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or more characters? Given a string, I want to do a regex on it. This is my current code: final Pattern pattern =



 

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. Jun 18, 2020 Assassins Creed 2. 9 CloneDVD PC DVD9 [WITH CRACK]. Action / Adventure 2010 1 DVD ISO Windows Crack Q: Java Regex: how to match 2 or more characters? Given a string, I want to do a regex on it. This is my current code: final Pattern pattern = Pattern.compile("[A-Z][0-9]+"); final Matcher matcher = pattern.matcher(s); It works for strings such as 12345, but when I add an extra character like so: final String s2 = "12345678"; If I then use my current regex, it will match 12345678, but I want to match 1234567. I know I can use \d{2} or even \d{2,4} but how can I combine these to make it so it matches 2 or more characters? I also want to match 0 or more non-alpha characters. I tried combining these like so: final Pattern pattern = Pattern.compile("[A-Z][0-9]\d{2}[^A-Z]"); But I get an error from the console that tells me: "Unknown or invalid modifier '^'". A: It seems you may be able to use alternation to accomplish your desired matching. I'm not positive, but you can also use a non-capturing group. Pattern: ([A-Z][0-9]{2})|([^A-Z]+) (...) Represents non-capturing group. Alternative: Pattern: ([A-Z][0-9]{2})+ (...) Represents non-capturing group. With either of these, your input might look like this: String in = "12345678"; Pattern p = Pattern.compile( "[A-Z][0-9]{2}|([^A-Z]+)"); Matcher m = p.matcher(in); while(m.find()){ // Matches 1 to 2 digits, 2 letters, or 1 to any non-alphanumeric character. } System.out.println(m.group(1)); Pattern p2




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