How can I replace multiple empty lines with a single empty line in bash?

Question

I have a file that contains:

something



something else

something else again

I need a bash command, sed/grep w.e that will produce the fo

Answer 1

A solution with awk, which replaces several blank lines with a single blank line:

awk 'BEGIN{bl=0}/^$/{bl++;if(bl==1)print;else next}/^..*$/{bl=0;print}' myfile

Answer 2

Use awk:

awk '{ /^\s*$/?b++:b=0; if (b<=1) print }' file

Breakdown:

/^\s*$/?b++:b=0
    - ? :       the ternary operator
    - /^\s*$/   matches a blank line
    - b         variable that counts consecutive blank lines (b++).
                however, if the current line is non-blank, b is reset to 0.


if (b<=1) print
    print if the current line is non-blank (b==0)
          or if there is only one blank line (b==1).

By adjusting the regex, you can generalize it to other scenarios like squeezing multiple blank lines (">") in email: https://stackoverflow.com/a/59189823/12483961

Answer 3

Usually, if I find that sed can't do something I need, I turn to awk:

awk '
BEGIN {
    blank = 0;
}

/^[[:blank:]]*$/ {
     if (!blank) {
          print;
     }
     blank = 1;
     next;
}

{
     print;
     blank = 0;
}' file

Answer 4

Pipelining it to |uniq may be solution (if other than empty lines don't duplicate)

Answer 5

I take it that you'll probably want to remove lines that only have whitespace.

That can be done with:

sed /^[:space:]*$/d FILE

Answer 6

If someone want use perl

perl -00pe0 < file

will do the same, as cat -s :)