这样的递归如何做？

程序员文章站 2024-02-05 12:01:28

...

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 = 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 ....

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这样的递归如何做？