Remove lines between patterns and print all if any of patterns doesn't exists

Question

I have next file:

G
H
A
B
C
D
N

Let's delete lines from A to D and we will get next output:

G
H
N

It's very easy to do with sed '/A/,/D/d , but if my file don't have D, then output will be empty. I want if there isn't second pattern (D) do not delete anything and show full file.

Second question - how to delete lines between patterns and next line after (N)? Kind of sed '/A/,+1d , but sed '/A/,/D/+1d will not work.

There is no different for me to use sed, awk or python/bash scripts.

Answer 1

With 2 pass awk you can do this:

# when 2nd pattern is not found
awk -v ps='A' -v pe='P' 'NR==FNR{if ($0 ~ ps) start=FNR; else if ($0 ~ pe) stop=FNR; if (stop) nextfile; else next} !stop || FNR<start || FNR>stop' file file
G
H
A
B
C
D
N

# when 2nd pattern is found
awk -v ps='A' -v pe='D' 'NR==FNR{if ($0 ~ ps) start=FNR; else if ($0 ~ pe) stop=FNR; if (stop) nextfile; else next} !stop || FNR<start || FNR>stop' file file
G
H
N

About your 2nd part you can tweak this awk a bit with another parameter:

awk -v n=2 -v ps='A' -v pe='D' 'NR==FNR {
   if ($0 ~ ps)
      start=FNR
   else if ($0 ~ pe)
      stop=FNR+n
   if (stop)
      nextfile
   else
      next
}
!stop || FNR<start || FNR>stop' file file

Answer 2

One option out of many that use perl: hold the text in an accumulator once you see A , then print them at the end if you didn't see D . That way you only make one pass through the file (although you use a lot of memory for big files!).

use strict; use warnings;
my $accumulator = '';  # Text we're holding while we wait for a "D"
my $printing = 1;      # Are we currently printing lines?

while(<>) {
    if(/A/) {  # stop printing; start accumulating
        $printing = 0;
        $accumulator .= $_;    # $_ is the current line
        next;
    }

    if(/D/) {  # we got a D, so we're back to printing
        $accumulator = '';   # discard the text we now know we're deleting
        $printing = 1;
        next;
    }

    if($printing) {
        print;
    } else {
        $accumulator .= $_;
    }
}

print $accumulator;  # which is empty if we had both A and D

I tried this on your testcase, and on your testcase with the D removed. It can also handle files with multiple A / D pairs. I have not tested it on files where the D comes before the A , or on files with a single line including both A and D .

Answer 3

With D :

$ awk '
/A/ || f { 
    f=1                   # flag up
    b=b (b==""?"":ORS)$0  # buffer...
    if(/D/) {             # until D
        print b           # output...
        b=f=""            # and reset buffer and flag
    }
    next
}
END {                     # output if rows runout before finding D
    if(b)
        print b
}1' file                  # output outside the range
G
H
A
B
C
D
N

With missing D :

$ cat file2
G
H
A
B
C
N
$ awk '/A/||f{f=1;b=b (b==""?"":ORS)$0;if(/D/){print b;b=f=""}next}END{if(b)print b}1' file2
G
H
A
B
C
N

Answer 4

This might work for you (GNU sed):

sed '/A/{:a;N;/D/!ba;d}' file

Match on A and then gather up lines in the pattern space until a match on D then delete the pattern space. If a D is not matched the N command will terminate sed processing and by default this prints whatever is found in the pattern space.

To delete as above +1, use:

sed '/A/{:a;N;/D/!ba;N;d}' file

NB If the +1 line does not exist, no lines are deleted.