How could I remove the last character from a string if it is a punctuation, in ruby?

Gah,开发者_开发技巧 regex is slightly confusing.

I'm trying to remove all possible punctuation characters at the end of a string:

if str[str.length-1] == '?' || str[str.length-1] == '.' || str[str.length-1] == '!' or str[str.length-1] == ',' || str[str.length-1] == ';' 
    str.chomp!
end

I'm sure there's a better way to do this. Any pointers?

str.sub!(/[?.!,;]?$/, '')

• [?.!,;] - Character class. Matches any of those 5 characters (note, . is not special in a character class)
• ? - Previous character or group is optional
• $ - End of string.

This basically replaces an optional punctuation character at the end of the string with the empty string. If the character there isn't punctuation, it's a no-op.

The original question stated 'Remove all possible punctuation characters at the end of a string," but the example only mentioned showed 5, "?", ".", "!", ",", ";". Presumably the other punctuation characters such as ":", '"', etc. should be included in "all possible punctuation characters," so use the :punct: character class as noted by kurumi:

str.sub!(/[[:punct:]]?$/,'')

all non-word characters.

text.gsub(/\W$/, "")

what this does is does is looks at the string, finds the punctuation at the end, and globally substitutes with nothing = thus removing it.

This really does work, and it's a ruby way to use regex.

You should be able to use a regular expression to do this. I don't have any experience with ruby, but here's a RegEx that should work:

/(.*)[^\w\d]?$/

You can use backreference #1 to get the string without the last punctuation character.

mystring.gsub(/[[:punct:]]+$/,"")

Without using the empty string:

str = /[?.!,;]?\z/.match(str).pre_match

or in the other order

str = str.match(/[?.!,;]?\z/).pre_match

str.chomp! won't do anything in this case. You'd have to do str.chop!