On Jan 2, 2008 11:11 PM, Murali <[EMAIL PROTECTED]> wrote: > Hi, > > I have a scalar variable $x whose value is "a(b" > > I am trying to replace the paranthesis with a backslash followed by > parenthesis, I mean, ( to \( > > $x =~ s/\(/\\(/g; > > However this is making $x as "a\\(b" which is not what I want. snip
One of those two statements is false. If $x holds only "a(b" then $x =~ s/\(/\\(/g will result in $x holding "a\(b" perl -le '$x = "a(b"; print $x; $x =~ s/\(/\\(/; print $x' prints a(b a\(b So, if you are seeing 'a\\(b' then one of two things is occurring: 1. you aren't really seeing 'a\\(b', you just think it is there because that is what is in the replacement string 2. $x doesn't just contain "a(b" I would suggest printing the variable to the screen to see if it really contains 'a\\(b'. If it doesn't (and I expect that it doesn't), I would suggest going back and rereading the Quote and Quote-like Operators section in perldoc perlop or http://perldoc.perl.org/perlop.html#Quote-and-Quote-like-Operators to understand why a \\ in the second part of the substitutionis turning into a \ in the output (hint: it has to do with interpolation and escapes). If you do indeed see a second backslash in the output then $x must not have contained what you said it contained. Go back an take a better look at it. If it contains something like 'a(b a\(b' then the substitution will turn it into 'a\(b a\\(b'. You can use the zero-width negative look-behind assertion* that John mentions in his email to prevent it from prepending a \ to a ( that is already preceded by a \. * read more in perldoc perlre or http://perldoc.perl.org/perlre.html#Look-Around-Assertions -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/
