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...
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,regex-lookarounds,regex-greedy
^(?!(?:^|.*,)([^,\n]*),.*\1(?:,|$)).*$ Try this.See demo. https://regex101.com/r/wU7sQ0/24...
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 &...
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 ...
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...
Note that even with \{-\}, you still have the greedy .* within your group. Try select.*\ncoalesce\_.\{-\}\n) x...
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); } ...
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....
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'....
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();...
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...
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...
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 /,...
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;...
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...
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}?$....
How about [0-9]+[ ]*lbs?[ ]*[0-9]+[ ]*ozs? In your attempt, you're making the units optional, so it'll probably match stuff you don't want it to match. Make the 's' optional instead. Cheers, Paulo To capture the numbers, you'd need ([0-9]+)[ ]*lbs?[ ]*([0-9]+)[ ]*ozs?. To convert into kg, in Python you'd have...
That is because your limiting the string using the ^$ 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,sitemap,regex-negation,regex-greedy,regedit
Keep things simple: .*/f/$ vs .*/flight/$ ...