sed command to replace a blank line with two lines of content

Question

How do I replace the first blank line with two lines of content? I did see a question on replacing multiple blank lines with a single blank line in vim sed but don't quite see how to adapt that. So, for example, if my input file is:

% abd
% def

% jkl

% mno

I would like to have a sed command that replaces just the first blank line with these two lines (one containing ghi and the other containing %):

% abd
% def
% ghi
%
% jkl

% mno

Answer 1

Sed matches entire lines but doesn't include the newline, so a blank line will just be an empty string. You can use ^ to match the beginning of a line and $ to match the end, so ^$ matches a blank line. Just replace that with % ghi\n%:

sed 's/^$/% ghi\n%/'

The newline that already existed will remain, so you'll end up with % ghi on one line and % on the next

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Edit: If it needs to only match once then the expression is a bit more complicated. The easiest way I know to do it in sed is:

sed '0,/^$/ s/^$/% ghi\n%/'

The replacement is wrapped in an address range 0,/^$/, which means "only apply the following to the lines between 0 and the first line that matches ^$". Since the replacement expression checks for ^$ as well, the only line that's actually going to change is the first that matches ^$ -- the first blank line

Answer 2

sed '/^$/{i\
% ghi\
%
d
}'

When sed finds a blank line, this inserts the two lines '% ghi' and '%', then deletes the blank line.

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From the comments

This produces a syntax error.

Use a real shell instead of a sea-shell; it will save everyone grief in the long run. Place the commands on 5 lines of a simple file and use sed -f sed.script. There are probably other ways to achieve this in C shell - this works, though:

echo '/^$/{\'  > sed.script
echo '% ghi\' >> sed.script
echo '%'      >> sed.script
echo 'd'      >> sed.script
echo '}'      >> sed.script
sed -f sed.script data.file
rm -f sed.script