Remove blank lines with grep

Question

Try the following:

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

The -e option allows regex patterns for matching.

The single quotes around ^$ makes it work for Cshell. Other shells will be happy with either single or double quotes.

UPDATE: This works for me for a file with blank lines or "all white space" (such as windows lines with "\r\n" style line endings), whereas the above only removes files with blank lines and unix style line endings:

grep -v -e '^[[:space:]]*$' foo.txt

Keep it simple.

grep . filename.txt

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"

grep -v "^[[:space:]]*$"

The -v makes it print lines that do not completely match

===Each part explained===
^             match start of line
[[:space:]]   match whitespace- spaces, tabs, carriage returns, etc.
*             previous match (whitespace) may exist from 0 to infinite times
$             match end of line

Running the code-

$ echo "
> hello
>       
> ok" |
> grep -v "^[[:space:]]*$"
hello
ok

To understand more about how/why this works, I recommend reading up on regular expressions. http://www.regular-expressions.info/tutorial.html

If you have sequences of multiple blank lines in a row, and would like only one blank line per sequence, try

grep -v "unwantedThing" foo.txt | cat -s

cat -s suppresses repeated empty output lines.

Your output would go from

match1



match2

to

match1

match2

The three blank lines in the original output would be compressed or "squeezed" into one blank line.