I am getting completely different reults from string.scan and several regex testers...
I am just trying to grab the domain from the string, it is the last word.
The regex in question:
/([a-zA-Z0-9\-]*\.)*\w{1,4}$/
The string (1 single line, verified in Ruby's runtime btw)
str = 'Show more results from software.informer.com'
Work fine, but in ruby....
irb(main):050:0> str.scan /([a-zA-Z0-9\-]*\.)*\w{1,4}$/
=> [["informer."]]
I would think that I would get a match on software.informer.com ,which is my goal.
Your regex is correct, the result has to do with the way String#scan behaves. From the official documentation:
"If the pattern contains groups, each individual result is itself an array containing one entry per group."
Basically, if you put parentheses around the whole regex, the first element of each array in your results will be what you expect.
It does not look as if you expect more than one result (especially as the regex is anchored). In that case there is no reason to use scan.
'Show more results from software.informer.com'[ /([a-zA-Z0-9\-]*\.)*\w{1,4}$/ ]
#=> "software.informer.com"
If you do need to use scan (in which case you obviously need to remove the anchor), you can use (?:) to create non-capturing groups.
'foo.bar.baz lala software.informer.com'.scan( /(?:[a-zA-Z0-9\-]*\.)*\w{1,4}/ )
#=> ["foo.bar.baz", "lala", "software.informer.com"]
You are getting a match on software.informer.com. Check the value of $&. The return of scan is an array of the captured groups. Add capturing parentheses around the suffix, and you'll get the .com as part of the return value from scan as well.
The regex testers and Ruby are not disagreeing about the fundamental issue (the regex itself). Rather, their interfaces are differing in what they are emphasizing. When you run scan in irb, the first thing you'll see is the return value from scan (an Array of the captured subpatterns), which is not the same thing as the matched text. Regex testers are most likely oriented toward displaying the matched text.
How about doing this :
/([a-zA-Z0-9\-]*\.*\w{1,4})$/
This returns
informer.com
On your test string.
http://rubular.com/regexes/13670
Related
I have ruby app that uses first matched string by regex. my_url.match(/my_regex/).first
As a strings I have a list of urls that contain adress or postcode and from each of them I need to extract postocode or adreess by using regex
Samples of urls:
http://www.adresses.co.uk/avon/bath-city
http://www.adresses.co.uk/postcode/rm107jj
My regex:
\.co\.uk\/postcode\/([^\/]*)|\.co\.uk\/(?!postcode)([^\/]*\/[^\/]*)
My problem is that for non postcode urls a first matched data by this regex is nil see_on_rubular
How to rewrite or change this reflex so it will skip nil matches or to make first matches non nils. I need to solve it with regex not in ruby coding please.
Here's a regex that captures in group #1 everything after postcode/ if it's present, or else everything after .co.uk/:
\.co\.uk\/(?:postcode\/)?([^\/\n]+(?:\/[^\/\n]+)?)
(DEMO)
Note that this will give unexpected results if there are unwanted path elements at the end of a postcode link, such as:
http://www.adresses.co.uk/postcode/rm107jj/oops
UPDATE: Based on the comments, it looks like you want to match just the last path element. But we can't simply capture the second element, because there might be only one:
http://www.adresses.co.uk/west-midlands
We can, however, make the first element optional:
\.co\.uk\/(?:[^\/\n]+\/)?([^\/\n]+)
Notice how I used a non-capturing group for the optional portion, so the part you want is still captured in group #1.
...
I'm trying to match some text if it does not have another block of text in its vicinity. For example, I would like to match "bar" if "foo" does not precede it. I can match "bar" if "foo" does not immediately precede it using negative look behind in this regex:
/(?<!foo)bar/
but I also like to not match "foo 12345 bar". I tried:
/(?<!foo.{1,10})bar/
but using a wildcard + a range appears to be an invalid regex in Ruby. Am I thinking about the problem wrong?
You are thinking about it the right way. But unfortunately lookbehinds usually have be of fixed-length. The only major exception to that is .NET's regex engine, which allows repetition quantifiers inside lookbehinds. But since you only need a negative lookbehind and not a lookahead, too. There is a hack for you. Reverse the string, then try to match:
/rab(?!.{0,10}oof)/
Then reverse the result of the match or subtract the matching position from the string's length, if that's what you are after.
Now from the regex you have given, I suppose that this was only a simplified version of what you actually need. Of course, if bar is a complex pattern itself, some more thought needs to go into how to reverse it correctly.
