How to strip leading and trailing quote from string, in Ruby - ruby

Question

I want to strip leading and trailing quotes, in Ruby, from a string. The quote character will occur 0 or 1 time. For example, all of the following should be converted to foo,bar:
"foo,bar"
"foo,bar
foo,bar"
foo,bar

Answer 1

You could also use the chomp function, but it unfortunately only works in the end of the string, assuming there was a reverse chomp, you could:
'"foo,bar"'.rchomp('"').chomp('"')
Implementing rchomp is straightforward:
class String
def rchomp(sep = $/)
self.start_with?(sep) ? self[sep.size..-1] : self
end
end
Note that you could also do it inline, with the slightly less efficient version:
'"foo,bar"'.chomp('"').reverse.chomp('"').reverse
EDIT: Since Ruby 2.5, rchomp(x) is available under the name delete_prefix, and chomp(x) is available as delete_suffix, meaning that you can use
'"foo,bar"'.delete_prefix('"').delete_suffix('"')

Answer 2

I can use gsub to search for the leading or trailing quote and replace it with an empty string:
s = "\"foo,bar\""
s.gsub!(/^\"|\"?$/, '')
As suggested by comments below, a better solution is:
s.gsub!(/\A"|"\Z/, '')

Answer 3

As usual everyone grabs regex from the toolbox first. :-)
As an alternate I'll recommend looking into .tr('"', '') (AKA "translate") which, in this use, is really stripping the quotes.

Answer 4

Another approach would be
remove_quotations('"foo,bar"')
def remove_quotations(str)
if str.start_with?('"')
str = str.slice(1..-1)
end
if str.end_with?('"')
str = str.slice(0..-2)
end
end
It is without RegExps and start_with?/end_with? are nicely readable.

Answer 5

It frustrates me that strip only works on whitespace. I need to strip all kinds of characters! Here's a String extension that will fix that:
class String
def trim sep=/\s/
sep_source = sep.is_a?(Regexp) ? sep.source : Regexp.escape(sep)
pattern = Regexp.new("\\A(#{sep_source})*(.*?)(#{sep_source})*\\z")
self[pattern, 2]
end
end
Output
'"foo,bar"'.trim '"' # => "foo,bar"
'"foo,bar'.trim '"' # => "foo,bar"
'foo,bar"'.trim '"' # => "foo,bar"
'foo,bar'.trim '"' # => "foo,bar"
' foo,bar'.trim # => "foo,bar"
'afoo,bare'.trim /[aeiou]/ # => "foo,bar"

Answer 6

Assuming that quotes can only appear at the beginning or end, you could just remove all quotes, without any custom method:
'"foo,bar"'.delete('"')

Answer 7

I wanted the same but for slashes in url path, which can be /test/test/test/ (so that it has the stripping characters in the middle) and eventually came up with something like this to avoid regexps:
'/test/test/test/'.split('/').reject(|i| i.empty?).join('/')
Which in this case translates obviously to:
'"foo,bar"'.split('"').select{|i| i != ""}.join('"')
or
'"foo,bar"'.split('"').reject{|i| i.empty?}.join('"')

Answer 8

Regexs can be pretty heavy and lead to some funky errors. If you are not dealing with massive strings and the data is pretty uniform you can use a simpler approach.
If you know the strings have starting and leading quotes you can splice the entire string:
string = "'This has quotes!'"
trimmed = string[1..-2]
puts trimmed # "This has quotes!"
This can also be turned into a simple function:
# In this case, 34 is \" and 39 is ', you can add other codes etc.
def trim_chars(string, char_codes=[34, 39])
if char_codes.include?(string[0]) && char_codes.include?(string[-1])
string[1..-2]
else
string
end
end

Answer 9

You can strip non-optional quotes with scan:
'"foo"bar"'.scan(/"(.*)"/)[0][0]
# => "foo\"bar"