Re: How to replace on a smooth way?

[Ratazong]

$s =~ s/(§[\S]*)/\[$1\]/g;

You search for a §, followed with anything but a blank and store it to $1. Then you replace it by itself surrounded by brackets. The modifier g means that you do it as often as possible.

Update: Thanks to i5513 and MidLifeXis for their improvements of my regex and for polishing my rusty knowledge! I like that in PerlMonks: by answering questions you often learn something for yourself, too :-)

[MidLifeXis]

In addition to i5513's response, you do not need to place \S within brackets — it is already a character class.

[i5513]

Hello,
Backslashes can be omited in the second operator of the substitution, there is not need to scape brackets there.

[bestsiteeditor]

Thank you so much, but here comes the tricky part: I have to match "§" and in the replacement to use "& # 0167;" otherwise - either does not match for "& # 0167;", or replacing with "§" in it - brings strange prints

[MidLifeXis]

The stuff within the parens is a capture - if you move the § outside of the capturing parenthesis, and then add your required entity to the replacement text, it should do what you need.

So, if you have a replacement s/(A\S+)/[$1]/ and want to replace A with something else, you would use s/A(\S+)/[B$1]/ instead, replacing A and B as necessary.

A read of perlre, perlrequick, and perlretut may be helpful. I also wonder if HTML::Entities may be a more appropriate solution to the second requirement.

[roboticus]

bestsiteeditor:

That sounds like you may have an encoding problem, then.

...roboticus

When your only tool is a hammer, all problems look like your thumb.