Recursive regex in Perl

Question

I've got strings like these:

| Released    = {{start-date|June 14, 1972}}
| Released    = {{Start date|1973|03|01|df=y}}

I'd like to replace all | within {{ }} with ^

| Released    = {{start-date^June 14, 1972}}
| Released    = {{Start date^1973^03^01^df=y}}

I can't use substring replacement because there are | symbols outside {{ }}, which must be left intact. And because I don't know exactly how many parts does the string in {{ }} have, I can't use something like s/{{(.+?)\|(.+?)}}/{{$1^$2}}/.

I suppose I need to use some kind of recursion here?

I'll work on an answer momentarily, but I just couldn't resist the comment "you are scraping something from Wikipedia, right?" — Ray Toal
– Ray Toal, Commented Jul 12, 2011 at 5:54
And if that's the case, it will probably behoove you to know that the only 100%-working parser for wikitext is MediaWiki itself (and even that can be buggy sometimes). — user149341
– user149341, Commented Jul 12, 2011 at 6:01
Yes. I know about dbpedia.org and some other similar resources but can't use them because data is too outdated for my task. — DEgorov
– DEgorov, Commented Jul 12, 2011 at 6:02
I don't need to parse all wikitext features and definitely don't need to render it to HTML… — DEgorov
– DEgorov, Commented Jul 12, 2011 at 6:03

Qtax · Accepted Answer · 2011-07-12 10:16:54Z

5

A simple solution:

s/\|(?=[^{}\n]*}})/^/g

Even simpler solution, but probably broken in many cases:

s/(?!^)\|/^/gm

Here is a bit more robust regex:

s/(?:\G(?!^)(?:(?>[^|]*?}})(?>.*?{{))*|^(?>.*?{{))(?>[^|]*?(?=}}|\|))\K\|(?=.*?}})/^/gs;

Commented:

s/
(?:
  \G(?!^)                       # inside of a {{}} tag
  (?: (?>[^|]*?}}) (?>.*?{{) )* # read till we find a | in another tag if none in current
  |
  ^(?>.*?{{)                    # outside of tag, parse till in
)
(?> [^|]*? (?=}}|\|) )          # eat till a | or end of tag
\K                              # don't include stuff to the left of \K in the match
\|                              # the |
(?=.*?}})                       # just to make sure the tag is closed
/^/gsx;

Input:

|}}
| Re|eased    = {{start-date|June 14^, {|1972}|x}}
| Released    = {{Start date}|1973|03|01}|df=y|}}
| || {{|}} {{ |

Output:

|}}
| Re|eased    = {{start-date^June 14^, {^1972}^x}}
| Released    = {{Start date}^1973^03^01}^df=y^}}
| || {{^}} {{ |

Example: http://ideone.com/fbY2W

edited Jul 12, 2011 at 10:16

answered Jul 12, 2011 at 9:36

Qtax

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2 Comments

DEgorov Over a year ago

Thanks a lot! Didn't know about lookaheads :( Slightly modified to my needs — works fine!

regilero Over a year ago

+1 for a nice obfuscated code exammple, 12 lines of comments and I still cannot understand it. I think one line per character would do it :D

Jonathan Hall · Accepted Answer · 2011-07-12 06:16:59Z

3

This may not be the most concise way to do it, but it's the first working method I came up with.

my $new;
for ( split /({{.*?}})/ ) {
    s/\|/^/g if /^{{/;
    $new .= $_;
}
$_ = $new;

edited Jul 12, 2011 at 6:16

answered Jul 12, 2011 at 5:57

Jonathan Hall

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4 Comments

Ray Toal Over a year ago

This does not work because you have a greedy .* in your split regex. When I apply it to "A {{b|c}} c|d {{e|f|}} g h" I get "A {{b^c}} c^d {{e^f^}} g h" which incorrectly turns the pipe between c and d into a caret.

Déjà vu Over a year ago

Replace the split with split /({{.*?}})/, that should do it.

Jonathan Hall Over a year ago

@Ray Toal: Good catch. Updated my answer.

Ray Toal Over a year ago

@Flimzy I knew you would. +1 now. Nice answer. I was sort of surprised it works without escaping the left brace; technically that is a metacharacter.

tadmc · Accepted Answer · 2011-07-12 11:40:05Z

2

s{({{.*?}})}
 {my $x = $1;
  $x =~ tr/|/^/;
  $x
 }ge;

answered Jul 12, 2011 at 11:40

tadmc

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