2

For a silly bbcode parser I wanted to add two definitions into one, my original definition was this for preg_replace:

'#\[s\](.*?)\[/s\]#si', '<strike>\\1</strike>'

And this works, I wished for the user to be able to use either [s] or [strike] to initiate text in that format, so I naturally added something like this thinking it would work:

'#\[(s|strike)\](.*?)\[/(s|strike)\]#si', '<strike>\\1</strike>'

Unfortunately that fails, instead of what you would expect, both [s] and [strike] (used properly) make: s and strike (my markdown is correct to show its real looking result, it shows s or strike regardless of what is inside it)

Why does it replace the inner text with the tag name instead? Is my adding parentheses around the s|strike the problem? I am probably doing this all wrong..

Improve this question
26
  • 2
    BBcode is not regular. Use a BBCode parser Commented Nov 22, 2010 at 11:39
  • 2
    @Gordon: You’re barking up the wrong tree. Modern regexes have next to nothing to do with REGULAR languages and compatibility classes. Regular expressions haven’t been REGULAR since Ken Thompson first put (.)\1 into his backtracking NFA code in grep: the language described by (.)\1 is not REGULAR in that st00pid textbook REGULARity definition that nobody uses and which does not apply to modern regexes. Commented Nov 22, 2010 at 13:47
  • 2
    @Gordon: That article is wrong! I can easily make a pattern he can’t break. He’s not talking about modern regexes, only about textbook REGULAR regular expressions, something that nobody uses. Even egrep can match (.)\1, which is not REGULAR. See here, here, and here — &c&c&c! Commented Nov 22, 2010 at 15:56
  • 4
    @Gordon: You don’t mean not feasible; you mean not practical, or perhaps not expedient. I certainly do not advise reïnventing perfectly good wheels. I am just sick and tired of people mindlessly parroting this slap-down refrain, “You cannot do X with regexes”, when they really mean one or more of “We do not know how to do so”, “Do not do so”, or “There are easier ways to accomplish your goal.” It’s dismissive and disingenuous, even dishonest. But the querents should understand there’s no moral superiority in fitting everything into a single regex; indeed, it has numerous drawbacks. Commented Nov 22, 2010 at 17:42
  • 3
    @Gordon: On the contrary, I vehemently and vociferously disagree with you. The misapplication of the high-brow term REGULAR has nothing to do with real pattern matching. It has a highly irregular and utterly counterintuitive meaning that deceives anybody but an ivory-tower egghead. I am sick and tired of hearing you and everybody else pretending that regular expressions are REGULAR. They are not, and it is even required that they not be: notice that even POSIX BREs must support backrefs, thereby putting the lie to all your REGULAR pontificating. \((?:[^()]*+|(?0))*\) is a beautiful regex. Commented Nov 22, 2010 at 19:35

1 Answer 1

Reset to default
3

The problem is that you added two new regex groups, (s|strike) in the opening tag and (s|strike) in the closing tag. So inside your resulting code you will get s or strike. You can fix that by simply using the correct group number, 2.

Another way would be to make that new groups non-referencing, by adding a ?: to the beginning, but I guess the first solution is easier to understand:

#\[(?:s|strike)\](.*?)\[/(?:s|strike)\]#si
Improve this answer
Sign up to request clarification or add additional context in comments.

4 Comments

Ah, thank you, this helps my understanding. I thought only (.*?) would capture a group, I totally forgot (anything) can too. EDIT: but does the first (s|strike) make a group too? why just the second one? is the first one \0? Confuses me, but I may get it after sleep :P
All (..) capture groups (unless it begins with ?:). But groups are numbered starting with 1 because the “group” 0 usually represents the whole matched string (in this case [s]some text[/s]).
Oh!!.. This makes complete sense to me now. Thank you :)
Named groups are also captivating (sorry ☺) as used in (?<GROUP_NAME> … ). They also number, but the preferred way to access them is \k<GROUP_NAME> from within the pattern and $+{GROUP_NAME} from without. There are a few situations where you can refer to numbered or named groups without backref notation. Mostly in the (CONDITION) YES_PART | NO_PART) condition test of the conditional pattern. You can write ((2)…|…) or (<GROUP_NAME>)…|…). There are also some recursion tests where you don’t use the backslash to talk about the group. Named groups are superior to numbered ones.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.