if (preg_match( ‘/(?<=\|\|) # Assert starting position directly after ||< [^,|]*, # Match any number of characters except , or |; then a , [^,|]*, # twice (so we have matched at least two commas) [^|]* # Then match anything except | characters (?=\|\|) # until we are right before || /x’, $subject, $regs)) { … Read more
Do a single regex replacement: string input = @”Hello World XXX Goodbye XXX Rabbit!”; Regex rgx = new Regex(@”\s*\w+\s+(?:XXX|xxx)”); // or maybe [Xx]{3} string result = rgx.Replace(input, “”, 1); Console.WriteLine(result); Hello Goodbye XXX Rabbit! Demo This replacement only would target XXX for removal if it be preceded by a word (one character or more). Explore … Read more
You can use regular expressions: import re s = [el for el in re.split(‘([\W+])’, ‘☕ Drink the ❶ best ☕coffee☕’) if el.strip()] print(s) output: [‘☕’, ‘Drink’, ‘the’, ‘❶’, ‘best’, ‘☕’, ‘coffee’, ‘☕’] 2 solved Separate string in Python [closed]
I didn’t really get what you mean, but if you want to replace a specific text in your file with a random integer then try: import fileinput from random import randint with fileinput.FileInput(fileToSearch, inplace=True, backup=’.bak’) as file: for line in file: print(line.replace(textToSearch, textToReplace), end=”) with textToReplace = randint(1990, 2020) Hope that helps 7 solved replace … Read more
Your pattern matches multiple positions. One before a character (including a character) and one right after. You can test it here. If you include a start string ancor it will work to match anything (even empty strings) and replace with Test ^.* solved Replace every value in a pandas dataframe series [duplicate]
No need to write a regex when the ability to parse URLs already exist. MDN: Web technology for developers > See Web APIs > URL const url = new URL(“https://stackoverflow.com/questions/ask”); console.log(`${url.origin}`); // Includes port console.log(`${url.protocol}//${url.hostname}`); // Alternatively… 2 solved How can I extract an actual URL from a string? [duplicate]
You want that $ in either case. Instead of ‘slash OR end’, it’s more ‘optional slash and then a very much not-optional end’. So.. /?$. You don’t need to normally escape slashes, especially in java regexes: Pattern.compile(“[-/](\\d+)/?$”) This reads: From minus or slash, a bunch of digits, then 0 or 1 slashes, then the end. … Read more
Regex is the wrong tool for handling HTML (or XML) 99.9% of the time. Instead, use a parser, like Nokogiri: require ‘nokogiri’ html=”<img src=”https://filin.mail.ru/pic?width=90&height=90&email=multicc%40multicc.mail.ru&version=4&build=7″ style=””>” doc = Nokogiri::HTML(html) url = doc.at(‘img’)[‘src’] # => “https://filin.mail.ru/pic?width=90&height=90&email=multicc%40multicc.mail.ru&version=4&build=7” doc.at(‘img’)[‘style’] # => “” Once you’ve retrieved the data you want, such as the src, use another “right” tool, such as … Read more
You can use this code: import re domain = “collabedge-123.dc-01.com” # preg_match(“/^(collabedge-|cb|ss)([0-9]+).dc-([0-9]+).com$/”, $domain, $matches); regex = r”^(collabedge-|cb|ss)([0-9]+).dc-([0-9]+).com$” res = re.match(regex,domain) print(res.group(0)) print(res.group(1)) print(res.group(2)) print(res.group(3)) Output: collabedge-123.dc-01.com collabedge- 123 01 1 solved How to split multiple string formats using regex and assigning its different groups to variable [closed]
Use this regex :(?<=[()\/*+-])(?=[0-9()])|(?<=[0-9()])(?=[()\/*+-]) It will match every position between a digit or a parenthesis and a operator or a parenthesis.(?<=[()\/*+-])(?=[0-9()]) matches the position with a parenthesis or an operator at the left and a digit or parenthesis at the right(?<=[0-9()])(?=[()\/*+-]) is the same but with left and right reversed. Demo here 5 solved Regex … Read more
You can try ^\+?[0-9-]+$ See it here on Regexr The important parts are the anchors ^ and $ for the start and the end of the string. I added also \+? at the start, to match an optional +. The + needs to be escaped since it is a special character in regex, the ? … Read more
If you must do it with regular expressions, try something like this: a = re.finditer(‘<a.+?question-hyperlink”>(.+?)</a>’, html) for m in a: print m.group(1) Just for the reference, this code does the same, but in a far more robust way: doc = BeautifulSoup(html) for a in doc.findAll(‘a’, ‘question-hyperlink’): print a.text solved Python: Find a Sentence between some … Read more
Search for (with regex setting turned on) BeanUtils\.copyProperties\s*\(\s*([\w\_]+)\s*\,\s*([\w\_]+)\s*\)\s*\; And replace with: BeanUtils.copyProperties($2, $1); First escape all literal characters with backslash \ Wherever a space can be found when writing code, match it with 0 or more spaces. That by using \s* Could use [ ]* but \s might be sufficient in this case. Then add … Read more
The problem you’re facing here is that your regex is replacing newline characters with a space before the final replacement of newline characters by <br /> elements occurs. As a result, in cases where your hashed string is preceded by a newline, the highlight will appear on the previous line, rather than correctly being placed … Read more
You can use negative lookbehind: String regex = “(?<!GoodClass\\.)\\bNiceMethod\\b”; 1 solved Regular expression to fetch lines evaluating only one part of a string, java