MYSQL多字段COUNT如何处理？

[siemenliu]

SELECT parent.*, COUNT(node.CategoryID), COUNT(domain.PortfolioID)
FROM categories_new AS parent
LEFT JOIN categories_new AS node ON node.lft > parent.lft AND node.rgt < parent.rgt
LEFT JOIN portfolios as domain ON domain.CategoryID = parent.CategoryID
GROUP BY parent.CategoryNameCN
ORDER BY parent.lft

语句如上
其中两个count字段统计两个LEFT进来的数据，但是两个COUNT数据之间有相互影响
不用嵌套SQL可以实现这样的语句吗？

[phpboy]

SELECT parent.*, COUNT(node.CategoryID), COUNT(domain.PortfolioID) 
FROM categories_new AS parent ORDER BY parent.lft 
LEFT JOIN categories_new AS node ON node.lft > parent.lft AND node.rgt < parent.rgt 
LEFT JOIN portfolios as domain ON domain.CategoryID = parent.CategoryID 
GROUP BY parent.CategoryNameCN 

[siemenliu]

谢谢楼上，方法不成功

用了楼上的SQL提示一下错误

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'LEFT JOIN categories_new AS node ON node.lft > parent.lft AND node.rgt < parent.' at line 3

我记得LEFT JOIN不能跟在ORDER BY后面来着？

[siemenliu]

补充，我帖子里的那句SQL执行之后两个COUNT字段的值都等于其本身与另一个COUNT字段值的乘积。。。
难道SQL语句里面多COUNT就是会出问题的么？

[xuzuning]

用 sum 函数

[YHL27]

学习！！

[siemenliu]

sum函数结果离奇大
在手册看SUM函数的时候发现一个DISTINCT关键字，用在COUNT里面，问题可以解决
但效率异常低，比嵌套起来写要低得多。。饿。。谢谢4楼，结了