• You could add the character you're binding with to your replace

  s/([^?])'/$1\'\n/g
  [download]

• You could use negative lookbehind

  s/(?<!\?)'/\'\n/g
  [download]

Also, instead of -ne, I'd use -pe and get rid of the print statement. I'd also look at making it -pi.bak -e to do in-place editing with a backup file.

Dragonchild (and all who followed) - thank you. This line did the trick for me:

perl -pe "s/([^?])'/$1'/g" input.txt

Things you folks taught me:

• grouping in a regex - see perlrequick
• the special variable $1 - ditto
• the -p switch
• the -i switch
• check code before posting. I had two errors in mine (extra double quote and unnecessary escape)

  Cheers - Winter

Final Update: this is what I used in the end...

Parses a string in a file. The string uses an apostrophe as line terminator. Ignores apostrophes predeeded by the escape character, "?". Clever enough not to parse the same file more than once.

perl -p -i.bak -e "s/([^?])'([^\n])/$1'\n$2/g" input.txt

• perl: Invokes the command interpreter.
• -p: p switch is 'assume loop like -n but print line also'.
• -i.bak: i switch is edit in place, .bak is the extension of the backup file.
• -e: e switch is 'one line of program'.
• "...": Use double quotes for Windows OS.
• s/.../.../g: Substitute. Match 1st and substitute 2nd. g means global.
• (...)...(...): Groups - whatever the value between brackets winds up in $1, $2 etc.
• ([^?])'([^\n]): Match any apostrophes neither preceeded by the escape character "?" nor followed by a new line "\n" - three characters total.
• $1'\n$2: Replace matches with the value of group one (see above) followed by an apostrophe and a new line(\n) and the value of group two - four characters total.
• input.txt: File to parse.

Well, the version you have is very close. Try:

s/([^?]')/$1\n/g;
[download]

The $1 is replaced by the contents of the first group. Groups are designated by () in the match portion. So, it matches one non-? and one ' and replaces that with what it found and a \n.

This may be good enough, but keep in mind that it won't work if a ' is at the beginning of the string. A more complete solution would be:

s/'(?<!\?)/'\n/g;
[download]

Note that this is untested, but should work! :-)

s/([^?]\')/$1\n/g;
[download]

Looking from that angle, might it be easier to substitute on the (apparently unambiguous) +eol\d'
update: belated example:

#!usr/bin/perl -w

$foo = <DATA>;

$foo =~ s/\+eol\d'/\n/g;
    print $foo;


print "\n\n Done\n";

__DATA__
AAA+XXXX+1234++here?'s some text+eol1'BBB+XXXX+1234++here?'s some text
++eol2'CCC+XXXX+1234++here?'s some text+eol3'etc.
[download]

# OUTPUT:
#
# AAA+XXXX+1234++here?'s some text
# BBB+XXXX+1234++here?'s some text
# CCC+XXXX+1234++here?'s some text
# etc.
# 
#  Done

The following notion is mine, and may not be wise (CORRECTIVE comments welcome!):
It's usually worthwhile to match against the largest possible chunk of data, to minimize ambiguity for the regex.

$_ = "AAA+XXXX+1234++here?'s some text+eol1'BBB+XXXX+1234++here?'s som
+e text+eol2'CCC+XXXX+1234++here?'s some text+eol3'";
s/(?<!\?)'/\n/g;
print;
[download]

I think you might be having a shell quoting problem. The backslashes are swallowed by the shell if this is under sh. Try this:

perl -wne 's/[^?]\47/\47\n/g; print' filename
[download]

This solution won't work if you have two consecutive (unescaped) apostrophes, or if the file has no newlines and is too long to read in memory.

Thanks ambrus, well done for spotting the unnecessary double quote after the substitution - s/.../.../g". The Windows command environment requires double quotes, but I mistakenly included an extra one in the code I posted.