How to grep a section of a file in bash shell

grep is not well suited for this task, you need to go one tool "up":

sed -n '/^B/,/^E/p' infile

Output:

B
C
D
E
B
C
E

With regards to the Nth requirement, I think its easiest if you again advance one tool "up", namely awk:

awk '/^B/ { f = 1; n++ } f && n == wanted; /^E/ { f = 0 }' wanted=2 infile

Output:

B
C
E

The flag f will be set when /^B/ is encountered and unset when /^E/ occurs, much in the same way the sed notation works. n keeps a tally of how many blocks have passed and when f == 1 && n == wanted is true, the default block will be executed ({ print $0 }).

@Thor's sed command cannot be beaten, but with the following perl script I try to address the part of your question in parenthesis: "... the (Nth) occurrence ...".

Usage:

./script <start-regex> <end-regex> [N]

Examples with the file in your question:

$ ./script "B" "E" < examplefile
B
C
D
E
B
C
E

$ ./script "B" "E" 2 < examplefile
B
C
D
E
F
G
B
C
E

There is no error checking or whatsoever and the script is non-greedy, i.e. from A B C D E E F only B C D E will be grep'ed with N=1.

---

#!/usr/bin/perl

if ($ARGV[2] != "") { $n = $ARGV[2] } else { $n = 1 }
$begin_str = $ARGV[0];
$end_str = $ARGV[1];

while(<STDIN>) {
  if($_ =~ $begin_str) { $flag=1 }             # beginning of match, set flag    
  if($_ =~ $end_str && $flag eq 1) { $i++ }    # i-th occurence of end string

  if($i eq $n) {                               # end of match after n occurences of end string
    $flag=2;
    $i=0; 
  }

  if ($flag ge 1) {                            # append currrent line to matching part
    $out.=$_;
  }

  if($flag eq 2) {                             # after detection of end of match, print complete match
    print $out;
    # print "---\n";                           # separator after a match
    $out="";
    $flag=0;
  }

}