On Jul 27, John W. Krahn said:
>> No, $1 is either "&" or undef. Perhaps you want:
>>
>> s[([&<>])|\\($rx)][<$rep{$+}/>]go;
>>
>> That $+ means "the last () that matched". But that still replaces & with
>> <&/>. So you'd need to make the '<' and '/>' part of the %rep hash's
>> values. I would suggest:
>
>I thought '<' and '/>' were already in the hash values? If so, wouldn't
>this work?
They could be, sure, but he didn't seem to have done it, which is why I
then modified it to assume that.
> s[([&<>]|(?<=\\)$rx)][$rep{$1}]go;
That doesn't remove the \ from the string, though.
--
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RPI Acacia brother #734 http://www.perlmonks.org/ http://www.cpan.org/
** Look for "Regular Expressions in Perl" published by Manning, in 2002 **
<stu> what does y/// stand for? <tenderpuss> why, yansliterate of course.
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