Shell: simple way to get all lines before first blank line

Question

What's the best shell command to output the lines of a file until you encounter the first blank line? For example:

output these
lines

but do not output anything after the above blank line
(or the blank line itself)

awk? something else?

Answer 1

sed -e '/^$/,$d' <<EOF
this is text
so is this

but not this
or this
EOF

Answer 2

More awk:

awk -v 'RS=\n\n' '1;{exit}'

More sed:

sed -n -e '/./p;/./!q'
sed -e '/./!{d;q}'
sed -e '/./!Q'   # thanks to Jefromi

How about directly in shell?

while read line; do [ -z "$line" ] && break; echo "$line"; done

(Or printf '%s\n' instead of echo, if your shell is buggy and always handles escapes.)

Answer 3

# awk '!NF{exit}1' file
output these
lines

Answer 4

With sed:

sed '/^$/Q' <file>

Edit: sed is way, way, way faster. See ephemient's answer for the fastest version.

To do this in awk, you could use:

awk '{if ($0 == "") exit; else print}' <file>

Note that I intentionally wrote this to avoid using regular expressions. I don't know what awk's internal optimizations are like, but I suspect direct string comparison would be faster.

Answer 5

Awk solution

awk '/^$/{exit} {print} ' <filename>

Answer 6

Here's a solution using Perl:

#! perl

use strict;
use warnings;

while (<DATA>) {
    last if length == 1;
    print;
}

__DATA__
output these
lines

but don't output anything after the above blank line
(or the blank line itself)

Answer 7

A couple of Perl one-liners

$ perl -pe'last if /^$/' file..

$ perl -lpe'last unless length' file..

Answer 8

Another Perl solution:

perl -00 -ne 'print;exit' file
perl -00 -pe 'exit if $. == 2' file