Re: Parsing and adding to a "variable".

Replies are listed 'Best First'.
Re: Re: Parsing and adding to a "variable". by ysth (Canon) on Jan 25, 2004 at 20:29 UTC
After thinking about it some, I don't like the way some of the solutions behave when you have \n at the end or beginning. To me, it would make more sense to ignore these. You can do that as a separate step, or integrate it in to the regex: `my $paragraphs = $comments_to_add =~ s&(?=.\S\|\A)\s(.?)\s?(?:\n(?!\r?\n)\|\z)&<p>$1</p>&g;` [download] This assumes you want `<p></p>` even around an empty comment. If that's not the case, it's easier: `my $paragraphs = $comments_to_add =~ s&\s(.+?)\s?(?:\n(?!\r?\n)\|\z)&<p>$1</p>&g;` [download] I was dissatisfied with how complicated this turned out. If anyone has suggestions for making it simpler, I'd love to see.	[reply] [d/l] [select]
Re: Re: Parsing and adding to a "variable". by ysth (Canon) on Jan 25, 2004 at 18:16 UTC
That's exactly the problem with split in scalar context. It does place the split up values into an array, @_, even though it doesn't appear to. Since @_ is also used for subs to access passed parameters, this is often a bug waiting to happen. Even if, in a particular case, you know overwriting @_ isn't a problem, it's a bad idiom to get into the habit of using. In your case, you could count paragraphs with m// instead (untested): `my $paragraphs = 1 + (() = $_content_to_add =~ m/(?:\r?\n)+/g);` [download] or substitute <p>-stuff and count at the same time (also untested): `my $paragraphs = 1 + $_content_to_add =~ s!(?:\r?\n)+!</p>\n<p>!g; $_content_to_add = "<p>$_content_to_add</p>";` [download]	[reply] [d/l] [select]