regex find and replace 0x0D, 0x0A characters

Question

I have a text file of a database dump with some line break characters (0x0A0x0D) in the middle of lines. I want to replace them with commas, but I can't do it simply, because those characters are the actual line break characters where I do want line breaks!

But I noticed that the line break sequences I want to keep are surrouned by space characters ( 0x20), so I was thinking a regex to find and replace any 0x0A0x0D sequence without a leading or trailing space.

How can I do this?

Answer 1

The regex for a whitespace character is, of course, \s. However, since you want a non-whitespace character, you can use \S! Therefore, your regex to replace would be \S\n\r\S.

EDIT:

#!/usr/bin/perl
use strict; use warnings;
my $pattern = "xxxxxxxxxxxxxxxxxxxy\n\ryxxxxxxxxxxxxxxxxxxx \n\r xxxxxxxxxxxxxxxxxxxy\n\ryxxxxxxxxxxxxxxxxxxx";
$pattern =~ s/(\S)(\n\r)(\S)/$1$3/g;
print "$pattern\n";
exit;

result:

xxxxxxxxxxxxxxxxxxxyyxxxxxxxxxxxxxxxxxxx 

 xxxxxxxxxxxxxxxxxxxyyxxxxxxxxxxxxxxxxxxx

I changed the regex to replace with $1$3 so you retain the characters that \S matches.

Answer 2

Here's a way with GNU awk. Set the record separator RS to match the separators you want to keep, and weed out the other \r\n sequences.

gawk -vRS=' \r\n ' '{gsub(/\r\n/, ""); printf "%s%s", $0, RT}'