Regular Expression Question

[evdoxos]

You know that math expressions may have many parentheses '(' and ')' and there are parentheses in other parentheses:

...(...(...(...)...(...)...(...(...(...)...)...(...)...)...)...)...
   [COLOR=Red]1[/COLOR]   2   3   2   3   2   3   4   5   4   3   4     3   2   1   [COLOR=Red]0[/COLOR]

I want to get all matches that start with '(' and finish with the corresponding ')'.
In the above example the match should be from 1 to 0.
Is there a regular expression -in PHP- that can do this?

 

[un4given[MAD]]

This one should work:

$subj = "123(ab(cde)fg(ghijklmn)sdf)jerhk"; //test string
$pattern = "/\((.*)\)/"; //our regexp

if (preg_match($pattern, $subj, $matches))
	echo $matches[0];

After execution $matches[0] will contain "(ab(cde)fg(ghijklmn)sdf)" and $matches[1] -- "ab(cde)fg(ghijklmn)sdf" (without open and close parentheses)

 

[evdoxos]

It dosn't work. Inserting one more parenthesis in your test string:

"123(ab(cde)fg(ghijklmn)s)df)jerhk"

We shoud take (ab(cde)fg(ghijklmn)s) but we take (ab(cde)fg(ghijklmn)s)df)

There is a solution for a similar problem but in Perl : http://perl.plover.com/yak/regex/samples/slide083.html
But I want in PHP.

 

[Peter]

It will be possible to do but will not be easy imho.

I will have a look tomorrow see if I can come up with a regexp for it. Cant do it atm as had a few drinks.

On a side note however what was previously posted should do it. If the maths expression has mismatching parenthesis then it is an invalid expression

 

[evdoxos]

On a side note however what was previously posted should do it. If the maths expression has mismatching parenthesis then it is an invalid expression

You are right.
I should give this example: 123(ab(cde)fg(ghijklmn)sdf)je(rh)k

The result then should be (ab(cde)fg(ghijklmn)sdf) but it is (ab(cde)fg(ghijklmn)sdf)je(rh)

 

[un4given[MAD]]

I apologize for my inattention to details...
So, if we have, for example, "ab(cd)ef(ghi(jkl)mn(o)prs)" string, then result should be "(cd)"?

And is it really necessary to perform this task with regexp?

 

[MJ]

And is it really necessary to perform this task with regexp?

Agreed. Probably not the best way to do this. Some string functions could do the job with less of a headache.

Solution: Have a loop. As you parse through the string, add 1 to an $open variable when you hit '(' and add one to $close when you hit ')'. When $open == $close you are in business and know the start and stop positions of the string to take.

Problem solved.

 

[evdoxos]

So, if we have, for example, "ab(cd)ef(ghi(jkl)mn(o)prs)" string, then result should be "(cd)"?

Yes it should be (cd)

Solution: Have a loop. As you parse through the string, add 1 to an $open variable when you hit '(' and add one to $close when you hit ')'. When $open == $close you are in business and know the start and stop positions of the string to take.Problem solved.

I'm looking for to match more complicated cases as:

alphaNumericFunctionName(...(...(...)...(...)...(...(...(...)...)...(...)...)...)...(...)...)

and using preg_replace_callback() to replace some things in it (speed is very important).

 

[Kate]

Doing it with regular expressions is the right approach IMO but you need to use recursive backreferences as the number of nested parentheses is not fixed.

<?php
$expr = "123(ab(cde)fg(ghijklmn)sdf)je(rh)k";
$pattern="/\(([^()]+|(?R))*\)/";
if (preg_match($pattern,$expr,$matches)) {
echo $matches[0];
}
?>

For this expression to work the parentheses must be nested properly.

 

[un4given[MAD]]

$pattern="/\(([^()]+|(?R))*\)/";

Wow, that was really amazing!
Could you please, explain, what is (?R) in this expression?

 

[Kate]

?R is a special pattern that performs a recursive substitution of the whole regular expression which allows unlimited nesting depth

 

[evdoxos]

"/\(([^()]+|(?R))*\)/"

You are GOD!!! It works perfectly!!!