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
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/^$/% ghi\n%/'
The newline that already existed will remain, so you'll end up with % ghi on one line and % on the next
Edit: If it needs to only match once then the expression is a bit more complicated. The easiest way I know to do it in sed is:
sed '0,/^$/ 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
Here's another way with sed that replaces only the first empty line without using \n in the RHS or the gnu sed 0 address extension:
sed '/^$/{ # if line is empty
x # exchange pattern space w. hold space
//{ # if pattern space is empty (means the hold buffer was empty)
s//%/ # replace it with a % character
h # overwrite hold space (so now hold space looks like this: ^%$)
s/$/ ghi/ # add a space and the string ghi after the %, then
G # append content of hold buffer to pattern space so now the
} # pattern space looks like this: ^% ghi\n%$
//!x # if pattern space is not empty it means a change was
} # already made so exchange back (do nothing)
' infile
one liner:
sed -e'/^$/{x;//{s//%/;h;s/$/ ghi/;G' -e'}' -e'//!x' -e'}' infile
Though really, this is piece of cake for ed:
ed -s infile <<< $'/^$/s//% ghi\\\n%/\n,p\nq'
replace ,p with w to edit the file in-place.
NB: This answers the original question, which asked for:
How do I replace a single blank line with two lines of content?
This does not answer the 'chameleon' question.
sed '/^$/{i\
% ghi\
%
d
}'
When sed finds a blank line, this inserts the two lines '% ghi' and '%', then deletes the blank line.
From the comments
This produces a syntax error.
Use a real shell instead of a sea-shell; it will save everyone grief in the long run. Place the commands on 5 lines of a simple file and use sed -f sed.script. There are probably other ways to achieve this in C shell - this works, though:
echo '/^$/{\' > 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
sed -e1\!b -e:n -e"/^$/c$(printf '\\\n%%%s' \ ghi '')" -en\;bn
sed understands three primitives for scripted, fixed-string writes to stdout. All three begin after the immediately following, backslash-escaped newline in the script and end either at the first occurring unescaped newline in the script or the first occurring end-of-file in the script:
i
inserts the fixed-string to standard out right nowa
append to stdout in the order of scripted occurrence before the next line-cycle, or, for the last line, at the end of this onec
(1(,2)?)?, ends the current line-cycle, and changes output to the fixed-string for the last of any address[es]printf '\n\n\n\n\n' |
sed -e1\!b -e:n \
-e"/^$/c$(printf '\\\n%%%s' \ ghi '')" \
-en\;bn
% ghi %
So this script branches out of the script for every line which is ! not the 1st, but from the first line until it can change a blank line to the fixed-string, it auto-prints every not-blank line before overwriting it with the next and looping back to the :next label to go again.
