Using sed to delete all lines between two matching patterns - regex

Question

I have a file something like:
# ID 1
blah blah
blah blah
$ description 1
blah blah
# ID 2
blah
$ description 2
blah blah
blah blah
How can I use a sed command to delete all lines between the # and $ line? So the result will become:
# ID 1
$ description 1
blah blah
# ID 2
$ description 2
blah blah
blah blah
Can you please kindly give an explanation as well?

Answer 1

Use this sed command to achieve that:
sed '/^#/,/^\$/{/^#/!{/^\$/!d}}' file.txt
Mac users (to prevent extra characters at the end of d command error) need to add semicolons before the closing brackets
sed '/^#/,/^\$/{/^#/!{/^\$/!d;};}' file.txt
OUTPUT
# ID 1
$ description 1
blah blah
# ID 2
$ description 2
blah blah
blah blah
Explanation:
/^#/,/^\$/ will match all the text between lines starting with # to lines starting with $. ^ is used for start of line character. $ is a special character so needs to be escaped.
/^#/! means do following if start of line is not #
/^$/! means do following if start of line is not $
d means delete
So overall it is first matching all the lines from ^# to ^\$ then from those matched lines finding lines that don't match ^# and don't match ^\$ and deleting them using d.

Answer 2

$ cat test
1
start
2
end
3
$ sed -n '1,/start/p;/end/,$p' test
1
start
end
3
$ sed '/start/,/end/d' test
1
3

Answer 3

In general form, if you have a file with contents of form abcde, where section a precedes pattern b, then section c precedes pattern d, then section e follows, and you apply the following sed commands, you get the following results.
In this demonstration, the output is represented by => abcde, where the letters show which sections would be in the output. Thus, ae shows an output of only sections a and e, ace would be sections a, c, and e, etc.
Note that if b or d appear in the output, those are the patterns appearing (i.e., they're treated as if they're sections in the output).
Also don't confuse the /d/ pattern with the command d. The command is always at the end in these demonstrations. The pattern is always between the //.
sed -n -e '/b/,/d/!p' abcde => ae
sed -n -e '/b/,/d/p' abcde => bcd
sed -n -e '/b/,/d/{//!p}' abcde => c
sed -n -e '/b/,/d/{//p}' abcde => bd
sed -e '/b/,/d/!d' abcde => bcd
sed -e '/b/,/d/d' abcde => ae
sed -e '/b/,/d/{//!d}' abcde => abde
sed -e '/b/,/d/{//d}' abcde => ace

Answer 4

Another approach with sed:
sed '/^#/,/^\$/{//!d;};' file
/^#/,/^\$/: from line starting with # up to next line starting with $
//!d: delete all lines except those matching the address patterns

Answer 5

I did something like this long time ago and it was something like:
sed -n -e "1,/# ID 1/ p" -e "/\$ description 1/,$ p"
Which is something like:
-n suppress all output
-e "1,/# ID 1/ p" execute from the first line until your pattern and p (print)
-e "/\$ description 1/,$ p" execute from the second pattern until the end and p (print).
I might be wrong with some of the escaping on the strings, so please double check.

Answer 6

The example below removes lines between "if" and "end if".
All files are scanned, and lines between the two matching patterns are removed ( including them ).
IFS='
'
PATTERN_1="^if"
PATTERN_2="end if"
# Search for the 1st pattern in all files under the current directory.
GREP_RESULTS=(`grep -nRi "$PATTERN_1" .`)
# Go through each result
for line in "${GREP_RESULTS[#]}"; do
# Save the file and line number where the match was found.
FILE=${line%%:*}
START_LINE=`echo "$line" | cut -f2 -d:`
# Search on the same file for a match of the 2nd pattern. The search
# starts from the line where the 1st pattern was matched.
GREP_RESULT=(`tail -n +${START_LINE} $FILE | grep -in "$PATTERN_2" | head -n1`)
END_LINE="$(( $START_LINE + `echo "$GREP_RESULT" | cut -f1 -d:` - 1 ))"
# Remove lines between first and second match from file
sed -e "${START_LINE},${END_LINE}d;" $FILE > $FILE
done