To be precise

Some text
Some text goes here.
Some more text

and I want to extract entire block that starts from "begin" till "end".

with awk we can do like awk '/begin/,/end/' text.

How to do with grep?


Updated 18-Nov-2016 (since grep behavior is changed: grep with -P parameter now doesn't support ^ and $ anchors [on Ubuntu 16.04 with kernel v:4.4.0-21-generic])(wrong (non-)fix)

$ grep -Pzo "begin(.|\n)*\nend" file
Some text goes here.  

note: for other commands just replace the '^' & '$' anchors with new-line anchor '\n' ______________________________

With grep command:

grep -Pzo "^begin\$(.|\n)*^end$" file

If you want don't include the patterns "begin" and "end" in result, use grep with Lookbehind and Lookahead support.

grep -Pzo "(?<=^begin$\n)(.|\n)*(?=\n^end$)" file

Also you can use \K notify instead of Lookbehind assertion.

grep -Pzo "^begin$\n\K(.|\n)*(?=\n^end$)" file

\K option ignore everything before pattern matching and ignore pattern itself.
\n used for avoid printing empty lines from output.

Or as @AvinashRaj suggests there are simple easy grep as following:

grep -Pzo "(?s)^begin$.*?^end$" file

grep -Pzo "^begin\$[\s\S]*?^end$" file

(?s) tells grep to allow the dot to match newline characters.
[\s\S] matches any character that is either whitespace or non-whitespace.

And their output without including "begin" and "end" is as following:

grep -Pzo "^begin$\n\K[\s\S]*?(?=\n^end$)" file # or grep -Pzo "(?<=^begin$\n)[\s\S]*?(?=\n^end$)"

grep -Pzo "(?s)(?<=^begin$\n).*?(?=\n^end$)" file

see the full test of all commands here (out of dated as grep behavior with -P parameter is changed)


^ point the beginning of a line and $ point the end of a line. these added to the around of "begin" and "end" to matching them if they are alone in a line.
In two commands I escaped $ because it also using for "Command Substitution"($(command)) that allows the output of a command to replace the command name.

From man grep:

-o, --only-matching
      Print only the matched (non-empty) parts of a matching line,
      with each such part on a separate output line.

-P, --perl-regexp
      Interpret PATTERN as a Perl compatible regular expression (PCRE)

-z, --null-data
      Treat the input as a set of lines, each terminated by a zero byte (the ASCII 
      NUL character) instead of a newline. Like the -Z or --null option, this option 
      can be used with commands like sort -z to process arbitrary file names.
  • 1
    change your grep like grep -Pzo "(?<=begin\n)(.|\n)*(?=\nend)" file to not to print \n character which exists on the line begin. Nov 19 '14 at 12:18
  • 1
    Use DOTALL modifier to make dot to match even newline chars also grep -Pzo "(?s)begin.*?end" file Nov 19 '14 at 12:19
  • Or Simply, grep -Pzo "begin[\s\S]*?end" file Nov 19 '14 at 12:20
  • 1
    The siólution doesn't work. It produces an error: grep: ein nicht geschütztes ^ oder $ wird mit -Pz nicht unterstützt The translation of the error is something like: grep: a not protected ^ or $ is not supported with -Pz
    – musbach
    Nov 15 '16 at 8:01
  • 1
    Yes, I know, that's in your answer. I'm sure it worked when you posted this, but try it again today. The behavior of grep seems to have changed.
    – terdon
    Nov 15 '16 at 9:24

In case your grep doesn't support perl syntax (-P), you can try joining the lines, matching the pattern, then expanding the lines again as below:

$ tr '\n' , < foo.txt | grep -o "begin.*end" | tr , '\n'
Some text goes here.

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