这样的递归怎么做？！
<br />select a,<br />(select b from c where ...) as d,<br />e<br />from f,(select j from h where ...) as i<br />where ....<br />
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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：
<br />select count(*)<br />from f,(select j from h where ...) as i<br />where ....<br />
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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<br />select a,<br />(select b from c where ...) as d,<br />e<br />from f,(select j from h where ...) as i<br />where ....<br />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);
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select count(*)
from f,(select j from h where ...) as i
where ....
851
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