这样的递归如何做？

select a,

(select b from c where ...) as d,

e

from f,(select j from h where ...) as i

where ....

select count(*)

from f,(select j from h where ...) as i

where ....

$s = <<< TXT

select a,

(select b from c where ...) as d,

e

from f,(select j from h where ...) as i

where ....

TXT;



$ar = preg_split('/(\(?\bselect\b
------解决方案--------------------
\bfrom\b)/i', $s, -1, PREG_SPLIT_NO_EMPTY 
------解决方案--------------------
 PREG_SPLIT_DELIM_CAPTURE);



$n = 0;

$st = array();

for($i=0; $i
  $t = strtolower($ar[$i]);

  if($t == 'select' 
------解决方案--------------------
 $t == '(select') {

    $st[] = $i;

  }

  if($t == 'from') {

    if(count($st) == 1) break;

    array_pop($st);

  }

}

for($i--; $i>$st[0]+1; $i--) unset($ar[$i]);

$ar[$st[0]+1] = " count(*)\n";

echo join('', $ar);