I have a string, which contains a bunch of HTML documents, tagged with #name:
string = "#one\n\n<html>\n</html>\n\n#two\n<html>\n</html>\n\n\n"
I want to get an array of two-element arrays, each of which with a tag as the first element and the HTML document as the second:
[ ["#one", "<html>\n</html>"], ["#two", "<html>\n</html>"] ]
In order to solve the problem, I crafted the following regular expression:
regex = /(#.+)\n+([^#]+)\n+/
and applied it in string.scan regex.
However, instead of the desired output, I get the following:
[ ["#one", "<html>\n</html>\n"], ["#two", "<html>\n</html>\n\n"] ]
There are trailing newline characters at the end of each document. It appears that only one newline character was removed from the documents, but others stayed at the place.
How can the aforementioned regular expression be changed in order to remove all the trailing characters from the resulting documents?
The reason only the last \n was thrown away is because the two relevant capturing parts in your regex: .+ and [^#]+ capture everything up to the last \n (in order to make matching possible at all). It does not matter that they are followed by \n+. Remember that regex works from the left to the right. If some substring (sequences of \n in this case) can fit in either the preceding part of the following part of a regex, it actually fits in the preceding part.
With generality, I would suggest doing this:
string.split(/\s+(?=#)/).map{|s| s.strip.split(/\s+/, 2)}
# => [["#one", "<html>\n</html>"], ["#two", "<html>\n</html>"]]
You can remove duplicated newlines first:
string.gsub(/\n+/, "\n").scan(regex)
=> [["#one", "<html>\n</html>"], ["#two", "<html>\n</html>"]]
Related
I am trying to create a regex that matches a pattern in some part of a string, but not in another part of the string.
I am trying to match a substring that
(i) is surrounded by a balanced pair of one or more consecutive backticks `
(ii) and does not include as many consecutive backticks as in the surrounding patterns
(iii) where the surrounding patterns (sequence of backticks) are not adjacent to other backticks.
This is some variant of the syntax of inline code notation in Markdown syntax.
Examples of matches are as follows:
"xxx`foo`yyy" # => matches "foo"
"xxx``foo`bar`baz``yyy" # => matches "foo`bar`baz"
"xxx```foo``bar``baz```yyy" # => matches "foo``bar``baz"
One regex to achieve this is:
/(?<!`)(?<backticks>`+)(?<inline>.+?)\k<backticks>(?!`)/
which uses a non-greedy match.
I was wondering if I can get rid of the non-greedy match.
The idea comes from when the prohibited pattern is a single character. When I want to match a substring that is surrounded by a single quote ' that does not include a single quote in it, I can do either:
/'.+?'/
/'[^']+'/
The first one uses non-greedy match, and the second one uses an explicit non-matching pattern [^'].
I am wondering if it is possible to have something like the second form when the prohibited pattern is not a single character.
Going back to the original issue, there is negative lookahead syntax(?!), but I cannot restrict its effective scope. If I make my regex like this:
/(?<!`)(?<backticks>`+)(?<inline>(?!.*\k<backticks>).*)\k<backticks>(?!`)/
then the effect of (?!.*\k<backticks>) will not be limited to within (?<inline>...), but will extend to the whole string. And since that contradicts with the \k<backticks> at the end, the regex fails to match.
Is there a regex technique to ensure non-matching of a pattern (not-necessarily a single character) within a certain scope?
You can search for one or more characters which aren't the first character of a delimiter:
/(?<!`)(?<backticks>`+)(?<inline>(?:(?!\k<backticks>).)+)\k<backticks>(?!`)/
So I have a string that looks like this:
#jackie#test.com, #mike#test.com
What I want to do is before any email in this comma separated list, I want to remove the #. The issue I keep running into is that if I try to do a regular \A flag like so /[\A#]+/, it finds all the instances of # in that string...including the middle crucial #.
The same thing happens if I do /[\s#]+/. I can't figure out how to just look at the beginning of each string, where each string is a complete email address.
Edit 1
Note that all I need is the regex, I already have the rest of the stuff I need to do what I want. Specifically, I am achieving everything else like this:
str.gsub(/#/, '').split(',').map(&:strip)
Where str is my string.
All I am looking for is the regex portion for my gsub.
You may use the below negative lookbehind based regex.
str.gsub(/(?<!\S)#/, '').split(',').map(&:strip)
(?<!\S) Negative lookbehind asserts that the character or substring we are going to match would be preceeded by any but not of a non-space character. So this matches the # which exists at the start or the # which exists next to a space character.
Difference between my answer and hwnd's str.gsub(/\B#/, '') is, mine won't match the # which exists in :# but hwnd's answer does. \B matches between two word characters or two non-word characters.
Here is one solution
str = "#jackie#test.com, #mike#test.com"
p str.split(/,[ ]+/).map{ |i| i.gsub(/^#/, '')}
Output
["jackie#test.com", "mike#test.com"]
I created a regexp to match the following scenerios: SomethingCool, HelloWorld, MyNameIsDonato, etc. However, it does not account for spaces:
> 'Something Cooler' =~ /([A-Z][a-z0-9]+)+/
=> 0
That passes and it should not pass. A space is not an alphanumeric character. So why does this pass and how can I fix it?
