这样的递归如何做？

这样的递归怎么做？！

select a,

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

e

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

where ....

我要把外层select ... from ...中，select后面的内容替换掉，而保留from后面的内容。最后变成：

select count(*)

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

where ....

其实这相当于xml/html节点的替换，类似递归问题，想了很久也没想到解决方法。
------解决方案--------------------
正则....不行

如果你只是想得到返回的行数, 你总是可以这样做:
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

如果你非要严格做替换, 要做语法分析, 考虑到单双引号, 括号等等....
------解决方案--------------------
不知道你是要写SQL指令，还是要做字符串替换
如果是做字符串替换，可以这么写

$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);

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