Question

以下查询将结果限制为20行，并且没有应用过滤器ID 6。

SELECT p.id, p.description
FROM   product p
LEFT JOIN product_filter pf 
ON pf.product_id = p.id
WHERE  p.status = TRUE
GROUP  BY p.id
HAVING SUM(pf.filter_id = '6') = 0 
ORDER  BY p.description ASC 
LIMIT  0, 20

这很好用，但现在我想计算具有相同条件的行数。

SELECT COUNT(DISTINCT p.id) AS total 
FROM product p
LEFT JOIN product_filter pf 
ON pf.product_id = p.id
WHERE p.status = TRUE
HAVING SUM(pf.filter_id = 6) = 0

此查询返回零结果。

编辑：根据要求， SQLFiddle 以更详细的方式帮助您查看问题。

Answer 1

你错过了GROUP BY p.id. 没有GROUP BY的情况不起作用。

SELECT COUNT(DISTINCT p.id) AS total 
FROM product p
LEFT JOIN product_filter pf 
ON pf.product_id = p.id
WHERE p.status = TRUE
GROUP BY p.id
HAVING SUM(pf.filter_id = 6) = 0

演示http://sqlfiddle.com/#!9/00b12/6

Answer 2

虽然@Raymond Nijland的答案是正确的，但在我的环境中，他建议的代码不起作用。

通过查看这篇文章COUNT(*) returning multiple rows instead of just one，我设法通过以下方式使其发挥作用：

SELECT COUNT(DISTINCT p.id) AS total 
FROM (SELECT p2.id 
      FROM product AS p2 
      LEFT JOIN product_filter pf 
        ON pf.product_id = p2.id 
      GROUP BY p2.id 
      HAVING SUM(pf.filter_id = '6') = 0) As p

使用HAVING条件获取总行数

2 个答案: