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

For example, this regex

(.*)<FooBar>

will match:

abcde<FooBar>

But how do I get it to match across multiple lines?

abcde
fghij<FooBar>

In Javascript you can use [^]* to search for zero to infinite characters, including line breaks.

$("#find_and_replace").click(function() {_x000D_
  var text = $("#textarea").val();_x000D_
  search_term = new RegExp("[^]*<Foobar>", "gi");;_x000D_
  replace_term = "Replacement term";_x000D_
  var new_text = text.replace(search_term, replace_term);_x000D_
  $("#textarea").val(new_text);_x000D_
});

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>_x000D_
<button id="find_and_replace">Find and replace</button>_x000D_
<br>_x000D_
<textarea ID="textarea">abcde_x000D_
fghij&lt;Foobar&gt;</textarea>

"." normally doesn't match line-breaks. Most regex engines allows you to add the S-flag (also called DOTALL and SINGLELINE) to make "." also match newlines. If that fails, you could do something like [\S\s].

In java based regular expression you can use [\s\S]

The question is, can . 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.

Special note about lua-patterns: they are not considered regular expressions, but . matches any char there, same as POSIX based engines.

Another note on matlab and octave: the . matches any char 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 boost'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 oracle (it is POSIX based), use 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, no need to use any modifiers, see bash (demo).

The tcl (demo), postgresql (demo), r (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:

• sed - 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 - 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.
• gnu-grep - 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 - pcregrep -Mi "(?si)abc\K.*?(?=<Foobar>)" file (M enables file slurping here). Note pcregrep is a good solution for Mac OS grep users.

See demos.

Non-POSIX-based engines:

• php - Use s modifier PCRE_DOTALL modifier: preg_match('~(.*)<Foobar>~s', $s, $m) (demo)
• c# - 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;
• powershell - Use (?s) inline option: $s = "abcde`nfghij<FooBar>"; $s -match "(?s)(.*)<Foobar>"; $matches[1]
• perl - Use s modifier (or (?s) inline version at the start) (demo): /(.*)<FooBar>/s
• python - Use re.DOTALL (or re.S) flags or (?s) inline modifier (demo): m = re.search(r"(.*)<FooBar>", s, flags=re.S) (and then if m:, print(m.group(1)))
• java - Use Pattern.DOTALL modifier (or inline (?s) flag) (demo): Pattern.compile("(.*)<FooBar>", Pattern.DOTALL)
• groovy - Use (?s) in-pattern modifier (demo): regex = /(?s)(.*)<FooBar>/
• scala - Use (?s) modifier (demo): "(?s)(.*)<Foobar>".r.findAllIn("abcde\n fghij<Foobar>").matchData foreach { m => println(m.group(1)) }
• javascript - Use [^] or workarounds [\d\D] / [\w\W] / [\s\S] (demo): s.match(/([\s\S]*)<FooBar>/)[1]
• c++ (std::regex) Use [\s\S] or the JS workarounds (demo): regex rex(R"(([\s\S]*)<FooBar>)");

• vba vbscript - 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, same as in JS regex) behavior.)

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

• rtrebase-r - Base R PCRE regexps - use (?s): regmatches(x, regexec("(?s)(.*)<FooBar>",x, perl=TRUE))[[1]][2] (demo)
• ricustringrstringi - in stringr/stringi regex funtions that are powered with ICU regex engine, also use (?s): stringr::str_match(x, "(?s)(.*)<FooBar>")[,2] (demo)
• go - Use the inline modifier (?s) at the start (demo): re: = regexp.MustCompile(`(?s)(.*)<FooBar>`)
• swift - Use dotMatchesLineSeparators or (easier) pass the (?s) inline modifier to the pattern: let rx = "(?s)(.*)<Foobar>"
• objective-c - 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];
• re2, google-apps-script - Use (?s) modifier (demo): "(?s)(.*)<Foobar>" (in Google Spreadsheets, =REGEXEXTRACT(A2,"(?s)(.*)<Foobar>"))

NOTES ON (?s):

In most non-POSIX engines, (?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 . will be affected that are located to the right of it unless this is a pattern passed to Python re. In Python re, regardless of the (?s) location, the whole pattern . are 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 char, [\s\S] / [\d\D] / [\w\W] constructs can be used.

In POSIX, [\s\S] is not matching any char (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 char, \ or s or S.

([\s\S]*)<FooBar>

The dot matches all except newlines (\r\n). So use \s\S, which will match ALL characters.

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

Often we have to modify a substring with a few keywords spread across lines preceding the substring. Consider an xml element:

<TASK>
  <UID>21</UID>
  <Name>Architectural design</Name>
  <PercentComplete>81</PercentComplete>
</TASK>

Suppose we want to modify the 81, to some other value, say 40. First identify .UID.21..UID., then skip all characters including \n till .PercentCompleted.. The regular expression pattern and the replace specification are:

String hw = new String("<TASK>\n  <UID>21</UID>\n  <Name>Architectural design</Name>\n  <PercentComplete>81</PercentComplete>\n</TASK>");
String pattern = new String ("(<UID>21</UID>)((.|\n)*?)(<PercentComplete>)(\\d+)(</PercentComplete>)");
String replaceSpec = new String ("$1$2$440$6");
//note that the group (<PercentComplete>) is $4 and the group ((.|\n)*?) is $2.

String  iw = hw.replaceFirst(pattern, replaceSpec);
System.out.println(iw);

<TASK>
  <UID>21</UID>
  <Name>Architectural design</Name>
  <PercentComplete>40</PercentComplete>
</TASK>

The subgroup (.|\n) is probably the missing group $3. If we make it non-capturing by (?:.|\n) then the $3 is (<PercentComplete>). So the pattern and replaceSpec can also be:

pattern = new String("(<UID>21</UID>)((?:.|\n)*?)(<PercentComplete>)(\\d+)(</PercentComplete>)");
replaceSpec = new String("$1$2$340$5")

and the replacement works correctly as before.

Note that (.|\n)* can be less efficient than (for example) [\s\S]* (if your language's regexes support such escapes) and than finding how to specify the modifier that makes . also match newlines. Or you can go with POSIXy alternatives like [[:space:][:^space:]]*.

For Eclipse worked following expression:

Foo

jadajada Bar"

Regular-Expression:

Foo[\S\s]{1,10}.*Bar*

/(.*)<FooBar>/s

the s causes Dot (.) to match carriage returns

generally . doesn't match newlines, so try ((.|\n)*)<foobar>

Try: .*\n*.*<FooBar> assuming you are also allowing blank newlines. As you are allowing any character including nothing before <FooBar>.

I wanted to match a particular if block in java

   ...
   ...
   if(isTrue){
       doAction();

   }
...
...
}

If I use the regExp

if \(isTrue(.|\n)*}

it included the closing brace for the method block so I used

if \(!isTrue([^}.]|\n)*}

to exclude the closing brace from the wildcard match.

In the context of use within languages, regular expressions act on strings, not lines. So you should be able to use the regex normally, assuming that the input string has multiple lines.

In this case, the given regex will match the entire string, since "<FooBar>" is present. Depending on the specifics of the regex implementation, the $1 value (obtained from the "(.*)") will either be "fghij" or "abcde\nfghij". As others have said, some implementations allow you to control whether the "." will match the newline, giving you the choice.

Line-based regular expression use is usually for command line things like egrep.

Try this:

((.|\n)*)<FooBar>

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

In Ruby ruby you can use the 'm' option (multiline):

/YOUR_REGEXP/m

See the Regexp documentation on ruby-doc.org for more information.

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:

(?s).*<FooBar>

we can also use

(.*?\n)*?

to match everything including newline without greedy

This will make the new line optional

(.*?|\n)*?

Typically searching for three consecutive lines in Powershell it would look like:

$file = get-content file.txt -raw

$pattern = 'lineone\r\nlinetwo\r\nlinethree\r\n'     # "windows" text
$pattern = 'lineone\nlinetwo\nlinethree\n'           # "unix" text
$pattern = 'lineone\r?\nlinetwo\r?\nlinethree\r?\n'  # both

$file -match $pattern

# output
True

Bizarrely, this would be unix text at the prompt, but windows text in a file:

$pattern = 'lineone
linetwo
linethree
'

Here's a way to print out the line endings:

'lineone
linetwo
linethree
' -replace "`r",'\r' -replace "`n",'\n'

# output
lineone\nlinetwo\nlinethree\n

I had the same problem and solved it in probably not the best way but it works. I replaced all line breaks before I did my real match:

mystring= Regex.Replace(mystring, "\r\n", "")

I am manipulating HTML so line breaks don't really matter to me in this case.

I tried all of the suggestions above with no luck, I am using .Net 3.5 FYI

Solution:

Use pattern modifier sU will get the desired matching in PHP.

example:

preg_match('/(.*)/sU',$content,$match);

Source:

http://dreamluverz.com/developers-tools/regex-match-all-including-new-line http://php.net/manual/en/reference.pcre.pattern.modifiers.php

Use RegexOptions.Singleline, it changes the meaning of . to include newlines

Regex.Replace(content, searchText, replaceText, RegexOptions.Singleline);

Option 1

One way would be to use the s flag (just like the accepted answer):

/(.*)<FooBar>/s

Demo 1

Option 2

A second way would be to use the m (multiline) flag and any of the following patterns:

/([\s\S]*)<FooBar>/m

or

/([\d\D]*)<FooBar>/m

or

/([\w\W]*)<FooBar>/m

Demo 2

RegEx Circuit

jex.im visualizes regular expressions: