Remove blank lines with grep

Question

I tried grep -v \'^$\' in Linux and that didn\'t work. This file came from a Windows file system.

Answer 1

The same as the previous answers:

grep -v -e '^$' foo.txt

Here, grep -e means the extended version of grep. '^$' means that there isn't any character between ^(Start of line) and $(end of line). '^' and '$' are regex characters.

So the command grep -v will print all the lines that do not match this pattern (No characters between ^ and $).

This way, empty blank lines are eliminated.

Answer 2

Do lines in the file have whitespace characters?

If so then

grep "\S" file.txt

Otherwise

grep . file.txt

Answer obtained from: https://serverfault.com/a/688789

Answer 3

Using Perl:

perl -ne 'print if /\S/'

\S means match non-blank characters.

Answer 4

Use:

$ dos2unix file
$ grep -v "^$" file

Or just simply awk:

awk 'NF' file

If you don't have dos2unix, then you can use tools like tr:

tr -d '\r' < "$file" > t ; mv t "$file"

Answer 5

awk 'NF' file-with-blank-lines > file-with-no-blank-lines

Answer 6

egrep -v "^\s\s+"

egrep already do regex, and the \s is white space.

The + duplicates current pattern.

The ^ is for the start