You need to anchor the regex to the beginning and end of the string, or it will just match one of the words:
^([A-Z][a-z0-9]+)+$
^ and $ anchor the beginnings and ends of lines, respectively. To anchor to the beginning and end of the string, use \A and \Z.
It's worth noting that this is useless if you're trying to find camelcase names within a larger string. For that, use your original regex.
I want to scrape data from some text and dump it into an array. Consider the following text as example data:
| Example Data
| Title: This is a sample title
| Content: This is sample content
| Date: 12/21/2012
I am currently using the following regex to scrape the data that is specified after the 'colon' character:
/((?=:).+)/
Unfortunately this regex also grabs the colon and the space after the colon. How do I only grab the data?
Also, I'm not sure if I'm doing this right.. but it appears as though the outside parens causes a match to return an array. Is this the function of the parens?
EDIT: I'm using Rubular to test out my regex expressions
You could change it to:
/: (.+)/
and grab the contents of group 1. A lookbehind works too, though, and does just what you're asking:
/(?<=: ).+/
In addition to #minitech's answer, you can also make a 3rd variation:
/(?<=: ?)(.+)/
The difference here being, you create/grab the group using a look-behind.
If you still prefer the look-ahead rather than look-behind concept. . .
/(?=: ?(.+))/
This will place a grouping around your existing regex where it will catch it within a group.
And yes, the outside parenthesis in your code will make a match. Compare that to the latter example I gave where the entire look-ahead is 'grouped' rather than needlessly using a /( ... )/ without the /(?= ... )/, since the first result in most regular expression engines return the entire matched string.
I know you are asking for regex but I just saw the regex solution and found that it is rather hard to read for those unfamiliar with regex.
I'm also using Ruby and I decided to do it with:
line_as_string.split(": ")[-1]
This does what you require and IMHO it's far more readable.
For a very long string it might be inefficient. But not for this purpose.
In Ruby, as in PCRE and Boost, you may make use of the \K match reset operator:
\K keeps the text matched so far out of the overall regex match. h\Kd matches only the second d in adhd.
So, you may use
/:[[:blank:]]*\K.+/ # To only match horizontal whitespaces with `[[:blank:]]`
/:\s*\K.+/ # To match any whitespace with `\s`
Seee the Rubular demo #1 and the Rubular demo #2 and
Details
: - a colon
[[:blank:]]* - 0 or more horizontal whitespace chars
\K - match reset operator discarding the text matched so far from the overall match memory buffer
.+ - matches and consumes any 1 or more chars other than line break chars (use /m modifier to match any chars including line break chars).
Given the following string, I'd like to match the elements of the list and parts of the rest after the colon:
foo,bar,baz:something
I.e. I am expecting the first three match groups to be "foo", "bar", "baz". No commas and no colon. The minimum number of elements is 1, and there can be arbitrarily many. Assume no whitespace and lower case.
I've tried this, which should work, but doesn't populate all the match groups for some reason:
^([a-z]+)(?:,([a-z]+))*:(something)
That matches foo in \1 and baz (or whatever the last element is) in \2. I don't understand why I don't get a match group for bar.
Any ideas?
EDIT: Ruby 1.9.3, if that matters.
EDIT2: Rubular link: http://rubular.com/r/pDhByoarbA
EDIT3: Add colon to the end, because I am not just trying to match the list. Sorry, oversimplified the problem.
This expression works for me: /(\w+)/i
If you want to do it with regex, how about this?
(?<=^|,)("[^"]*"|[^,]*)(?=,|$)
This matches comma-separated fields, including the possibility of commas appearing inside quoted strings like 123,"Yes, No". Regexr for this.
More verbosely:
(?<=^|,) # Must be preceded by start-of-line or comma
(
"[^"]*"| # A quote, followed by a bunch of non-quotes, followed by quote, OR
[^,]* # OR anything until the next comma
)
(?=,|$) # Must end with comma or end-of-line
Usage would be with something like Python's re.findall(), which returns all non-overlapping matches in the string (working from left to right, if that matters.) Don't use it with your equivalent of re.search() or re.match() which only return the first match found.
(NOTE: This actually doesn't work in Python because the lookbehind (?<=^|,) isn't fixed width. Grr. Open to suggestions on this one.)
Edit: Use a non-capturing group to consume start-of-line or comma, instead of a lookbehind, and it works in Python.
>>> test_str = '123,456,"String","String, with, commas","Zero-width fields next",,"",nyet,123'
>>> m = re.findall('(?:^|,)("[^"]*"|[^,]*)(?=,|$)',test_str)
>>> m
['123', '456', '"String"', '"String, with, commas"',
'"Zero-width fields next"', '', '""', 'nyet', '123']
Edit 2: The Ruby equivalent of Python's re.findall(needle, haystack) is haystack.scan(needle).
Maybe split will be better solution for this case?
'foo,bar,baz'.split(',')
=> ["foo", "bar", "baz"]
If I am interpreting your post correctly, you want everything separated by commas before the colon (:).
The appropriate regex for this would be:
[^\s:]*(,[^\s:]*)*(:.*)?
This should find everything you are looking for.