regex - How do I match any character across multiple lines in a regular expression?

ID : 5265

viewed : 466

Tags : regexmultilinelua-patternsmatlaboctaveboostoraclebashtclpostgresqlrsedperlgnu-greppcregrepphpc#powershellpythonjavagroovyscalajavascriptc++vbavbscriptrubyrtrebase-rricustringrstringigoswiftobjective-cre2google-apps-scriptregex

Top 5 Answer for regex - How do I match any character across multiple lines in a regular expression?

vote vote


Try this:


It basically says "any character or a newline" repeated zero or more times.

vote vote


It depends on the language, but there should be a modifier that you can add to the regex pattern. In PHP it is:


The s at the end causes the dot to match all characters including newlines.

vote vote


The question is, can the . pattern match any character? The answer varies from engine to engine. The main difference is whether the pattern is used by a POSIX or non-POSIX regex library.

A special note about : they are not considered regular expressions, but . matches any character there, the same as POSIX-based engines.

Another note on and : the . matches any character by default (demo): str = "abcde\n fghij<Foobar>"; expression = '(.*)<Foobar>*'; [tokens,matches] = regexp(str,expression,'tokens','match'); (tokens contain a abcde\n fghij item).

Also, in all of 's regex grammars the dot matches line breaks by default. Boost's ECMAScript grammar allows you to turn this off with regex_constants::no_mod_m (source).

As for (it is POSIX based), use the n option (demo): select regexp_substr('abcde' || chr(10) ||' fghij<Foobar>', '(.*)<Foobar>', 1, 1, 'n', 1) as results from dual

POSIX-based engines:

A mere . already matches line breaks, so there isn't a need to use any modifiers, see (demo).

The (demo), (demo), (TRE, base R default engine with no perl=TRUE, for base R with perl=TRUE or for stringr/stringi patterns, use the (?s) inline modifier) (demo) also treat . the same way.

However, most POSIX-based tools process input line by line. Hence, . does not match the line breaks just because they are not in scope. Here are some examples how to override this:

  • - There are multiple workarounds. The most precise, but not very safe, is sed 'H;1h;$!d;x; s/\(.*\)><Foobar>/\1/' (H;1h;$!d;x; slurps the file into memory). If whole lines must be included, sed '/start_pattern/,/end_pattern/d' file (removing from start will end with matched lines included) or sed '/start_pattern/,/end_pattern/{{//!d;};}' file (with matching lines excluded) can be considered.
  • - perl -0pe 's/(.*)<FooBar>/$1/gs' <<< "$str" (-0 slurps the whole file into memory, -p prints the file after applying the script given by -e). Note that using -000pe will slurp the file and activate 'paragraph mode' where Perl uses consecutive newlines (\n\n) as the record separator.
  • - grep -Poz '(?si)abc\K.*?(?=<Foobar>)' file. Here, z enables file slurping, (?s) enables the DOTALL mode for the . pattern, (?i) enables case insensitive mode, \K omits the text matched so far, *? is a lazy quantifier, (?=<Foobar>) matches the location before <Foobar>.
  • - pcregrep -Mi "(?si)abc\K.*?(?=<Foobar>)" file (M enables file slurping here). Note pcregrep is a good solution for macOS grep users.

See demos.

Non-POSIX-based engines:

  • - Use the s modifier PCRE_DOTALL modifier: preg_match('~(.*)<Foobar>~s', $s, $m) (demo)

  • - Use RegexOptions.Singleline flag (demo):
    - var result = Regex.Match(s, @"(.*)<Foobar>", RegexOptions.Singleline).Groups[1].Value;
    - var result = Regex.Match(s, @"(?s)(.*)<Foobar>").Groups[1].Value;

  • - Use the (?s) inline option: $s = "abcde`nfghij<FooBar>"; $s -match "(?s)(.*)<Foobar>"; $matches[1]

  • - Use the s modifier (or (?s) inline version at the start) (demo): /(.*)<FooBar>/s

  • - Use the re.DOTALL (or re.S) flags or (?s) inline modifier (demo): m ="(.*)<FooBar>", s, flags=re.S) (and then if m:, print(

  • - Use Pattern.DOTALL modifier (or inline (?s) flag) (demo): Pattern.compile("(.*)<FooBar>", Pattern.DOTALL)

  • - Use (?s) in-pattern modifier (demo): regex = /(?s)(.*)<FooBar>/

  • - Use (?s) modifier (demo): "(?s)(.*)<Foobar>".r.findAllIn("abcde\n fghij<Foobar>").matchData foreach { m => println( }

  • - Use [^] or workarounds [\d\D] / [\w\W] / [\s\S] (demo): s.match(/([\s\S]*)<FooBar>/)[1]

  • (std::regex) Use [\s\S] or the JavaScript workarounds (demo): regex rex(R"(([\s\S]*)<FooBar>)");

  • - Use the same approach as in JavaScript, ([\s\S]*)<Foobar>. (NOTE: The MultiLine property of the RegExp object is sometimes erroneously thought to be the option to allow . match across line breaks, while, in fact, it only changes the ^ and $ behavior to match start/end of lines rather than strings, the same as in JavaScript regex) behavior.)

  • - Use the /m MULTILINE modifier (demo): s[/(.*)<Foobar>/m, 1]

  • - Base R PCRE regexps - use (?s): regmatches(x, regexec("(?s)(.*)<FooBar>",x, perl=TRUE))[[1]][2] (demo)

  • - in stringr/stringi regex funtions that are powered with the ICU regex engine. Also use (?s): stringr::str_match(x, "(?s)(.*)<FooBar>")[,2] (demo)

  • - Use the inline modifier (?s) at the start (demo): re: = regexp.MustCompile(`(?s)(.*)<FooBar>`)

  • - Use dotMatchesLineSeparators or (easier) pass the (?s) inline modifier to the pattern: let rx = "(?s)(.*)<Foobar>"

  • - The same as Swift. (?s) works the easiest, but here is how the option can be used: NSRegularExpression* regex = [NSRegularExpression regularExpressionWithPattern:pattern options:NSRegularExpressionDotMatchesLineSeparators error:&regexError];

  • , - Use the (?s) modifier (demo): "(?s)(.*)<Foobar>" (in Google Spreadsheets, =REGEXEXTRACT(A2,"(?s)(.*)<Foobar>"))

NOTES ON (?s):

In most non-POSIX engines, the (?s) inline modifier (or embedded flag option) can be used to enforce . to match line breaks.

If placed at the start of the pattern, (?s) changes the bahavior of all . in the pattern. If the (?s) is placed somewhere after the beginning, only those .s will be affected that are located to the right of it unless this is a pattern passed to Python's re. In Python re, regardless of the (?s) location, the whole pattern . is affected. The (?s) effect is stopped using (?-s). A modified group can be used to only affect a specified range of a regex pattern (e.g., Delim1(?s:.*?)\nDelim2.* will make the first .*? match across newlines and the second .* will only match the rest of the line).

POSIX note:

In non-POSIX regex engines, to match any character, [\s\S] / [\d\D] / [\w\W] constructs can be used.

In POSIX, [\s\S] is not matching any character (as in JavaScript or any non-POSIX engine), because regex escape sequences are not supported inside bracket expressions. [\s\S] is parsed as bracket expressions that match a single character, \ or s or S.

vote vote


If you're using Eclipse search, you can enable the "DOTALL" option to make '.' match any character including line delimiters: just add "(?s)" at the beginning of your search string. Example:

vote vote


In many regex dialects, /[\S\s]*<Foobar>/ will do just what you want. Source

Top 3 video Explaining regex - How do I match any character across multiple lines in a regular expression?