[ad_1] I’ll bite. This should work: var re = /(^\d{2}\.\d{4}$)/; (^ – begin \d{2} – match two digits \. – the period between digit sets \d{4} – match four digits $) – end And if you need the parantheses: var re = /(^\(\d{2}\.\d{4}\)$)/; jsFiddle 0 [ad_2] solved Regular Expression javascript for cip code [closed]
[ad_1] You can use \[.*?\]\{.*?\} on your example which I tested out here. This breaks into two smaller regex: \[.*?\], which is the square brackets with any chars inside. \{.*?\}, which is the curly brackets with any chars inside. Both of these are using a non-greedy .* inside the brackets. This way if there is … Read more
[ad_1] You should use preg_replace function. Example: $str=”var=104&anothervar=10&page=14&name=stack”; $str = preg_replace(‘/&?page=\d+&?/’, ”, $str); echo $str; &page=\d+? [&]? means 0 or 1 occurence of & character. (in case it was first parametr ampersand with no ampresand in the begining) \d+? means at least 1 or more number after = Output: var=104&anothervar=104&name=stack 9 [ad_2] solved Regular expression … Read more
[ad_1] Based on the expected output, you want more than just digits. The digits can have /, – and ,. Maybe, perhaps, you wanted something that starts with and ends with a digit, but anything in the middle other than a space? import re a=”invoice β invoice # 2018-33167-2 β date 03/21/2018 β total due … Read more
[ad_1] Not entirely sure what you’re planning to do, but if you want to stop people from entering more than 26 non-space characters in a row, then the regex /\S{27}/ will check for that. If it matches, the string contains 27 or more non-space characters. 10 [ad_2] solved regex space validation
[ad_1] You can do something like this https://regex101.com/r/bUflp4/2 for you regex. Be care however, it won’t accept tag inside others. For the replacement, you can use something like this var elements = document.getElementsByClassName(“text-to-replace”); for (var i = 0; i < elements.length; i++) { var innerHTML = elements.item(i).innerHTML; var regex = new RegExp(‘([^ ,<]* [^ ,<\/]*)’, ‘ig’); … Read more
[ad_1] You haven’t described your requirements very well. It looks like a simple substring formula is all you need. val str = “[CS]v1|<bunch of alpha numberic text>[CE]” val extract = str.substring(7, str.length-4) But if you really want to use regex this might do. val str = “[CS]v1|<bunch of alpha numberic text>[CE]” val extractor = “[^|]+\\|(.*)\\[..]”.r … Read more
[ad_1] RegEx: When given an string that does not match to pattern how to define if exist larger string (containing smaller one) that does match? [ad_2] solved RegEx: When given an string that does not match to pattern how to define if exist larger string (containing smaller one) that does match?
[ad_1] The PCRE pattern (([^()]+)\(((?R)?)\)) matches your input as follows: group 1: Match1(Match2()) group 2: Match1 group 3: Match2() Note: Python’s regex module, Perl, .NET, Ruby 2.0 and PHP support recursive patterns. Other popular languages like JavaScript, Ruby <1.9 and Java don’t. See: https://www.regular-expressions.info/recurse.html Demo: https://regex101.com/r/mX2fG5/57 3 [ad_2] solved Match between brakets that may contain … Read more
[ad_1] You could find all substrings of length 4+, and then down select from those to find only the shortest possible combinations that contain one of each letter: s=”AAGTCCTAG” def get_shortest(s): l, b = len(s), set(‘ATCG’) options = [s[i:j+1] for i in range(l) for j in range(i,l) if (j+1)-i > 3] return [i for i … Read more
[ad_1] You can use: [1-7]|9|8[abcd] Depending on whether you want the whole string to match, so that nothing else follows or precedes the match, you may need to add anchors: ^([1-7]|9|8[abcd])$ Or, alternatively, if you just want to match digits that are not followed by a letter (a, b, c, d) except when it is … Read more
[ad_1] preg_replace(‘/(<span.*\/span>)/u’, ”, $str); 3 [ad_2] solved php regex replace clear span code
[ad_1] Look at xgettext also xgettest-js – it should do what you’re looking for! [ad_2] solved Need a way to extract string constants from source code [closed]
[ad_1] I think you can do it using a positive lookahead until you encounter INFO or DEBUG or WARN. ERROR[\S\s]*?(?=\s+INFO|DEBUG|WARN) Match ERROR Match any whitespace character or any non-whitespace character zero or more times non greedy [\S\s]* A positive lookahead (?= Assert that what follows is one or more whitespaces \s+ Followed by either INFO … Read more
[ad_1] Using https://regex101.com/ the following works: 192\.168\.71\.1\sGET\sHTTP\/1\.0\s\/test\/abc Remember to escape special characters like . and \ if you want to use them as literals. If you want to use any of these characters as a literal in a regex, you need to escape them with a backslash Take a look at http://www.regular-expressions.info/characters.html to see more … Read more