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How should I remove all the leading spaces from a string? - swift
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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()
}
}
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Hello there fellow Swift devs!
I am a junior dev, and I'm trying to figure out a best way to tokenize / parse Swift String as an exercise.
What I have is a string which looks like this:
let string = "This is a {B}string{/B} and this is a substring."
What I would like to do is, tokenize the string, and change the "strings / tokens" inside the tags you see.
I can see using NSRegularExpression and it's matches, but it feels too generic. I would like to have only say 2 of these tags, that change the text. What would be the best approach in Swift 5.2^?
if let regex = try? NSRegularExpression(pattern: "\\{[a-z0-9]+\}", options: .caseInsensitive) {
let string = self as NSString
return regex.matches(in: self, options: [], range: NSRange(location: 0, length: string.length)).map {
// now $0 is the result? but it won't work for enclosing the tags :/
}
}
If the option of using html tags instead of {B}{/B} is acceptable, then you can use the StringEx library that I wrote for this purpose.
You can select a substring inside the html tag and replace it with another string like this:
let string = "This is a <b>string</b> and this is a substring."
let ex = string.ex
ex[.tag("b")].replace(with: "some value")
print(ex.rawString) // This is a <b>some value</b> and this is a substring.
print(ex.string) // This is a some value and this is a substring.
if necessary, you can also style the selected substrings and get NSAttributedString:
ex[.tag("b")].style([
.font(.boldSystemFont(ofSize: 16)),
.color(.black)
])
myLabel.attributedText = ex.attributedString
Not sure if you have solved it with NLTokenizer or not, but you can certainly solve it with Regx here is how (I have implemented it as generic, in future if you have to handle different kinds of tags and substite different string for them small tweak to the logic should do the job )
override func viewDidLoad() {
super.viewDidLoad()
let regexStr = "(\\{B\\}(\\s*\\w+\\s*)*\\{\\/B\\})"
let regex = try! NSRegularExpression(pattern: regexStr)
var string = "Sandeep {B}Bhandaari{/B} is here{B}Sandeep{/B}"
var foundRanges = [NSRange]()
regex.enumerateMatches(in: string, options: [], range: NSMakeRange(0, string.count)) { (match, flag, stop) in
if let matchRange = match?.range(at: 1) {
foundRanges.append(matchRange)
}
}
let substituteString = "abcd"
var replacedString = string as NSString
let foundRangesCount = foundRanges.count
var currentRange = 0
while foundRangesCount > currentRange {
let range = foundRanges[currentRange]
replacedString = replacedString.replacingCharacters(in: range, with: substituteString) as NSString
reEvaluateAllRanges(ranges: &foundRanges, byOffset: range.length - substituteString.count)
currentRange += 1
}
debugPrint(replacedString)
}
func reEvaluateAllRanges(ranges: inout [NSRange], byOffset: Int) {
var newFoundRange = [NSRange]()
for range in ranges {
newFoundRange.append(NSMakeRange(range.location - byOffset, range.length))
}
ranges = newFoundRange
}
Input: "Sandeep {B}Bhandaari{/B} is here"
Output: Sandeep abcd is here
Input: "Sandeep {B}Bhandaari{/B} is here{B}Sandeep{/B}"
Output: Sandeep abcd is hereabcd
Look at the edge case handling Longer strings replaced by smaller substitute strings and vice versa also detection of string enclosed in tag with / without space
EDIT 1:
Regx (\\{B\\}(\\s*\\w+\\s*)*\\{\\/B\\}) should be self explanatory, incase you need help with understanding it use cheat sheet
regex.enumerateMatches(in: string, options: [], range: NSMakeRange(0, string.count)) { (match, flag, stop) in
if let matchRange = match?.range(at: 1) {
foundRanges.append(matchRange)
}
}
I could have modified substring here itself, but if you have more than one match and if you mutate string evaluated ranges will be corrupted hence am saving all found ranges into an array and apply replace on each one of them later
let substituteString = "abcd"
var replacedString = string as NSString
let foundRangesCount = foundRanges.count
var currentRange = 0
while foundRangesCount > currentRange {
let range = foundRanges[currentRange]
replacedString = replacedString.replacingCharacters(in: range, with: substituteString) as NSString
reEvaluateAllRanges(ranges: &foundRanges, byOffset: range.length - substituteString.count)
currentRange += 1
}
Here am iterating through all found match ranges and replace character in range with substitute string, you can always have a switch / if else ladder inside while loop to look for different types of tags and pass different substitute strings for each tags
func reEvaluateAllRanges(ranges: inout [NSRange], byOffset: Int) {
var newFoundRange = [NSRange]()
for range in ranges {
newFoundRange.append(NSMakeRange(range.location - byOffset, range.length))
}
ranges = newFoundRange
}
This function modifies all the ranges in array using the offset, remember you need to only modify range's location, length remains same
One bit of optimisation you can do is probably get rid of ranges from array for which you have already applied substitute strings
var hello = "hello, how are you?"
var hello2 = "hello, how are you #tom?"
i want to delete every letter behind the # sign.
result should be
var hello2 = "hello, how are you #tom?"
