regex problem

evo4ever

Hey. I have an experimental function which converts bb tags. It works fine, but to a certain extent....

function adv_bbcode(){

$str = "1: [ url=http://localhost ] Local [ /url ]";

print preg_replace("/\[url=(\w+)\](\w+)\[\/url\]/", "<a href=$1>$2</a>", $str);
}

Now the problem, as you might expect, it won't match characters like ": / . ? * " etc. Im aware that if you want to match any char you use the ".". But im not quite sure how to use this. Help appreciated. Thanks.

Edit: OK VB stripped my slashes. But Im sure you can figured out where they go. The main problem lies in the (\w+) groups.

bubblenut

"/[\wurl=(.)\w](.)[\w\/url\w]/"
HTH
Bubble

Shrike

Since all URL's starting http will have :// next it might be worth stating that directly. You can use .? to capture everything up until the closing [/URL] too.

$string = preg_replace("/\\[URL=(\\w+:\\/\\/.*?)\\](.*?)\\[\\/URL\\]/si", "<a href=\"$1\">$2</a>", $string);

I think .? amd . are fairly processor heavy though so it might be worth using character classes or something. Also the whole htttp(s):// bit should probably be optional 🙂

bubblenut

I agree with you one the [url]http://[/url] bit and I was using \w where I meant to use \s, however I don't understand your .?, ? makes the preceding character optional which means nothing when using the quantifier. Unless my understanding of the ? is wrong.
NB can be whitespace between the opening '[' and 'url' also between closing '"' and ']'.

"/[\sURL=(\w+:\/\/.?)\s](.?)[\s\/URL\s]/si"

Bubble

Shrike

? extends the meaning of (, also 0 or 1 quantifier, also quantifier minimizer

? means 0 or 1 in this case so .*? allows a non-greedy match, i.e. it can be anchored by the next piece of the regexp.

bubblenut

Oh, cool I didn't know "also quantifier minimizer". So, does that mean that
'.?\s' is the same as '[^\s]'?
Cheers
Bubble

Shrike

.?\s would mean anything followed by a whitespace. [^\s] means nothing any number of times?

bubblenut

Originally posted by Shrike
.?\s would mean anything followed by a whitespace.

If that's the case what's the point of the ? because .\s means anything followed by a whitespace

Originally posted by Shrike
[^\s] means nothing any number of times?

No it doesn't, it means anything except whitespace any numbr of times. So I should have written [^\s]\s I think.

Bubble

Shrike

Consider

$string = "foo bar ";
$string = preg_match("/.*\\s/", $string, $matches);
$string = preg_match("/.*?\\s/", $string, $matches2);
print_r($matches);
print_r($matches2);

.* is a greedy match, i.e. it won't stop on the first match, if it finds one afterward.

bubblenut

So in that case .?\s is the same as [^\s]\s

Bubble

Shrike

Yep I think so.