Sorry if the title is not clear.
What I mean is this:
If I have a variable, we'll call that a, with a value of "Hello\nWorld", it would be written as
var a = "Hello\nWorld
And if I were to print it, I'd get
Hello
World
How could I print it as:
Hello\nWorld
I know this is a little old however I was looking for a solution to the same problem and I figured out something easy.
If you're wanting to print out a string that shows the escape characters like "\nThis Thing\nAlso this"
print(myString.debugDescription)
Here's a more complete version of #Pedro Castilho's answer.
import Foundation
extension String {
static let escapeSequences = [
(original: "\0", escaped: "\\0"),
(original: "\\", escaped: "\\\\"),
(original: "\t", escaped: "\\t"),
(original: "\n", escaped: "\\n"),
(original: "\r", escaped: "\\r"),
(original: "\"", escaped: "\\\""),
(original: "\'", escaped: "\\'"),
]
mutating func literalize() {
self = self.literalized()
}
func literalized() -> String {
return String.escapeSequences.reduce(self) { string, seq in
string.replacingOccurrences(of: seq.original, with: seq.escaped)
}
}
}
let a = "Hello\0\\\t\n\r\"\'World"
print("Original: \(a)\r\n\r\n\r\n")
print("Literalized: \(a.literalized())")
You can't, not without changing the string itself. The \n character sequence only exists in your code as a representation of a newline character, the compiler will change it into an actual newline.
In other words, the issue here is that the "raw" string is the string with the actual newline.
If you want it to appear as an actual \n, you'll need to escape the backslash. (Change it into \\n)
You could also use the following function to automate this:
func literalize(_ string: String) -> String {
return string.replacingOccurrences(of: "\n", with: "\\n")
.replacingOccurrences(of: "\t", with: "\\t")
}
And so on. You can add more replacingOccurrences calls for every escape sequence you want to literalize.
If "Hello\nWorld" is literally the string you're trying to print, then all you do is this:
var str = "Hello\\nWorld"
print(str)
I tested this in the Swift Playgrounds!
Late to the party but the answer to this question is to map the String UnicodeScalarView Unicode.Scalar elements converting them to escaped ascii strings. Then you can simply join back the string:
extension Unicode.Scalar {
var asciiEscaped: String { escaped(asASCII: true) }
}
extension StringProtocol {
var asciiEscaped: String {
unicodeScalars.map(\.asciiEscaped).joined()
}
}
print("Hello\nWorld".asciiEscaped) // Hello\nWorld
Just use double \
var a = "Hello\\nWorld"
Related
I'm trying to replace some HTML codes in Swift with the appropriate characters. I used a String extension.
extension String {
mutating func fix_HTML_Codes() {
let originalString = self
let newString = originalString.replacingOccurrences(of: "'", with: "\'")
let newString2 = newString.replacingOccurrences(of: """, with: "\"")
self = newString2
}
}
However, instead of replacing my escaped single quote with a single quote, it actually replaces it with \', anyone know why?
Here's an example of what I'm getting:
"On which Beatles album would you find the song \'Eleanor Rigby\'?"
It's including the escape character.
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:
â\tâ (U+0009 CHARACTER TABULATION)
â â (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
}
}
While I was trying to use Swift's String(format: format, args) method, I found out that I cannot print the formatted string directly without newline(\n) and brackets being added.
So for example, I have a code like this:
func someFunc(string: String...) {
print(String(format: "test %#", string))
}
let string = "string1"
someFunc(string, "string2")
the result would be:
"test (\n string1,\n string2\n)\n"
However, I intend to deliver the result like this:
"test string1 string2"
How can I make the brackets and \n not being printed?
Since string parameter is a sequence, you can use joinWithSeparator on it, like this:
func someFunc(string: String...) {
print(string.joinWithSeparator(" "))
}
someFunc("quick", "brown", "fox", "jumps", "over", "the", "lazy", "dog")
You will first need to concatenate your list of input strings, otherwise it will print as a list, hence the commas and parentheses. You can additionally strip out the newline characters and whitespace from the ends of the string using a character set.
var stringConcat : String = ''
for stringEntry in string {
stringConcat += stringEntry.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet())
}
Then just print out the stringConcat.
I am trying to get the contents of a string:
var oldString = "hi this \t is my \t string"
oldString = String(oldString.componentsSeparatedByString("\t"))
print(oldString[1])
I get the ERROR : " 'subscript' is unavailable:cannot subscript String
with an Int..."
How would I access a string that is contained in another string?
The only way I have come up with is to get character by character using:
for index in content.characters.indices{
print(String(oldString[index]))
}
The code above results in:
h,
i,
t,
h,
..
I need:
hi this,
is my,
string
Thank you in advance!
You should read the error message and figure out where does the error message come from.
Your String(oldString.componentsSeparatedByString("\t")) gives you a String not [String].
What you need to do is assigning oldString.componentsSeparatedByString("\t") to an array:
let stringArray = oldString.componentsSeparatedByString("\t")
for str in stringArray {
print(str)
}
In swift you can extend any type and add overloads for different operations. In the example below we've created an extension that allows you to subscript String returning each word and get an array from your string.
Simply paste this into a playground to test:
extension String {
func array() -> [String] {
return self.componentsSeparatedByString("\t")
}
subscript (i: Int) -> String {
return self.componentsSeparatedByString("\t")[i]
}
}
Once you've added your extension, you can use it throughout your application like so:
var str = "hi this \t is my \t string"
print(str[0]) //prints hi this
for i in str.array() {
print(i)
}
prints:
hi this
is my
string
var oldString = "hi this \t is my \t string"
let stringArray = oldString.componentsSeparatedByString("\t")
//In case you also need the index while iterating
for (index, value) in stringArray.enumerate(){
print("index is \(index) and string is \(value)")
}
for str in stringArray{
print(str)
}
Output will be as follows
index is 0 and string is hi this
index is 1 and string is is my
index is 2 and string is string
hi this
is my
string
I'm trying to remove the last punctuation of a string in swift 2.0
var str: String = "This is a string, but i need to remove this comma, \n"
var trimmedstr: String = str.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet())
First I'm removing the the white spaces and newline characters at the end, and then I need to check of the last character of trimmedstr if it is a punctuation. It can be a period, comma, dash, etc, and if it is i need to remove it it.
How can i accomplish this?
There are multiple ways to do it. You can use contains to check if the last character is in the set of expected characters, and use dropLast() on the String to construct a new string without the last character:
let str = "This is a string, but i need to remove this comma, \n"
let trimmedstr = str.trimmingCharacters(in: .whitespacesAndNewlines)
if let lastchar = trimmedstr.last {
if [",", ".", "-", "?"].contains(lastchar) {
let newstr = String(trimmedstr.dropLast())
print(newstr)
}
}
Could use .trimmingCharacters(in:.whitespacesAndNewlines) and .trimmingCharacters(in: .punctuationCharacters)
for example, to remove whitespaces and punctuations on both ends of the String-
let str = "\n This is a string, but i need to remove this comma and whitespaces, \t\n"
let trimmedStr = str.trimmingCharacters(in:
.whitespacesAndNewlines).trimmingCharacters(in: .punctuationCharacters)
Result -
This is a string, but i need to remove this comma and whitespaces