I am currently trying to create a regex that will find this pattern /- - [/. However, since the [ is a special character, I am getting this error:
premature end of char-class: /- - [/ (SyntaxError)
I know this is happening because the compiler is expecting the [ in the regex to close, but I need it to match that in my pattern. How can I do this?
Yes, brackets are part of regex syntax. If you want to match a literal bracket (or any other special symbol, for that matter), escape with a backslash.
'foo[bar]' =~ /\[/ # => 3
Related
From a string:
"(book(1:3000))"
I need to exclude opening and closing brackets and match:
"book(1:3000)"
using regular expression.
I tried this regular expression:
/([^',()]+)|'([^']*)'/
which detects all characters and integers excluding brackets. The string detected by this regex is:
"book 1:3000"
Is there any regex that disregards the opening and closing brackets, and gives the entire string?
Build the regexp that explicitly states exactly what you want to extract: alphanumerics, followed by the opening parenthesis, followed by digits, followed by a colon, followed by digits, followed by closing parenthesis:
'(book(1:3000))'[/\w+\(\d+:\d+\)/]
#āĀ "book(1:3000)"
"(book(1:3000))"[/^\(?(.+?\))\)?/, 1]
=> "book(1:3000)"
"book(1:3000)"[/^\(?(.+?\))\)?/, 1]
=> "book(1:3000)"
The regex split on multiple lines for easier reading:
/
^ # start of string
\(? # character (, possibly (?)
( # start capturing
.+? # any characters forward until..
\) # ..a closing bracket
) # stop capturing
/x # end regex with /x modifier (allows splitting to lines)
1. Look for a possible ( in the beginning of string and ignore it.
2. Start capturing
3. Capture until and including the first )
But this is where it fails:
"book(1:(10*30))"[/^\(?(.+?\))\)?/, 1]
=> "book(1:(10*30)"
If you need something like that, you probably need to use a recursive regex as
described in another stackoverflow answer.
I am trying to match mathematica expressions like 1+2 and 1*2/3.... to infinity. Can someone explain why my regex matches the final case below, and how to fix it so that it matches only valid expressions (that might stretch forever)?
perms=["12+2*4","2+2","-2+","12+34-"]
perms.each do |line|
puts "#{line}=#{eval(line)}" if line =~ /^\d+([+-\/*]\d+){1,}/
end
I expected the output to be:
12+2*4=20
2+2=4
Inside a [character set], the - character defines a range of characters -- think of [a-z] or [0-9]. If you want to match a literal -, it must be the first or last character.
/^\d+(?:[+\/*-]\d+)+$/
Other things: {1,} is exactly +; and you need to anchor at the end too, so you don't match 1+2+
You should finalize your expression with $ to match the entire input string:
/^\d+([-+\/*]\d+){1,}$/
The wrong position of the hyphen - is one source of error in your expression. The missing $ the other.
I often see the gsub function being called with the pattern parameter enclosed in forward slashes. For example:
>> phrase = "*** and *** ran to the ###."
>> phrase.gsub(/\*\*\*/, "WOOF")
=> "WOOF and WOOF ran to the ###."
I thought maybe it had something to do with escaping asterisks, but using single quotes and double quotes works just as well:
>> phrase = "*** and *** ran to the ###."
>> phrase.gsub('***', "WOOF")
=> "WOOF and WOOF ran to the ###."
>> phrase.gsub("***", "WOOF")
=> "WOOF and WOOF ran to the ###."
Is it just convention to use forward slash? What am I missing?
Use forward slashes if you need to use regular expressions.
If you use a string argument with gsub, it will just do a plain character match.
In your example, you need backslashes to escape the asterisks when using a regular expression, because asterisks have a special meaning in regex (optionally match something any number of times). They are not necessary when using a string, because they are just matched exactly.
In your example, you probably don't need to use a regular expression, since it is a simple pattern. However, if you wanted to match *** only when it was at the beginning of a string (e.g. the first bunch in your example), then you would want to use a regex, for example:
phrase.gsub(/^\*{3}/, "WOOF")
For more information on regular expressions, see: http://www.regular-expressions.info/.
For more information on using regular expressions in Ruby, see: http://ruby-doc.org/core-2.2.0/Regexp.html.
To play with regular expressions as they work in Ruby, try: http://rubular.com/.
You are missing reading the documentation:
The pattern is typically a Regexp; if given as a String, any regular expression metacharacters it contains will be interpreted literally, e.g. '\d' will match a backlash followed by ādā, instead of a digit.
http://ruby-doc.org/core-2.1.4/String.html#method-i-gsub
In other words, you can give a string or a regular expression. Regular expressions can be delimited several ways:
Regexps are created using the /.../ and %r{...} literals, and by the Regexp::new constructor.
http://ruby-doc.org/core-2.2.2/Regexp.html
The benefit of %r and of the alternate %r delimiters is you can usually find a delimiter that doesn't collide with characters in the pattern, which would force escaping them, as in your example.
