Shell Programming and Scripting

how is this way?

Code:

awk '/START/{p++}/./{a[p]=a[p]"\n"$0}END{for(i=1;i<=p;i++) if(a[i]~/END \(VIO\)/) print a[i]}' urfile

Code:

sed -n '
/^START:/ {
    h
    :next
    n
    H
    /^END/ {
        /^END (VIO)/ {g; p;}
        d
    }
    b next
}' filename

Regards,
Alister

Another approach:

Code:

awk '/START/{s=x} {s=s?s RS $0:$0} /END \(VIO\)/{print s;s=x}' file

Hi Franklin52,

Your code worked. Can you please explain your code?

Is there anyway we can search above a line number in SED.

For example, I have a file like this,

START
.....
.....
.....
END
START
.....
.....
END
START
.....
xyz
.....
END
START
.....
.....
END

I know the line number of xyz which is say 12, and I want to fetch the block (START to END) that contains xyz. I can search the below lines of line number 12 like the following way,

sed -n '12,/END/p' <filename>

But is there any way we can search above line 12 to get the lines above starting from line 12 to START.