Note that if your pattern required both variable-length lookbehinds and lookaheads, you would have a harder time solving this. Also, in your case, it would be possible to deconstruct your lookbehind into multiple variable length ones (because you use neither + nor *):
/(?<!foo)(?<!foo.)(?<!foo.{2})(?<!foo.{3})(?<!foo.{4})(?<!foo.{5})(?<!foo.{6})(?<!foo.{7})(?<!foo.{8})(?<!foo.{9})(?<!foo.{10})bar/
But that's not all that nice, is it?
As m.buettner already mentions, lookbehind in Ruby regex has to be of fixed length, and is described so in the document. So, you cannot put a quantifier within a lookbehind.
You don't need to check all in one step. Try doing multiple steps of regex matches to get what you want. Assuming that existence of foo in front of a single instance of bar breaks the condition regardless of whether there is another bar, then
string.match(/bar/) and !string.match(/foo.*bar/)
will give you what you want for the example.
If you rather want the match to succeed with bar foo bar, then you can do this
string.scan(/foo|bar/).first == "bar"
I want to transform the following text
This is a , here is another 
into
This is a , here is another 
In other words I want to find all the image paths that are enclosed between brackets (the text is in Markdown syntax) and replace them with other paths. The string containing the new path is returned by a separate real_path function.
I would like to do this using String#gsub in its block version. Currently my code looks like this:
re = /!\[.*?\]\((.*?)\)/
rel_content = content.gsub(re) do |path|
real_path(path)
end
The problem with this regex is that it will match  instead of just foto.jpeg. I also tried other regexen like (?>\!\[.*?\]\()(.*?)(?>\)) but to no avail.
My current workaround is to split the path and reassemble it later.
Is there a Ruby regex that matches only the path inside the brackets and not all the contextual required characters?
Post-answers update: The main problem here is that Ruby's regexen have no way to specify zero-width lookbehinds. The most generic solution is to group what the part of regexp before and the one after the real matching part, i.e. /(pre)(matching-part)(post)/, and reconstruct the full string afterwards.
In this case the solution would be
re = /(!\[.*?\]\()(.*?)(\))/
rel_content = content.gsub(re) do
$1 + real_path($2) + $3
end
A quick solution (adjust as necessary):
s = 'This is a '
s.sub!(/!(\[.*?\])\((.*?)\)/, '\1(/folder1/\2)' )
p s # This is a [foto](/folder1/foto.jpeg)
You can always do it in two steps - first extract the whole image expression out and then second replace the link:
str = "This is a , here is another "
str.gsub(/\!\[[^\]]*\]\(([^)]*)\)/) do |image|
image.gsub(/(?<=\()(.*)(?=\))/) do |link|
"/a/new/path/" + link
end
end
#=> "This is a , here is another "
I changed the first regex a bit, but you can use the same one you had before in its place. image is the image expression like , and link is just the path like foto.jpeg.
[EDIT] Clarification: Ruby does have lookbehinds (and they are used in my answer):
You can create lookbehinds with (?<=regex) for positive and (?<!regex) for negative, where regex is an arbitrary regex expression subject to the following condition. Regexp expressions in lookbehinds they have to be fixed width due to limitations on the regex implementation, which means that they can't include expressions with an unknown number of repetitions or alternations with different-width choices. If you try to do that, you'll get an error. (The restriction doesn't apply to lookaheads though).
In your case, the [foto] part has a variable width (foto can be any string) so it can't go into a lookbehind due to the above. However, lookbehind is exactly what we need since it's a zero-width match, and we take advantage of that in the second regex which only needs to worry about (fixed-length) compulsory open parentheses.
Obviously you can put real_path in from here, but I just wanted a test-able example.
I think that this approach is more flexible and more readable than reconstructing the string through the match group variables
In your block, use $1 to access the first capture group ($2 for the second and so on).
From the documentation:
In the block form, the current match string is passed in as a parameter, and variables such as $1, $2, $`, $&, and $' will be set appropriately. The value returned by the block will be substituted for the match on each call.
As a side note, some people think '\1' inappropriate for situations where an unconfirmed number of characters are matched. For example, if you want to match and modify the middle content, how can you protect the characters on both sides?
It's easy. Put a bracket around something else.
For example, I hope replace a-ruby-porgramming-book-531070.png to a-ruby-porgramming-book.png. Remove context between last "-" and last ".".
I can use /.*(-.*?)\./ match -531070. Now how should I replace it? Notice
everything else does not have a definite format.
