how to match groups in regex?
java,regex,regex-lookarounds,regex-greedy
date, time is nothing in regex, you should parse this string for date, time. ([0-9]{4}-[0-9]{2}-[0-9]{2})\s([0-9]{2}:[0-9]{2}:[0-9]{2}) first group is date, second group is time. For example, this link...
Regex for URL to sites
regex,sitemap,regex-negation,regex-greedy,regedit
Keep things simple: .*/f/$ vs .*/flight/$ ...
Why is non-greedy character not acting 'non-greedy'?
Regular expressions will always try to match left to right. Even though .+? is non-greedy, the regex will still try to match from the beginning of the string if possible and only advance the starting position when the match fails. You have a couple of options to fix this: Include...
Java string split regex non-capturing group
regex-greedy,capturing-group,regex-group
To do that, the best option is to use the "find" method instead of split with this pattern: (?<=\\[)[^\\]]*(?=\\])|[^\\][.]+ Note that the order of the alternatives is important because the first win. So (?<=\\[)[^\\]]*(?=\\]) must be before [^\\][.]+ demo...
Regex lazy quantifier asserted beginning and end
regex,match,lazy-evaluation,regex-greedy
Both won't match your test string because your input contains 5 characters where your regex only matches the string which has 2 to 4 characters. This ? won't be needed when anchors are used. ^.*?$ does the same as ^.*$. And i suggest you to use ^\w{2,4}$ instead of ^\w{2,4}?$....
Regex - match a word and capture that word
That is because your limiting the string using the ^$ characters. Just do it as follows: (pic(ture)?s?\d+).*?event,'(\d+)' And the \1 and \3 captured groups will contain the picture name and event id respectively Demo The regex above will also match cases in your example such as pic1 and pictures10, and...
Regex to avoid data duplication in delimited string?
regex,regex-lookarounds,regex-greedy
^(?!(?:^|.*,)([^,\n]*),.*\1(?:,|$)).*$ Try this.See demo. https://regex101.com/r/wU7sQ0/24...
Regex to get lbs and ozs i.e. 16lb 4ozs and it's variations
How about [0-9]+[ ]*lbs?[ ]*[0-9]+[ ]*ozs? In your attempt, you're making the units optional, so it'll probably match stuff you don't want it to match. Make the 's' optional instead. Cheers, Paulo To capture the numbers, you'd need ([0-9]+)[ ]*lbs?[ ]*([0-9]+)[ ]*ozs?. To convert into kg, in Python you'd have...
How can I create a regex to match the inner-most match or work from right-to-left?
regex,regex-negation,regex-lookarounds,regex-greedy,grunt-usemin
You can isolate the information you want by using ["'](images\S+.svg)['"] or ["'](\S+.svg)['"] The first solution looks at 'images\anything-but-a-non-whitespace.svg' whereas the second solution looks at 'anything-but-a-non-whitespace.svg'....
Non-greedy match in vim
Note that even with \{-\}, you still have the greedy .* within your group. Try select.*\ncoalesce\_.\{-\}\n) x...
Java regex expression trouble in split function call
java,regex,string,split,regex-greedy
You could use the following expression: \b;\b. An example is available here. String str = "today;i;;drank water"; String[] split = str.split("\\b;\\b"); for(String s : split) System.out.println(s); Yields: today i;;drank water ...
how to code correct URI regex
This should do it: [^\./]+\.[^\./]+\.[^\./]+(?:/(.*))? That is: [^\./]+ = (anything but . and /) \. = dot ...? = Zero or one occurrence(s) of ... (?:...)? = Zero or one of ..., which is more than one character, but without capturing .... (?:/(.*))? = Capture everything after the last /,...
regex to ignore a character in between two string?
Try this, if (s.contains("<td>")) { String first = s.substring(0, s.indexOf("<td>")); String last = s.substring(s.indexOf("<td>"), s.indexOf("</td>") + 5); System.out.println("result : "+first + last.replace("**", "")); } else { System.out.println("result : "+s); } ...
Regex Or and Greedy
regex,regex-lookarounds,regex-greedy
You can use just this negation regex: some_text=(?<value>[^&]*) [^&]* will match 0 or more of any character that is not &...
Tcl greedy subexpression difference between + and *
Regarding the non-greediness. Tcl regular expressions have a quirk: the first quantifier in the expression sets the greediness for the whole expression. (See the "Matching" section of the re_syntax manual page, paying close attention to the word "preference"): A branch has the same preference as the first quantified atom in...
Shortest match in regex from end
Use negative lookahead assertion. foo(?:(?!foo).)*?boo DEMO (?:(?!foo).)*? - Non-greedy match of any character but not of foo zero or more times. That is, before matching each character, it would check that the character is not the letter f followed by two o's. If yes, then only the corresponding character will...
Regex first occurrence
I'm reposting as an answer since you said the comment solved your problem, but I need to reiterate Joeytje50's comment first don't parse html with regex's. Now that we've got that out of the way and you promise to only use this for educational purposes and never ever in production;...
C++ Regex: non-greedy match
c++,regex,c++11,c++14,regex-greedy
As Casimir et Hippolyte explained in his comment, I just need to: remove the quantifier Use std::regex_iterator It gives me the following code: std::regex paramsRegex("[\\?&]([^=]+)=([^&]+)"); std::string url_params = "?key1=val1&key2=val2&key3=val3&key4=val4"; std::smatch sm; auto params_it = std::sregex_iterator(url_params.cbegin(), url_params.cend(), paramsRegex); auto params_end = std::sregex_iterator(); while (params_it != params_end) { auto param = params_it->str();...
Optional match pattern after wildcard
regex,pattern-matching,regex-greedy
Like i said in my comment, you need to change the first .* to .*?. Because .* is greedy and it matches all the characters as much as possible. Changing it to .*?, will do a non-greedy match. ^(.*?) (\d*)x(\d*)(?:\s*&\s*\d*x\d*)?\s*(.*) (\(\d*\))$ DEMO...
Java regular expression matching two consecutive consonants
java,regex,string,regex-greedy
[^aeiou] matches non-letter characters as well, so you should use a different pattern: Pattern rx = Pattern.compile("[bcdfghjklmnpqrstuvwxyz]{2}", Pattern.CASE_INSENSITIVE); if (rx.matches(myString)) {...} If you would like to use && for an intersection, you can do it like this: "[a-z&&[^aeiou]]{2}" Demo....