Pattern matching exclusion 2

[CJason]

Ok, this SHOULD be easy, but for some reason I'm having a hard time with it. I want to match on '"' (double quote) but not when it's preceded by a backslash ('\"').

I tried this:

$string =~ /[^\\]\"/

but, it doesn't work. I've also tried all sorts of variations on this...but, can't get it to work!?!?! Any ideas are greatly appreciated!

 

[CJason]

FYI: What I'm actually trying to do is to count the number of double quotes in a string...excluding those double quotes that have a backslash preceding it. I was just trying to simplify matters with my example...and they kind of changed the problem. Here is what I'm REALLY trying to do:

$count =~ tr/[^\\]\"//;

 

[KevinADC]

You want a "negated look-behind asserttion": (?<!regexp)

$var  = 'test\" test "test" test \"test\" test "test" ';
$c++ while $var =~ m/(?<!\\)"/g;
print $c;

http://perldoc.perl.org/perlretut.html#Looking-ahead-and-looking-behind

tr/// has no concept of character classes or patterns so it can't be used in this case.

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- Kevin, perl coder unexceptional!

 

[CJason]

That is a new concept to me!! Thanks...it's been awhile since I've been able to learn a new perl concept!!!