, as follows:
while myString.last?.isWhitespace == true { myString = String(myString.dropLast()) }
This has the benefit (or disadvantage, depending on your point of view) that this removes all types of whitespace, not just spaces but (according to Apple docs) also including newlines, and specifically the following characters:
“\t” (U+0009 CHARACTER TABULATION)
“ “ (U+0020 SPACE)
U+2029 PARAGRAPH SEPARATOR
U+3000 IDEOGRAPHIC SPACE
Note: Even though .isWhitespace is a Boolean it can't be used directly in the while loop as it ends up being optional ? due to the chaining of the optional .last property, which returns nil if the String (or collection) is empty. The == true logic gets around this since nil != true.
I'd love to get some feedback on this, esp. in case anyone sees any issues or drawbacks with this simple single line approach.
Swift 5
extension String {
func trimTrailingWhiteSpace() -> String {
guard self.last == " " else { return self }
var tmp = self
repeat {
tmp = String(tmp.dropLast())
} while tmp.last == " "
return tmp
}
}