Swift remove ONLY trailing spaces from string

Question

many examples in SO are fixing both sides, the leading and trailing. My request is only about the trailing. My input text is: " keep my left side " Desired output: " keep my left side"

Of course this command will remove both ends:

let cleansed = messageText.trimmingCharacters(in: .whitespacesAndNewlines)

Which won't work for me.

How can I do it?

How about rangeOfCharacter(from: .whitespaceAndNewlines.inverted)? — jtbandes
– jtbandes, Commented Dec 31, 2016 at 22:42

vadian · Accepted Answer · 2016-12-31 22:46:30Z

11

A quite simple solution is regular expression, the pattern is one or more(+) whitespace characters(\s) at the end of the string($)

let string = " keep my left side "
let cleansed = string.replacingOccurrences(of: "\\s+$", 
                                         with: "", 
                                      options: .regularExpression)

answered Dec 31, 2016 at 22:46

vadian

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Comments

Obliquely · Accepted Answer · 2017-05-22 15:59:08Z

2

You can use the rangeOfCharacter function on string with a characterSet. This extension then uses recursion of there are multiple spaces to trim. This will be efficient if you only usually have a small number of spaces.

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
    }
}

"1234 ".trailingTrim(.whitespaces)

returns

"1234"

answered May 22, 2017 at 15:59

Obliquely

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Comments

dijipiji · Accepted Answer · 2017-06-19 09:37:34Z

1

Building on vadian's answer I found for Swift 3 at the time of writing that I had to include a range parameter. So:

func trailingTrim(with string : String) -> String {

    let start = string.startIndex
    let end = string.endIndex
    let range: Range<String.Index> = Range<String.Index>(start: start, end: end)


    let cleansed:String = string.stringByReplacingOccurrencesOfString("\\s+$",
                                                                      withString: "",
                                                                      options: .RegularExpressionSearch,
                                                                      range: range)

    return cleansed
}

answered Jun 19, 2017 at 9:37

dijipiji

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Comments

user187676 · Accepted Answer · 2020-07-19 13:56:13Z

0

Simple. No regular expressions needed.

extension String {

    func trimRight() -> String {
        let c = reversed().drop(while: { $0.isWhitespace }).reversed()
        return String(c)
    }
}

answered Jul 19, 2020 at 13:56

user187676

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