The answer is to put brackets around something else, then protect them:
"a-ruby-porgramming-book-531070.png".sub(/(.*)(-.*?)\./, '\1.')
# => "a-ruby-porgramming-book.png"
If you want add something before matched content, you can use:
"a-ruby-porgramming-book-531070.png".sub(/(.*)(-.*?)\./, '\1-2019\2.')
# => "a-ruby-porgramming-book-2019-531070.png"
At the moment I have a regular expression that looks like this:
^(cat|dog|bird){1}(cat|dog|bird)?(cat|dog|bird)?$
It matches at least 1, and at most 3 instances of a long list of words and makes the matching words for each group available via the corresponding variable.
Is there a way to revise this so that I can return the result for each word in the string without specifying the number of groups beforehand?
^(cat|dog|bird)+$
works but only returns the last match separately , because there is only one group.
OK, so I found a solution to this.
It doesn't look like it is possible to create an unknown number of groups, so I went digging for another way of achieving the desired outcome: To be able to tell if a string was made up of words in a given list; and to match the longest words possible in each position.
I have been reading Mastering Regular Expressions by Jeffrey E. F. Friedl and it shed some light on things for me. It turns out that NFA based Regexp engines (like the one used in Ruby) are sequential as well as lazy/greedy. This means that you can dictate how a pattern is matched using the order in which you give it choices. This explains why scan was returning variable results, it was looking for the first word in the list that matched the criteria and then moved on to the next match. By design it was not looking for the longest match, but the first one. So in order to rectify this all I needed to do was reorder the array of words used to generate the regular expression from alphabetical order, to length order (longest to shortest).
array = %w[ as ascarid car id ]
list = array.sort_by {|word| -word.length }
regexp = Regexp.union(list)
Now the first match found by scan will be the longest word available. It is also pretty simple to tell if a string contains only words in the list using scan:
if "ascarid".scan(regexp).join.length == word.length
return true
else
return false
end
Thanks to everyone that posted in response to this question, I hope that this will help others in the future.
You could do it in two steps:
Use /^(cat|dog|bird)+$/ (or better /\A(cat|dog|bird)+\z/) to make sure it matches.
Then string.scan(/cat|dog|bird/) to get the pieces.
You could also use split and a Set to do both at once. Suppose you have your words in the array a and your string in s, then:
words = Set.new(a)
re = /(#{a.map{|w| Regexp.quote(w)}.join('|')})/
parts = s.split(re).reject(&:empty?)
if(parts.any? {|w| !words.include?(w) })
# 's' didn't match what you expected so throw a
# hissy fit, format the hard drive, set fire to
# the backups, or whatever is appropriate.
else
# Everything you were looking for is in 'parts'
# so you can check the length (if you care about
# how many matches there were) or something useful
# and productive.
end
When you use split with a pattern that contains groups then
the respective matches will be returned in the array as well.
In this case, the split will hand us something like ["", "cat", "", "dog"] and the empty strings will only occur between the separators that we're looking for and so we can reject them and pretend they don't exist. This may be an unexpected use of split since we're more interested in the delimiters more than what is being delimited (except to make sure that nothing is being delimited) but it gets the job done.
Based on your comments, it looks like you want an ordered alternation so that (ascarid|car|as|id) would try to match from left to right. I can't find anything in the Ruby Oniguruma (the Ruby 1.9 regex engine) docs that says that | is ordered or unordered; Perl's alternation appears to be specified (or at least strongly implied) to be ordered and Ruby's certainly behaves as though it is ordered:
>> 'pancakes' =~ /(pan|pancakes)/; puts $1
pan
So you could sort your words from longest to shortest when building your regex:
re = /(#{a.sort_by{|w| -w.length}.map{|w| Regexp.quote(w)}.join('|')})/
and hope that Oniguruma really will match alternations from left to right. AFAIK, Ruby's regexes will be eager because they support backreferences and lazy/non-greedy matching so this approach should be safe.
Or you could be properly paranoid and parse it in steps; first you'd make sure your string looks like what you want:
if(s !~ /\A(#{a.map{|w| Regexp.quote(w)}.join('|')})+\z/)
# Bail out and complain that 's' doesn't look right
end
The group your words by length:
by_length = a.group_by(&:length)
and scan for the groups from the longest words to the shortest words:
# This loses the order of the substrings within 's'...
matches = [ ]
by_length.keys.sort_by { |k| -k }.each do |group|
re = /(#{a.map{|w| Regexp.quote(w)}.join('|')})/
s.gsub!(re) { |w| matches.push(w); '' }
end
# 's' should now be empty and the matched substrings will be
# in 'matches'
There is still room for possible overlaps in these approaches but at least you'd be extracting the longest matches.
