Find specific pattern and print complete text block using awk or sed

Question

How can find a specific number in a text block and print the complete text block beginning with the key word \"BEGIN\" and ending with \"

Answer 1

$ awk -v RS= -v ORS='\n\n' '/\nB: 567/' file
BEGIN
A: xyz
B: 56789
C: abc
END

BEGIN
A: ghi
B: 56712
C: pqr
END

Note the \n before B to ensure it occurs at the start of a line.This is in place of the ^ start-of-string character you had originally since now each line isn't it's own string. You need to set ORS above to re-insert the blank line between records.

Answer 2

This awk should work:

awk -v s='B: 567' '$0~s' RS= file
BEGIN
A: xyz
B: 56789
C: abc
END
BEGIN
A: ghi
B: 56712
C: pqr
END

Answer 3

A bit lenghty but the RS-trick was already posted :-)

BEGIN {found=0;start=0;i=0}


/BEGIN/ {
    start=1
    delete a
}

/.*567.*/ {found=1}

{
    if (start==1) {
        a[i++]=$0
    }
}

/END/ {
    if (found) {
        for (i in a)
            print a[i]
    }
    found=0
    start=0
    delete a
}

Output:

$ awk -f s.awk input
BEGIN
A: xyz
B: 56789
C: abc
END
BEGIN
A: ghi
B: 56712
C: pqr
END

Answer 4

This might work for you (GNU sed):

sed -n '/^BEGIN/{x;d};H;/^END/{x;s/^B: 567/&/mp}' file

or this:

sed -n '/^BEGIN/!b;:a;$!{N;/\nEND/!ba};/\nB: 567/p' file

Answer 5

perl -lne 'if(/56789/){$f=1}
           push @a,$_;
           if(/END/){
              if($f){print join "\n",@a}
           undef @a;$f=0}' your_file

Answer 6

You can undef RS to split records in blank lines and check if the string matches in the whole block:

awk 'BEGIN { RS = "" } /\nB:[[:space:]]+567/ { print $0 ORS }' infile

It yields:

BEGIN
A: xyz
B: 56789
C: abc
END 

BEGIN
A: ghi
B: 56712
C: pqr
END