->
hello2.trimmed()
print(hello2.trimmed())
-> "hello, how are you"
Update
As i want to use it to link multiple users and replace the space behind #sign with the correct name, I always need the reference to the latest occurrence of the #sign to replace it.
text3 = "hey i love you #Tom #Marcus #Peter"
Example what the final version should look like
to start off
var text = "hello #tom #mark #mathias"
i want to always get the index of the latest # sign in the text
Expanding on #appzYourLife answer, the following will also trim off the whitespace characters after removing everything after the # symbol.
import Foundation
var str = "hello, how are you #tom"
if str.contains("#") {
let endIndex = str.range(of: "#")!.lowerBound
str = str.substring(to: endIndex).trimmingCharacters(in: .whitespacesAndNewlines)
}
print(str) // Output - "hello, how are you"
UPDATE:
In response to finding the last occurance of the # symbol in the string and removing it, here is how I would approach it:
var str = "hello, how are you #tom #tim?"
if str.contains("#") {
//Reverse the string
var reversedStr = String(str.characters.reversed())
//Find the first (last) occurance of #
let endIndex = reversedStr.range(of: "#")!.upperBound
//Get the string up to and after the # symbol
let newStr = reversedStr.substring(from: endIndex).trimmingCharacters(in: .whitespacesAndNewlines)
//Store the new string over the original
str = String(newStr.characters.reversed())
//str = "hello, how are you #tom"
}
Or looking at #appzYourLife answer use range(of:options:range:locale:) instead of literally reversing the characters
var str = "hello, how are you #tom #tim?"
if str.contains("#") {
//Find the last occurrence of #
let endIndex = str.range(of: "#", options: .backwards, range: nil, locale: nil)!.lowerBound
//Get the string up to and after the # symbol
let newStr = str.substring(from: endIndex).trimmingCharacters(in: .whitespacesAndNewlines)
//Store the new string over the original
str = newStr
//str = "hello, how are you #tom"
}
As an added bonus, here is how I would approach removing every # starting with the last and working forward:
var str = "hello, how are you #tom and #tim?"
if str.contains("#") {
while str.contains("#") {
//Reverse the string
var reversedStr = String(str.characters.reversed())
//Find the first (last) occurance of #
let endIndex = reversedStr.range(of: "#")!.upperBound
//Get the string up to and after the # symbol
let newStr = reversedStr.substring(from: endIndex).trimmingCharacters(in: .whitespacesAndNewlines)
//Store the new string over the original
str = String(newStr.characters.reversed())
}
//after while loop, str = "hello, how are you"
}
let text = "hello, how are you #tom?"
let trimSpot = text.index(of: "#") ?? text.endIndex
let trimmed = text[..<trimSpot]
Since a string is a collection of Character type, it can be accessed as such. The second line finds the index of the # sign and assigns its value to trimSpot, but if it is not there, the endIndex of the string is assigned through the use of the nil coalescing operator
??
The string, or collection of Characters, can be provided a range that will tell it what characters to get. The expression inside of the brackets,
..<trimSpot
is a range from 0 to trimSpot-1. So,
text[..<trimSpot]
returns an instance of type Substring, which points at the original String instance.
You need to find the range of the "#" and then use it to create a substring up to the index before.
import Foundation
let text = "hello, how are you #tom?"
if let range = text.range(of: "#") {
let result = text.substring(to: range.lowerBound)
print(result) // "hello, how are you "
}
Considerations
Please note that, following the logic you described and using the input text you provided, the output string will have a blank space as last character
Also note that if multiple # are presente in the input text, then the first occurrence will be used.
Last index [Update]
I am adding this new section to answer the question you posted in the comments.
If you have a text like this
let text = "hello #tom #mark #mathias"
and you want the index of the last occurrency of "#" you can write
if let index = text.range(of: "#", options: .backwards)?.lowerBound {
print(index)
}
Try regular expressions, they are much safer (if you know what you are doing...)
let hello2 = "hello, how are you #tom, #my #next #victim?"
let deletedStringsAfterAtSign = hello2.replacingOccurrences(of: "#\\w+", with: "", options: .regularExpression, range: nil)
print(deletedStringsAfterAtSign)
//prints "hello, how are you , ?"
