這樣的遞迴如何做？

這樣的遞迴怎麼做？！

select a,
(select b from c where ...) as d,
e
from f,(select j from h where ...) as i
where ....

我要把外層select ... from ...中，select後面的內容替換掉，而保留from後面的內容。最後變成：

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

其實這相當於xml/html節點的替換，類似遞迴問題，想了很久也沒想到解決方案。
------解決方案--------------------
正則....不行

如果你只是想得到返回的行數, 你總是可以這樣做:
select count(*) from (

-- 你的sql --
select a,
(select b from c where ...) as d,
e
from f,(select j from h where ...) as i
where ....

) tmp

如果你非要嚴格做替換, 要做文法分析, 考慮到單雙引號, 括弧等等....
------解決方案--------------------
不知道你是要寫SQL指令，還是要做字串替換
如果是做字串替換，可以這麼寫

$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
------解決方案--------------------
\bfrom\b)/i', $s, -1, PREG_SPLIT_NO_EMPTY 
------解決方案--------------------
 PREG_SPLIT_DELIM_CAPTURE);

$n = 0;
$st = array();
for($i=0; $i  $t = strtolower($ar[$i]);
  if($t == 'select' 
------解決方案--------------------
 $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);

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

