How to give a pattern for new line in grep? New line at beginning, new line at end. Not the regular expression way. Something like \n.

3

6 Answers

try pcregrep instead of regular grep:

pcregrep -M "pattern1.*\n.*pattern2" filename

the -M option allows it to match across multiple lines, so you can search for newlines as \n.

0

grep patterns are matched against individual lines so there is no way for a pattern to match a newline found in the input.

However you can find empty lines like this:

grep '^$' file
grep '^[[:space:]]*$' file # include white spaces 

2

Thanks to @jarno I know about the -z option and I found out that when using GNU grep with the -P option, matching against \n is possible. :)

Example:

grep -zoP 'foo\n\K.*'<<<$'foo\nbar'

Result:

bar

Example that involves matching everything including newlines:

.* will not match newlines. To match everything including newlines, use¹ (.|\n)*:

grep -zoP 'foo\n\K(.|\n)*'<<<$'foo\nbar\nqux'

Result:

bar
qux

¹ Seen here:

3

You can use this way...

grep -P '^\s$' file

• -P is used for Perl regular expressions (an extension to POSIX grep).
• \s match the white space characters; if followed by *, it matches an empty line also.
• ^ matches the beginning of the line. $ matches the end of the line.

5

As for the workaround (without using non-portable -P), you can temporary replace a new-line character with the different one and change it back, e.g.:

grep -o "_foo_" <(paste -sd_ file) | tr -d '_'

Basically it's looking for exact match _foo_ where _ means \n (so __ = \n\n). You don't have to translate it back by tr '_' '\n', as each pattern would be printed in the new line anyway, so removing _ is enough.

1

just found

grep $'\r'

It's using $'\r' for c-style escape in Bash.

in this article

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