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
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.
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.*'<<<$'foo\nbar'
Prints bar
You can use this way...
grep -P '^\s$' 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. 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
.
Source: Stackoverflow.com