Which characters you must and which you mustn’t escape indeed depends on the regex flavor you’re working with. For PCRE, and most other so-called Perl-compatible flavors, escape these outside character classes: .^$*+?()[{\| and these inside character classes: ^-]\ For POSIX extended regexes (ERE), escape these outside character classes (same as PCRE): .^$*+?()[{\| Escaping any other … Read more
This tested snippet should do it: import re line = re.sub(r”</?\[\d+>”, “”, line) Edit: Here’s a commented version explaining how it works: line = re.sub(r””” (?x) # Use free-spacing mode. < # Match a literal ‘<‘ /? # Optionally match a “https://stackoverflow.com/” \[ # Match a literal ‘[‘ \d+ # Match one or more digits … Read more
This is also possible with Notepad++: Go to the search menu, Ctrl + F, and open the Mark tab. Check Bookmark line (if there is no Mark tab update to the current version). Enter your search term and click Mark All All lines containing the search term are bookmarked. Now go to the menu Search … Read more
You need to escape the dot if you want to split on a literal dot: String extensionRemoved = filename.split(“\\.”)[0]; Otherwise you are splitting on the regex ., which means “any character”. Note the double backslash needed to create a single backslash in the regex. You’re getting an ArrayIndexOutOfBoundsException because your input string is just a … Read more
The non-greedy ? works perfectly fine. It’s just that you need to select dot matches all option in the regex engines (regexpal, the engine you used, also has this option) you are testing with. This is because, regex engines generally don’t match line breaks when you use .. You need to tell them explicitly that … Read more
Assuming you want the part between single quotes, use this regular expression with a Matcher: “‘(.*?)'” Example: String mydata = “some string with ‘the data i want’ inside”; Pattern pattern = Pattern.compile(“‘(.*?)'”); Matcher matcher = pattern.matcher(mydata); if (matcher.find()) { System.out.println(matcher.group(1)); } Result: the data i want
Use a HTML parser instead of regex. This is dead simple with Jsoup. public static String html2text(String html) { return Jsoup.parse(html).text(); } Jsoup also supports removing HTML tags against a customizable whitelist, which is very useful if you want to allow only e.g. <b>, <i> and <u>. See also: RegEx match open tags except XHTML … Read more
import re pattern = re.compile(“^([A-Z][0-9]+)+$”) pattern.match(string)
Regex: match everything but: a string starting with a specific pattern (e.g. any – empty, too – string not starting with foo): Lookahead-based solution for NFAs: ^(?!foo).*$ ^(?!foo) Negated character class based solution for regex engines not supporting lookarounds: ^(([^f].{2}|.[^o].|.{2}[^o]).*|.{0,2})$ ^([^f].{2}|.[^o].|.{2}[^o])|^.{0,2}$ a string ending with a specific pattern (say, no world. at the end): Lookbehind-based … Read more
Neither basic nor extended Posix/GNU regex recognizes the non-greedy quantifier; you need a later regex. Fortunately, Perl regex for this context is pretty easy to get: perl -pe ‘s|(http://.*?/).*|\1|’