I'm getting a regex headache, so hopefully someone can help me here. I'm doing some file syntax conversion and I've got this situation in the files:
OpenMarker
keyword some expression
keyword some expression
keyword some expression
keyword some expression
keyword some expression
CloseMarker
I want to match all instances of "keyword" inside the markers. The marker areas are repeated and the keyword can appear in other places, but I don't want to match outside of the markers. What I don't seem to be able to work out is how to get a regex to pull out all the matches. I can get one to do the first or the last, but not to get all of them. I believe it should be possible and it's something to do with repeated capture groups -- can someone show me the light?
I'm using grepWin, which seems to support all the bells and whistles.
You could use:
(?<=OpenMarker((?!CloseMarker).)*)keyword(?=.*CloseMarker)
this will match the keyword inside OpenMarker and CloseMarker (using the option "dot matches newline").
sed -n -e '/OpenMarker[[:space:]]*CloseMarker/p' /path/to/file | grep keyword should work. Not sure if grep alone could do this.
There are only a few regex engines that support separate captures of a repeated group (.NET for example). So your best bet is to do this in two steps:
First match the section you're interested in: OpenMarker(.*?)CloseMarker (using the option "dot matches newline").
Then apply another regex to the match repeatedly: keyword (.*) (this time without the option "dot matches newline").
Related
In my elastic search setup, I would like to create tokens separated by either " " or "-" and greater than 3 chars.
I believe pattern tokenizer can work but I am not able to create the regular expression.
Please help me in regular expression
You should be able to use the following regex in the pattern field of your pattern tokenizer:
([^\s-]{3,})
The \s means any whitespace character.
The - means the literal dash character.
Putting the two of them between [^ and ] means match any character that isn't the ones in the list (in this case, anything not whitespace and not a dash)
The {3,} means the previous match has to occur 3 times or more.
The parenthesis around the entire statement means you want to capture what is inside, and the pattern tokenizer pulls its tokens from the matching groups of the regex.
You can play with this regex here and see how it will split your string:
https://regex101.com/r/2e9p34/1
On a side note, there may be other better ways to do this that will better handle edge cases you aren't thinking of, but I decided to answer your question exactly as you asked it. I highly recommend exploring all of the options ElasticSearch provides for its analyzers for your use case to see which one best fits your needs.
Hope this helps!
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).
I'm having an issue trying to capture a group on a string:
"type=gist\nYou need to gist this though\nbecause its awesome\nright now\n</code></p>\n\n<script src=\"https://gist.github.com/3931634.js\"> </script>\n\n\n<p><code>Not code</code></p>\n"
My regex currently looks like this:
/<code>([\s\S]*)<\/code>/
My goal is to get everything in between the code brackets. Unfortunately, it's matching up to the 2nd closing code bracket Is there a way to match everything inside the code brackets up until the first occurrence of ending code bracket?
All repetition quantifiers in regular expressions are greedy by default (matching as many characters as possible). Make the * ungreedy, like this:
/<code>([\s\S]*?)<\/code>/
But please consider using a DOM parser instead. Regex is just not the right tool to parse HTML.
And I just learned that for going through multiple parts, the
String.scan( /<code>(.*?)<\/code>/ ){
puts $1
}
is a very nice way of going through all occurences of code - but yes, getting a proper parser is better...
I am trying to construct a regular expression to detect a keyword in google search string. i.e. a string from google for a search term "amazing car" is
https://www.google.pl/#hl=pl&output=search&sclient=psy-ab&q=amazing+car&oq=amazing+car&aq=f& ... etc
I tried with this regular expression to detect a keyword car:
(google\.).+(&|\?)q=(car)
But this does not seem to work correctly. Am I missing something?
Thank you very much for advice
Your expression would match only if the query started with "car". If you use ".*" in the group, the greedy .+ will make the "q=" match the "oq=" later in the URL.
This may work for you:
(google\.).+(&|\?)q=([^&]*car)
Or, safer though more complex, apply this regexp which will capture the keyword in the only capture group:
https?://(?:[^/]+\.)?google\.[^/]+/[^?]*[?#](?:.*&)?q=([^&]*)
Or, if your regexp engine doesn't support, non-capture groups, use this:
https?://([^/]+\.)?google\.[^/]+/[^?]*[?#](.*&)?q=([^&]*)
and read your keyword in the third group.
can any body tell me how to use regex for negation of string?
I wanna find all line that start with public class and then any thing except first,second and finally any thing else.
for example in the result i expect to see public class base but not public class myfirst:base
can any body help me please??
Use a negative lookahead:
public\s+class\s+(?!first|second).+
If Peter is correct and you're using Visual Studio's Find feature, this should work:
^:b*public:b+class:b+~(first|second):i.*$
:b matches a space or tab
~(...) is how VS does a negative lookahead
:i matches a C/C++ identifier
The rest is standard regex syntax:
^ for beginning of line
$ for end of line
. for any character
* for zero or more
+ for one or more
| for alternation
Both the other two answers come close, but probably fail for different reasons.
public\s+class\s+(?:(?!first|second).)+
Note how there is a (non-capturing) group around the negative lookahead, to ensure it applies to more than just the first position.
And that group is less restrictive - since . excludes newline, it's using that instead of \S, and the $ is not necessary - this will exclude the specified words and match others.
No slashes wrapping the expression since those aren't required in everything and may confuse people that have only encountered string-based regex use.
If this still fails, post the exact content that is wrongly matched or missed, and what language/ide you are using.
Update:
Turns out you're using Visual Studio, which has it's own special regex implementation, for some unfathomable reason. So, you'll be wanting to try this instead:
public:b+class:b+~(first|second)+$
I have no way of testing that - if it doesn't work, try dropping the $, but otherwise you'll have to find a VS user. Or better still, the VS engineer(s) responsible for this stupid non-standard regex.
Here is something that should work for you
/public\sclass\s(?:[^fs\s]+|(?!first|second)\S)+(?=\s|$)/
The second look a head could be changed to a $(end of line) or another anchor that works for your particular use case, like maybe a '{'
Edit: Try changing the last part to:
(?=\s|$)