Using sed to delete all lines between two matching patterns

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 (ie, 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