$ matches the point between \n and the character before it.

print "a\n\nc\n" =~ / a\n ^ $    c\n /mx ?1:0, "\n";  # 0
print "a\n\nc\n" =~ / a\n ^ $ \n c\n /mx ?1:0, "\n";  # 1
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Since $ matches before the newline, it doesn't ever actually match the newline, and thus the newline is not getting substituted. What you want to do is remove newlines that start lines.

$_ = "abc\n"
   . "def\n"
   . "\n"
   . "jkl\n"
   . "\n"
   . "\n"
   . "stu\n";

s/^\n//mg;
print;
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abc
def
jkl
stu
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That did the trick, thank you.


Thanks,
Dru

Perl, the Leatherman of Programming languages. - qazwart

You're matching the blank line, but not the newline you're trying to remove. Use s/\n^$//mg. (/g in case there is more than one.)

Another way to do what you want:

$vulns =~ tr/\n//s;
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