Remove the range of lines from file using shell script - shell

Question

My test file test.txt is given below:
destination mailerr { file("/var/log/mail.err" fsync(yes)); };
log { source(src); filter(f_mailerr); destination(mailerr); };
#
# and also all in one file:
#
destination mail { file("/var/log/mail"); };
log { source(src); filter(f_mail); destination(mail); };
destination mailwarn { file("/var/log/mail.warn"); };
log
{
#source(src);
filter(f_mailwarn); destination(mailwarn); };
I want to remove below lines using shell script
log
{
#source(src);
filter(f_mailwarn); destination(mailwarn); };
This lines might be comes different structure like
log{ source(src); filter(f_mailwarn); destination(mailwarn); };
(or)
log
{
source(src);
filter(f_mailwarn);
destination(mailwarn);
};
(or)
log
{ source(src); filter(f_mailwarn); destination(mailwarn); };
(or)
# log { source(src); filter(f_mailwarn); destination(mailwarn); };
};
These are the possibilities. I am using sed command:
sed '/log/{:a;/destination(mailwarn);.*}/d;N;ba}' test.txt
But it'll remove all line because first line it self "log" comes. So many of "log" comes in this file so how to remove the particular lines using shell script.

Answer 1

You can try this sed:
sed '/^[# ]*log/{:loop; /destination(.*}/d; /}/{n}; N; b loop;}' file
Explanation:
/^[# ]*log/ - Starts to process the block {} when regex matches.
/destination(.*}/d - Deletes the pattern space when regex matches. Starts a new cycle.
/}/{n} - When it finds }, then prints pattern space and reads next line of input. (for printing out an non log...destination() line.
N - This appends next line input to pattern space.
b loop - This transfers flow control to loop.

Answer 2

If the file is not too big, use the -z option to process entire file in one shot (#SLePort notes that this is a GNU specific option)
sed -z 's/log[ \t\n]*{[^}]*destination(mailwarn);[ \t\n]*};//g' test.txt
log start of match
[ \t\n]*{ zero or more non space/tab/newline characters followed by {
to avoid false matching log/mail.warn"); }; if [^{]*{ was used instead
[^}]* zero or more non } characters
destination(mailwarn); string to match
[ \t\n]*}; zero or more non space/tab/newline characters followed by };
the matched pattern gets deleted as replacement string is empty
Similar with perl
perl -0777 -pe 's/log\s*\{[^}]*destination\(mailwarn\);\s*};//g' test.txt

Answer 3

Here is my somewhat janky way of getting the job done. It seems to work in all the cases you posted, but is in no way a universal solution to this problem.
tr '\n' '\0' < test.txt | sed 's/log[^{}]*{[^}]*destination(mailwarn);[^}]*};//g' | tr '\0' '\n'
If test.txt contains null characters, it may not work properly. This command essentially looks for log followed by a { at some point, then until a } is reached destination(mailwarn);, eventually followed by };, and deletes this match if found.
I hope someone else posts a more robust solution to this problem, but this is the only quick solution I was able to come up with.

Answer 4

With sed:
sed '/^#*[[:blank:]]*log/{:a;/};/!{N;ba;s/\n//;};/destination(mailwarn)/d;}' file
Adds lines from log up to }; to the pattern space, search in it for destination(mailwarn) and if found, deletes lines.