I’m guessing that here we wish to validate these emails, for which we can use an expression similar to: ^([a-z0-9]+((?=\..+@.+\..+)[a-z0-9.]*@[a-z0-9]*\.[a-z0-9]*))$ with an i flag and we can allow more chars, if we like so. Demo RegEx Circuit jex.im visualizes regular expressions: solved RegEx to allow at least one dot and all character
You’re asking to match any number of lines of text, followed by only 1 additional newline at the end, simply: ^(.+\n)+\n(?!\n) will do what you’d like. Example here: https://regex101.com/r/Hy3buP/1 Explanation: ^ – Assert position at start of string (.+\n)+ – Match any positive number of lines of text ending in newline \n – Match the … Read more
As you guessed correctly, this has to do with capturing groups. According to the documentation for re.findall: If one or more groups are present in the pattern, return a list of groups Therefore, you need to make all your groups () non-capturing using the (?:) specifier. If there are no captured groups, it will return … Read more
I believe you are looking for a regex pattern (?<!%%)(?<=%)\w+(?=%)(?!%%) That would find variables that are surrounded by a single % character on each side. Test the regex here. Java code: final Pattern pattern = Pattern.compile(“(?<!%%)(?<=%)\\w+(?=%)(?!%%)”); final Matcher matcher = pattern.matcher(input); while (matcher.find()) { System.out.println(matcher.group(0)); } Test the Java code here. UPDATE: If you want … Read more