* has to be escaped because it has special meaning in a regex, but in a string it does not.
I'm working on some text processing in Ruby 1.8.7 to support some custom shortcodes that I've created. Here are some examples of my shortcode:
[CODE first-part]
[CODE first-part second-part]
I'm using the following RegEx to grab the
text.gsub!( /\[CODE (\S+)\s?(\S?)\]/i, replacementText )
The problem is this: the regex doesn't work on the following text:
[CODE first-part][CODE first-part-again]
The results are as follows:
1. first-part][CODE
2. first-part-again
It seems that the \s? is the problematic part of the regex that is searching on until it hits the last space, not the first one. When I change the regex to the following:
\[CODE ([\w-]+)\s?(\S*)\]/i
It works fine. The only concern I have is what all \w vs \s as I want to make sure the \w will match URL-safe characters.
I'm sure there's a perfectly valid explanation, but it's eluding me. Any ideas? Thanks!
Actually, thinking about it, just using [^\]] might not be enough, as it will swallow up all spaces as well. You also need to exclude those:
/\[CODE[ ]([^\]\s]+)\s?([^\]\s]*)\]/i
Note the [ ] - I just think it makes literal spaces more readable.
Working demo.
Explained in free-spacing mode:
\[CODE[ ] # match your identifier
( # capturing group 1
[^\]\s]+ # match one or more non-], non-whitespace characters
) # end of group 1
\s? # match an optional whitespace character
( # capturing group 2
[^\]\s]+ # match zero or more non-], non-whitespace characters
) # end of group 2
\] # match the closing ]
As none of the character classes in the pattern includes ], you can never possibly go beyond the end of the square bracketed expression.
By the way, if you find unnecessary escapes in regex as obscuring as I do, here is the minimal version:
/\[CODE[ ]([^]\s]+)\s?([^]\s]*)]/i
But that is definitely a matter of taste.
The problem was with the greedy \S+ in this
/\[CODE (\S+)\s?(\S?)\]/i
You could try:
/\[CODE (\S+?)\s?(\S?)\]/i
but actually your new character class is IMO superiror.
Even better might be:
/\[CODE ([^\]]+?)\s?([^\]]*)\]/i
I'm trying to learn RegEx in Ruby, based on what I'm reading in "The Rails Way". But, even this simple example has me stumped. I can't tell if it is a typo or not:
text.gsub(/\s/, "-").gsub([^\W-], '').downcase
It seems to me that this would replace all spaces with -, then anywhere a string starts with a non letter or number followed by a dash, replace that with ''. But, using irb, it fails first on ^:
syntax error, unexpected '^', expecting ']'
If I take out the ^, it fails again on the W.
>> text = "I love spaces"
=> "I love spaces"
>> text.gsub(/\s/, "-").gsub(/[^\W-]/, '').downcase
=> "--"
Missing //
Although this makes a little more sense :-)
>> text.gsub(/\s/, "-").gsub(/([^\W-])/, '\1').downcase
=> "i-love-spaces"
And this is probably what is meant
>> text.gsub(/\s/, "-").gsub(/[^\w-]/, '').downcase
=> "i-love-spaces"
\W means "not a word"
\w means "a word"
The // generate a regexp object
/[^\W-]/.class
=> Regexp
Step 1: Add this to your bookmarks. Whenever I need to look up regexes, it's my first stop
Step 2: Let's walk through your code
text.gsub(/\s/, "-")
You're calling the gsub function, and giving it 2 parameters.
The first parameter is /\s/, which is ruby for "create a new regexp containing \s (the // are like special "" for regexes).
The second parameter is the string "-".
This will therefore replace all whitespace characters with hyphens. So far, so good.
.gsub([^\W-], '').downcase
Next you call gsub again, passing it 2 parameters.
The first parameter is [^\W-]. Because we didn't quote it in forward-slashes, ruby will literally try run that code. [] creates an array, then it tries to put ^\W- into the array, which is not valid code, so it breaks.
Changing it to /[^\W-]/ gives us a valid regex.
Looking at the regex, the [] says 'match any character in this group. The group contains \W (which means non-word character) and -, so the regex should match any non-word character, or any hyphen.
As the second thing you pass to gsub is an empty string, it should end up replacing all the non-word characters and hyphens with empty string (thereby stripping them out )
.downcase
Which just converts the string to lower case.
Hope this helps :-)
You forgot the slashes. It should be /[^\W-]/
Well, .gsub(/[^\W-]/,'') says replace anything that's a not word nor a - for nothing.
You probably want
>> text.gsub(/\s/, "-").gsub(/[^\w-]/, '').downcase
=> "i-love-spaces"
Lower case \w (\W is just the opposite)
The slashes are to say that the thing between them is a regular expression, much like quotes say the thing between them is a string.