Just trying to do a play around with Swift String. I want to replace all the characters in a String to a blank space.
inputString = "This is a String"
outputString = " "
How could I achieve this in swift?
You can use the map function to replace any character with another:
String("foo".characters.map { _ in Character(" ") })
One possible way:
outputString = String(count: inputString.characters.count, repeatedValue: (" " as Character))
Another way:
let outputString = inputString.replacingOccurrences(of: "[^\\s]",
with: " ",
options: .regularExpression,
range: inputString.startIndex..<inputString.endIndex)
So given an input String of n chars you want another String of n blank spaces, right?
let inputString = "This is a String"
let outputString = String([Character](count: inputString.characters.count, repeatedValue: " "))
Related
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 multiple lines in an String, for example:
let str = "Mieter: Hannes Tester \nVerwalter: Michael Karner \n"
Now i want to remove the whole sentence between "Mieter" and the line break. So the result should be:
let str = "Verwalter: Michael Karner \n"
I could check with Regex, but i am only able to get the string between 2 words. For example with:
if let match = str.rangeOfString("(?<=Mieter)[^\n]+", options: .RegularExpressionSearch) {
print(str.substringWithRange(match)) // between
}
But how can i replace a whole line?
Edit:
It is not working anymore when its between a string:
let str = "Test \n Mieter: \n Hausverwalter: \n Firma: \n"
str.stringByReplacingOccurrencesOfString("^Mieter[^\n]+\\s", withString: "", options: .RegularExpressionSearch, range: nil)
// so "Mieter \n" is not being replaced.
Use replacingOccurrencesOf
let str = "Mieter: Hannes Tester \nVerwalter: Michael Karner \n"
str.replacingOccurrences(of: "^Mieter[^\n]+\\s", with: "", options: .regularExpression)
This is Swift 3 code
Edit:
If the substring is not at the beginning of the string remove the leading caret (^).
a string such as ! !! yuahl! ! , I want delete ! and , when I code like this
for index in InputName.characters.indices {
if String(InputName[index]) == "" || InputName.substringToIndex(index) == "!" {
InputName.removeAtIndex(index)
}
}
have this error " fatal error: subscript: subRange extends past String end ", how should I do? THX :D
Swift 5+
let myString = "aaaaaaaabbbb"
let replaced = myString.replacingOccurrences(of: "bbbb", with: "") // "aaaaaaaa"
If you need to remove characters only on both ends, you can use stringByTrimmingCharactersInSet(_:)
let delCharSet = NSCharacterSet(charactersInString: "! ")
let s1 = "! aString! !"
let s1Del = s1.stringByTrimmingCharactersInSet(delCharSet)
print(s1Del) //->aString
let s2 = "! anotherString !! aString! !"
let s2Del = s2.stringByTrimmingCharactersInSet(delCharSet)
print(s2Del) //->anotherString !! aString
If you need to remove characters also in the middle, "reconstruct from the filtered output" would be a little bit more efficient than repeating single character removal.
var tempUSView = String.UnicodeScalarView()
tempUSView.appendContentsOf(s2.unicodeScalars.lazy.filter{!delCharSet.longCharacterIsMember($0.value)})
let s2DelAll = String(tempUSView)
print(s2DelAll) //->anotherStringaString
If you don't mind generating many intermediate Strings and Arrays, this single liner can generate the expected output:
let s2DelAll2 = s2.componentsSeparatedByCharactersInSet(delCharSet).joinWithSeparator("")
print(s2DelAll2) //->anotherStringaString
I find that the filter method is a good way to go for this sort of thing:
let unfiltered = "! !! yuahl! !"
// Array of Characters to remove
let removal: [Character] = ["!"," "]
// turn the string into an Array
let unfilteredCharacters = unfiltered.characters
// return an Array without the removal Characters
let filteredCharacters = unfilteredCharacters.filter { !removal.contains($0) }
// build a String with the filtered Array
let filtered = String(filteredCharacters)
print(filtered) // => "yeah"
// combined to a single line
print(String(unfiltered.characters.filter { !removal.contains($0) })) // => "yuahl"
Swift 3
In Swift 3, the syntax is a bit nicer. As a result of the Great Swiftification of the old APIs, the factory method is now called trimmingCharacters(in:). Also, you can construct the CharacterSet as a Set of single-character Strings:
let string = "! !! yuahl! !"
string.trimmingCharacters(in: [" ", "!"]) // "yuahl"
If you have characters in the middle of the string you would like to remove as well, you can use components(separatedBy:).joined():
let string = "! !! yu !ahl! !"
string.components(separatedBy: ["!", " "]).joined() // "yuahl"
H/T #OOPer for the Swift 2 version
func trimLast(character chars: Set<Character>) -> String {
let str: String = String(self.reversed())
guard let index = str.index(where: {!chars.contains($0)}) else {
return self
}
return String((str[index..<str.endIndex]).reversed())
}
Note:
By adding this function in String extension, you can delete the specific character of string at last.
for index in InputName.characters.indices.reversed() {
if String(InputName[index]) == "" || InputName.substringToIndex(index) == "!" {
InputName.removeAtIndex(index)
}
}
Also you can add such very helpful extension :
import Foundation
extension String{
func exclude(find:String) -> String {
return stringByReplacingOccurrencesOfString(find, withString: "", options: .CaseInsensitiveSearch, range: nil)
}
func replaceAll(find:String, with:String) -> String {
return stringByReplacingOccurrencesOfString(find, withString: with, options: .CaseInsensitiveSearch, range: nil)
}
}
you can use this:
for example if you want to remove "%" the percent from 10%
if let i = text.firstIndex(of: "%") {
text.remove(at: i) //10
}
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
How would I split a string to include the separators?
Lets say I had a string such as...
let myString = "apple banana orange grapes"
If I used
let separatedString = myString.componentsSeparatedByString(" ")
my resulting array would be
["apple","banana","orange","grapes"]
How would I achieve a result of
["apple ","banana ","orange ","grapes"]
array.map lets you process the resulting array an add the space back in.
let separatedString = myString
.componentsSeparatedByString(" ")
.map { "\($0) " }
That last line iterates over all strings in the split up array and puts them in $0, and returns a new string with the space added back in which gets used as the replacement for the original string.
Alternative using regular expression:
let myString = "apple banana orange grapes"
let pattern = "\\w+\\s?"
let regex = try! NSRegularExpression(pattern: pattern, options: [])
let matches = regex.matchesInString(myString, options:[], range: NSMakeRange(0, myString.characters.count))
.map { (myString as NSString).substringWithRange($0.range)}
print(matches) // -> ["apple ", "banana ", "orange ", "grapes"]
Solution
Since you updated your question, it looks now you no longer want a new space on the last word.
So here's my updated code
let text = "apple banana orange grapes"
let chunks: [String] = text
.componentsSeparatedByString(" ")
.reverse()
.enumerate()
.map { $0.element + ( $0.index == 0 ? "" : " ") }
.reverse()
print(chunks) // ["apple ", "banana ", "orange ", "grapes"]
Multiple separators
Thank to #vadian for the suggestion
let text = "apple banana\norange grapes"
let chunks: [String] = text
.componentsSeparatedByCharactersInSet(.whitespaceAndNewlineCharacterSet())
.reverse()
.enumerate()
.map { $0.element + ( $0.index == 0 ? "" : " ") }
.reverse()
print(chunks) // ["apple ", "banana ", "orange ", "grapes"]