If you need to repeat parts of a regex, one option is to store the repeated part in a variable and just reference that, for example:
r = "(cat|dog|bird)"
str.match(/#{r}#{r}?#{r}?/)
You can do it with .Net regular expressions. If I write the following in PowerShell
$pat = [regex] "^(cat|dog|bird)+$"
$m = $pat.match('birddogcatbird')
$m.groups[1].captures | %{$_.value}
then I get
bird
dog
cat
bird
when I run it. I know even less about IronRuby than I do about PowerShell, but perhaps this means you can do it in IronRuby as well.
I have the following
address.gsub(/^\d*/, "").gsub(/\d*-?\d*$/, "").gsub(/\# ?\d*/,"")
Can this be done in one gsub? I would like to pass a list of patterns rather then just one pattern - they are all being replaced by the same thing.
You could combine them with an alternation operator (|):
address = '6 66-666 #99 11-23'
address.gsub(/^\d*|\d*-?\d*$|\# ?\d*/, "")
# " 66-666 "
address = 'pancakes 6 66-666 # pancakes #99 11-23'
address.gsub(/^\d*|\d*-?\d*$|\# ?\d*/,"")
# "pancakes 6 66-666 pancakes "
You might want to add little more whitespace cleanup. And you might want to switch to one of:
/\A\d*|\d*-?\d*\z|\# ?\d*/
/\A\d*|\d*-?\d*\Z|\# ?\d*/
depending on what your data really looks like and how you need to handle newlines.
Combining the regexes is a good idea--and relatively simple--but I'd like to recommend some additional changes. To wit:
address.gsub(/^\d+|\d+(?:-\d+)?$|\# *\d+/, "")
Of your original regexes, ^\d* and \d*-?\d*$ will always match, because they don't have to consume any characters. So you're guaranteed to perform two replacements on every line, even if that's just replacing empty strings with empty strings. Of my regexes, ^\d+ doesn't bother to match unless there's at least one digit at the beginning of the line, and \d+(?:-\d+)?$ matches what looks like an integer-or-range expression at the end of the line.
Your third regex, \# ?\d*, will match any # character, and if the # is followed by a space and some digits, it'll take those as well. Judging by your other regexes and my experience with other questions, I suspect you meant to match a # only if it's followed by one or more digits, with optional spaces intervening. That's what my third regex does.
If any of my guesses are wrong, please describe what you were trying to do, and I'll do my best to come up with the right regex. But I really don't think those first two regexes, at least, are what you want.
EDIT (in answer to the comment): When working with regexes, you should always be aware of the distinction between a regex the matches nothing and a regex that doesn't match. You say you're applying the regexes to street addresses. If an address doesn't happen to start with a house number, ^\d* will match nothing--that is, it will report a successful match, said match consisting of the empty string preceding the first character in the address.
That doesn't matter to you, you're just replacing it with another empty string anyway. But why bother doing the replacement at all? If you change the regex to ^\d+, it will report a failed match and no replacement will be performed. The result is the same either way, but the "matches noting" scenario (^\d*) results in a lot of extra work that the "doesn't match" scenario avoids. In a high-throughput situation, that could be a life-saver.
The other two regexes bring additional complications: \d*-?\d*$ could match a hyphen at the end of the string (e.g. "123-", or even "-"); and \# ?\d* could match a hash symbol anywhere in string, not just as part of an apartment/office number. You know your data, so you probably know neither of those problems will ever arise; I'm just making sure you're aware of them. My regex \d+(?:-\d+)?$ deals with the trailing-hyphen issue, and \# *\d+ at least makes sure there are digits after the hash symbol.
I think that if you combine them together in a single gsub() regex, as an alternation,
it changes the context of the starting search position.
Example, each of these lines start at the beginning of the result of the previous
regex substitution.
s/^\d*//g
s/\d*-?\d*$//g
s/\# ?\d*//g
and this
s/^\d*|\d*-?\d*$|\# ?\d*//g
resumes search/replace where the last match left off and could potentially produce a different overall output, especially since a lot of the subexpressions search for similar
if not the same characters, distinguished only by line anchors.
I think your regex's are unique enough in this case, and of course changing the order
changes the result.