And this code removes exactly what you need and leaves the characters after the strings clear, so you can see the , and ? still being there. :)
EDIT: what you asked in comments to this answer:
let hello2 = "hello, how are you #tom, #my #next #victim?"
if let elementIwannaAfterEveryAtSign = hello2.components(separatedBy: " #").last
{
let deletedStringsAfterAtSign = hello2.replacingOccurrences(of: "#\\w+", with: elementIwannaAfterEveryAtSign, options: .regularExpression, range: nil)
print(deletedStringsAfterAtSign)
//prints hello, how are you victim?, victim? victim? victim??
}
I have a question. I write tests for a mobile app. And at this moment, I compare the prices in different currencies like Pound, Euro, India, etc. A possible comparison should be "now £1.678,95". It's no problem for me to cut the "now", to cut the whitespace - short, to get the string in a possible Int or Double formation. BUT now I'm in France. In France the formation is "maintenant 2 500,00 €". No problem with the "maintenant", no problem with the whitespace outside the price and no problem with "€".
BUT there is a space IN the price between 2 and 500. If I run my test, I have only the "2" the rest is gone!
How can I do it, that the whitespace should not cut here. It should backspaced from "2 500,00" to "2500,00".
Hope you have an idea :) Thanks!
My code at this moment is:
var firstPrice = XCUIApplication().collectionViews.cells.element(boundBy: 0).staticTexts.element(boundBy: 2).label
firstPrice = firstPrice.replacingOccurrences(of: "£", with: "")
firstPrice = firstPrice.replacingOccurrences(of: "€", with: "")
firstPrice = firstPrice.replacingOccurrences(of: "₹", with: "")
let firstPriceArray = firstPrice.components(separatedBy: .whitespaces).filter { !$0.isEmpty }
firstPrice = firstPriceArray[1]
let firstPriceTrimmedDouble = firstPrice.toDouble(with: SiteIDHelper.locale(from: SiteIDHelper.SiteID(rawValue: Int(sideIDS))!))!
print(firstPriceTrimmedDouble)
Tough, because you don't know the locale of the string. 1234 Euro can be written as € 1,234.00 in English style, or 1.234,00 € in French and German styles. (Euro is very common in England too).
From the limited examples you provided, you can remove the first word, then deleting all spaces, commas, dots and currency signs from the remaining before converting it to Double:
let priceString = "maintenant 2 500,00 €"
let unwanted = " ,.£€₹"
var doubleValue : Double?
if let range = priceString.range(of: " ") {
let chars = priceString[range.upperBound..<priceString.endIndex]
.characters.filter({ !unwanted.characters.contains($0) })
doubleValue = Double(String(chars))
}
// run your asserts here
Not really clear what you are after... but to remove all whitespaces, try this:
var price: String = ...
while let range = price.rangeOfCharacter(from: .whitespaces) {
price.removeSubrange(range)
}
What's the best way to go about removing the first six characters of a string? Through Stack Overflow, I've found a couple of ways that were supposed to be solutions but I noticed an error with them. For instance,
extension String {
func removing(charactersOf string: String) -> String {
let characterSet = CharacterSet(charactersIn: string)
let components = self.components(separatedBy: characterSet)
return components.joined(separator: "")
}
If I type in a website like https://www.example.com, and store it as a variable named website, then type in the following
website.removing(charactersOf: "https://")
it removes the https:// portion but it also removes all h's, all t's, :'s, etc. from the text.
How can I just delete the first characters?
In Swift 4 it is really simple, just use dropFirst(n: Int)
let myString = "Hello World"
myString.dropFirst(6)
//World
In your case: website.dropFirst(6)
Why not :
let stripped = String(website.characters.dropFirst(6))
Seems more concise and straightforward to me.
(it won't work with multi-char emojis either mind you)
[EDIT] Swift 4 made this even shorter:
let stripped = String(website.dropFirst(6))
length is the number of characters you want to remove (6 in your case)
extension String {
func toLengthOf(length:Int) -> String {
if length <= 0 {
return self
} else if let to = self.index(self.startIndex, offsetBy: length, limitedBy: self.endIndex) {
return self.substring(from: to)
} else {
return ""
}
}
}
It will remove first 6 characters from a string
var str = "Hello-World"
let range1 = str.characters.index(str.startIndex, offsetBy: 6)..<str.endIndex
str = str[range1]
print("the end time is : \(str)")
I have a string like "http://example.com/a/b/c". I want to transform the string to "httpexamplecomabc" in order to save it as the file name. I have tried
let result = str.stringByTrimmingCharactersInSet(NSCharacterSet(charactersInString: "//"))
and
let result = str.stringByTrimmingCharactersInSet(NSCharacterSet(charactersInString: "/"))
but neither works. Any Idea on how to remove "/"? Thanks
You can replace occurrence of "/ " to "" by using
let mm = "http://example.com/a/b/c"
let newString = mm.stringByReplacingOccurrencesOfString("/", withString: "")
print(newString) // http:example.comabc