Escaping quantifier braces after hash in regex - ruby

What is the correct way to escape a hash character followed by quantification with braces in a regex? I want to detect a sequence of hash characters repeated within a given number of times. I want something like:
/#{1, 6}/
without interpolation. I cannot turn off interpolation entirely for the regex because I use it in other portions of the regex.

Use back-slash \ as the escape character e.g.
/\#{1, 6}/

Related

Ruby - Simple way to escape regex characters

I have an expression like this:
s.gsub! /[\?\/\\]/, ''
In fact, the list of forbidden characters is longer than that. Some require escaping, some don't. Is there a way I could just put the characters in some literal structure (?/\) and say "remove all these from the string". I'm aware of Regexp.quote but not sure how to use it in this context.
You can interpolate a statement inside a Ruby regular expression, exactly like you do for strings.
/#{...}/
In your case
s.gsub! /[#{Regexp.escape('?/\')}]/, ''

Double escape characters in elisp regex patterns

(regexp-opt '("this" "that"))
returns,
"\\(?:th\\(?:at\\|is\\)\\)
Why there are double backward slashes everywhere in this elisp regex. Doesn't elisp regex use single backward slash?
And, ? symbol is a postfix operator in regex patterns which means it acts upon the characters that precedes it..(http://www.gnu.org/software/emacs/manual/html_node/elisp/Regexp-Special.html#Regexp-Special). but here, there are no expressions before the ? operator. so, what does
(?:th\\
part mean in this regex.
The backslash is part of the regexp syntax. But to preserve it as part of a regexp string, you need to protect it with another backslash, as documented in the syntax for strings documentation:
'Likewise, you can include a backslash by preceding it with another backslash, like this: "this \\ is a single embedded backslash".'
As for the ?: construct, it's how you specify a non-capturing or "shy" group:
"A shy group serves the first two purposes of an ordinary group (controlling the nesting of other operators), but it does not get a number, so you cannot refer back to its value with ‘\digit’. Shy groups are particularly useful for mechanically-constructed regular expressions, because they can be added automatically without altering the numbering of ordinary, non-shy groups."
It's documented as part of the regexp backslash documentation. As the passage quoted above explains, it's useful in functions like regexp-opt for grouping patterns without creating capture groups.

What's the point of escaping a single backslash in single quotes in Ruby?

I don't quite understand what's the point of escaping a single backslash when you have a string in single quotes in Ruby. Why does Ruby treats backslashes 'differently'?
backslashes are an escape character so if you were to write '\' would think you were trying to escape the '.
Otherwise if it treated single character strings differentls you wanted to write ' you would have to use double quotes, which will quickly get harder to maintain when you need to remember which quotes to use when.
If your question is actually "What is the point of the language design requiring us to escape a backslash in single (as opposed to double) quotes", then that is to allow single quotes to appear within a string literal written with single quotes. In order to do that, there must be an escape character for single quotes, which is the backslash, and then, the escape character itself needs to be escaped.

Backslash + captured group within Ruby regular expression

How do I excape a backslash before a captured group?
Example:
"foo+bar".gsub(/(\+)/, '\\\1')
What I expect (and want):
foo\+bar
what I unfortunately get:
foo\\1bar
How do I escape here correctly?
As others have said, you need to escape everything in that string twice. So in your case the solution is to use '\\\\\1' or '\\\\\\1'. But since you asked why, I'll try to explain that part.
The reason is that replacement sequence is being parsed twice--once by Ruby and once by the underlying regular expression engine, for whom \1 is its own escape sequence. (It's probably easier to understand with double-quoted strings, since single quotes introduce an ambiguity where '\\1' and '\1' are equivalent but '\' and '\\' are not.)
So for example, a simple replacement here with a captured group and a double quoted string would be:
"foo+bar".gsub(/(\+)/, "\\1") #=> "foo+bar"
This passes the string \1 to the regexp engine, which it understands as a reference to a capture group. In Ruby string literals, "\1" means something else entirely (ASCII character 1).
What we actually want in this case is for the regexp engine to receive \\\1. It also understands \ as an escape character, so \\1 is not sufficient and will simply evaluate to the literal output \1. So, we need \\\1 in the regexp engine, but to get to that point we need to also make it past Ruby's string literal parser.
To do that, we take our desired regexp input and double every backslash again to get through Ruby's string literal parser. \\\1 therefore requires "\\\\\\1". In the case of single quotes one slash can be omitted as \1 is not a valid escape sequence in single quotes and is treated literally.
Addendum
One of the reasons this problem is usually hidden is thanks to the use of /.+/ style regexp quotes, which Ruby treats in a special way to avoid the need to double escape everything. (Of course, this doesn't apply to gsub replacement strings.) But you can still see it in action if you use a string literal instead of a regexp literal in Regexp.new:
Regexp.new("\.").match("a") #=> #<MatchData "a">
Regexp.new("\\.").match("a") #=> nil
As you can see, we had to double-escape the . for it to be understood as a literal . by the regexp engine, since "." and "\." both evaluate to . in double-quoted strings, but we need the engine itself to receive \..
This happens due to a double string escaping. You should use 5 slashes in this case.
"foo+bar".gsub(/([+])/, '\\\\\1')
Adding \ two more times escapes this properly.
irb(main):011:0> puts "foo+bar".gsub(/(\+)/, '\\\\\1')
foo\+bar
=> nil

Another way instead of escaping regex patterns?

Usually when my regex patterns look like this:
http://www.microsoft.com/
Then i have to escape it like this:
string.match(/http:\/\/www\.microsoft\.com\//)
Is there another way instead of escaping it like that?
I want to be able to just use it like this http://www.microsoft.com, cause I don't want to escape all the special characters in all my patterns.
Regexp.new(Regexp.quote('http://www.microsoft.com/'))
Regexp.quote simply escapes any characters that have special regexp meaning; it takes and returns a string. Note that . is also special. After quoting, you can append to the regexp as needed before passing to the constructor. A simple example:
Regexp.new(Regexp.quote('http://www.microsoft.com/') + '(.*)')
This adds a capturing group for the rest of the path.
You can also use arbitrary delimiters in Ruby for regular expressions by using %r and defining a character before the regular expression, for example:
%r!http://www.microsoft.com/!
Regexp.quote or Regexp.escape can be used to automatically escape things for you:
https://ruby-doc.org/core/Regexp.html#method-c-escape
The result can be passed to Regexp.new to create a Regexp object, and then you can call the object's .match method and pass it the string to match against (the opposite order from string.match(/regex/)).
You can simply use single quotes for escaping.
string.match('http://www.microsoft.com/')
you can also use %q{} if you need single quotes in the text itself. If you need to have variables extrapolated inside the string, then use %Q{}. That's equivalent to double quotes ".
If the string contains regex expressions (eg: .*?()[]^$) that you want extrapolated, use // or %r{}
For convenience I just define
def regexcape(s)
Regexp.new(Regexp.escape(s))
end

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