adding carriage return two lines above pattern

Question

I have database log file which looks like this

tbl
---
tmp1
col1|col2
---------
  22|  33
  32|  45
tbl
---
tmp2
col1|col2| col3
---------------
  22|  33|  123
  32|  45|  456

I am trying to write a sed command which find lines which begin with '---' and adds a carriage return two lines above. So that the output looks like:

tbl
---
tmp1

col1|col2
---------
  22|  33
  32|  45

tbl
---
tmp2

col1|col2| col3
---------------
  22|  33|  123
  32|  45|  456

Is it possible to do this using sed? If so how?

Answer 1

Method 1

Reversing the input using tac as suggested by Sundeep makes this problem a whole lot easier:

tac data.txt | sed '/^----*$/ {N;a\

}' | tac | sed '2,$p'

Method 2

This is a difficult sed-only problem (at least for me). I was close, but opted for perl. sed is turing complete, so this can definitely be translated. Intuitively, I'd say this could be completed with one extra line buffer and I hope someone answers with a sed-only implementation. There were times where it simply didn't hold. There were so many weird cases that I opted for queuing the whole way through:

my @queue = ();

while (<>) {
    if (/----*/) {
        do {
            print $queue[0];
            print "\n" if scalar @queue == 2;
            shift @queue;
        } while (scalar @queue);
        print;
    } else {
        push @queue, $_;
    }
}
do print shift @queue while scalar @queue;

I started looking at the problem like:

1
0
2
1
0

The numbers represent offsets from the following ---* line where 0 is the ----* line. When the offset is 2, print a new line.

The problem statement is clearly expressed as:
print \n before previous line if line ~ /----*/.

Anyway, here's the invocation / output:

$ perl script.pl data.txt

tbl
---
tmp1

col1|col2
---------
  22|  33
  32|  45

tbl
---
tmp2

col1|col2| col3
---------------
  22|  33|  123
  32|  45|  456