Question

I have a file that is in the following format:

Preamble

---------------------
Section 1
...
---------------------

---------------------
Section 2
...
---------------------

---------------------
Section 3
...
---------------------

Afterwords

And I want to extract each section by the separator so that I’ll have a result in:

file0:

Section 1
...

file1:

Section 2
...

file2:

Section 3
...

…

Is there a simple way to do this? Thanks.

Answer 1

[Update] Using chomp and $_ makes this even shorter.

This should do it:

If your input record separator is a sequence of 21 -‘s, this is easy with perl -ne:

perl -ne 'BEGIN{ $/=("-"x21)."\n"; $i=0; } 
  do { open F, ">file".($i++); 
       chomp;
       print F; 
       close F; 
  } if /^Section/' yourfile.txt

should work, and create files file0.. fileN.

Explanation

Easier to explain as a stand-alone Perl-script perhaps?

$/=("-"x21)."\n"; # Set the input-record-separator to "-" x 21 times
my $i = 0;        # output file number

open IN, "<yourfile.txt" or die "$!";

while (<IN>) {  # Each "record" will be available as $_ 
  do { open F, ">file".($i++); 
       chomp;     # remove the trailing "---..."
       print F;   # write the record to the file
       close F;   #
  } if /^Section/  # do all this only it this is a Section
}

Perl’s awk lineage was useful here, so let’s show an awk version for comparion:

awk 'BEGIN{RS="\n-+\n";i=0} 
  /Section/ {chomp; print > "file_"(i++)".txt" 
}' yourfile.txt

Not too bad compared to the perl version, it’s actually shorter. The $/ in Perl is the RS variable in awk. Awk has an upper hand here: RS may be a regular expression!