Matching nested Patterns with Regex (using PHP's recursion)

Question

I am currently trying to write a regular expression in PHP that allows me to match a specific pattern containing itself indefinetely nested. I know that per default regular expressions are not capable of doing that, but PHP's Recursive Patterns (http://php.net/manual/de/regexp.reference.recursive.php) should make it possible.

I have nested structures like this:

<a=5>
    <a=3>
        Foo
        <b>Bar</b>
    </a>
    Baz
</a>

Now I want to match the content of the outmost tag. In order to correctly match up the first opening tag with the last closing tag, I need PHP's recursion item (?R).

I tried a pattern like this:

/<a=5>((?R)|[^<]|<\/?[^a]|<\/?a[a-zA-Z0-9-])*<\/a>/s

Which basically means <a=5>, followed by as many as possible of the following, followed by </a>:

another tag (recursively)
any not-opening-tag character
any opening tag, followed by an optional slash, not followed by an "a"
the before WITH an a, but not finished (followed by at least 1 more character)

The last 2 cases could be just one case [tag not namend "a"], but I heard this should be avoided in regular expressions, because it needs lookarounds and would have bad performance.

However, I see no mistake in my RegEx, but it does not match the given string. I want the following match:

    <a=3>
        Foo
        <b>Bar</b>
    </a>
    Baz

Here's a link to play around with the RegEx: https://www.regex101.com/r/lO1wA6/1

If you're really just looking for the opening < marker, then it might indeed suffice. One note about the (?R): it doesn't recurse to the first group, but the whole pattern. Use R1. But still, try /x for readability and inline comments, and also give a more basic example where matching succeeded. — mario
– mario, Commented Dec 19, 2014 at 1:56
Thanks, I just realized my mistake. I want the outer tag to match a=5, but the inner ones to match a.*?. — Felk
– Felk, Commented Dec 19, 2014 at 2:00
This looks like xml/html. You should be using a xml/html parser for this. — pguardiario
– pguardiario, Commented Dec 19, 2014 at 2:33
I don't understand what you are doing, but the above seems to be a better solution from what I what I think you want to do. See here: php.net/manual/en/class.domnode.php ---- You can use the childNodes property of that class to pick up all the children recursively. — Kohjah Breese
– Kohjah Breese, Commented Dec 19, 2014 at 2:54
If it suitably matches the task without long-winded workarounds, regex is perfectly apt. What @pguardiario/Kohjah allude to is just a context-insensitive SO meme; libxml or its DOM traverser would choke on things like <a=5>. (Neither would a proper SGML toolkit make sense.) — mario
– mario, Commented Dec 19, 2014 at 20:54

nhahtdh · Accepted Answer · 2014-12-19 08:00:41Z

3

You can use this regex to match what you want (the regex placed in a string literal for sake of convenience):

'~<a=5>(<([a-zA-Z0-9]+)[^>]*>(?1)*</\2>|[^<>]++)*</a>~'

Here is a break down of the regex above:

<a=5>
(
  <([a-zA-Z0-9]+)[^>]*>
  (?1)*
  </\2>
  |
  [^<>]++
)*
</a>

The first part <([a-zA-Z0-9]+)[^>]*>(?1)*</\2> matches pair of matching tags and all its content. It assumes that the name of the tag consists of the characters [a-zA-Z0-9]. The name of the tag is captured ([a-zA-Z0-9]+) and backreference when matching the closing tag </\2>.

The second part [^<>]++ matches whatever else outside the tags. Note that there is no handling of quoted string, so depending on your input it may not work.

Then back to the routine call which recursively calls the first capturing group. You would notice that a tag can contain 0 or more instances of other tags or non-tag contents. Due to the way the regex is written, this property is also shared by the outer most <a=5>...</a> pair.

Demo on regex101

answered Dec 19, 2014 at 8:00

nhahtdh

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1 Comment

Felk Over a year ago

Thanks, this works the way I want it to work. I just needed to encapsulate the outer (...)* in another group and modify the group counters to be able to get the outermost tag as a match.

2 revs · Accepted Answer · 2014-12-19 07:04:32Z

try this:

PHP

$re = "/(<[^\\/>]+(\\/?)>)*([^<]+)(<\\/\\w+>)*/m";
$str = "<a=5>\n <a=3>\n Foo\n <b/>Bar</b>\n </a>\n Baz\n</a>";

preg_match_all($re, $str, $matches);
var_dump($matches);
// here  

 $matches[1];  //for open tag array
 $matches[2];  //for single tag mark array by ( />)
 $matches[3];  //for inner data array
 $matches[4];  //for close tag array

output

array (size=5)
  0 => 
    array (size=5)
      0 => string '<a=5>
 ' (length=7)
      1 => string '<a=3>
 Foo
 ' (length=12)
      2 => string '<b/>Bar</b>' (length=11)
      3 => string '
 </a>' (length=6)
      4 => string '
 Baz
</a>' (length=10)
  1 => 
    array (size=5)
      0 => string '<a=5>' (length=5)
      1 => string '<a=3>' (length=5)
      2 => string '<b/>' (length=4)
      3 => string '' (length=0)
      4 => string '' (length=0)
  2 => 
    array (size=5)
      0 => string '' (length=0)
      1 => string '' (length=0)
      2 => string '/' (length=1)
      3 => string '' (length=0)
      4 => string '' (length=0)
  3 => 
    array (size=5)
      0 => string '
 ' (length=2)
      1 => string '
 Foo
 ' (length=7)
      2 => string 'Bar' (length=3)
      3 => string '
 ' (length=2)
      4 => string '
 Baz
' (length=6)
  4 => 
    array (size=5)
      0 => string '' (length=0)
      1 => string '' (length=0)
      2 => string '</b>' (length=4)
      3 => string '</a>' (length=4)
      4 => string '</a>' (length=4)

Live Demo

OR

    $re = "/(<[^\\/>]+\\/?>)*([^<]+)(<\\/\\w+>)*/m";
    $str = "<a=5>fff\n <a=3>\n Foo\n <b/>Bar</b>\n </a>\n Baz\n</a>";

    preg_match_all($re, $str, $matches);
    //var_dump($matches);
    $md="";
    $c=count($matches[1]);
    foreach($matches[1] as $k=>$v){
        if($k!=0){
            $md.=$v.$matches[2][$k].$matches[3][$k];
        }
        else if ($c!=$k+1){
            $md.=$matches[2][$k].$matches[3][$k];
        }
    }   
var_dump($md);

Live

Output

 string 'fff
 <a=3>
 Foo
 <b/>Bar</b>
 </a>
 Baz
</a>' (length=44)

I don't understand what you did. Your demos seem to result in a lot of random matches.

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