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);