Question

我正在努力使用以下SQL：我有一个表有一个site_id和一个站点的引用域。我试图通过引荐来源和网站简单计算出现次数，然后计算一个网站的所有引荐来源的总数。

在我的结果中，计数和网站总数不正确。计数应该是一半，例如对于网站ID 1，Google和亚马逊的计数应为10，网站ID 2的总数应为18。我有什么想法可能做错了吗？

SELECT site_id, I.referrer_domain AS referrer_domain, COUNT(*) AS items, t.site_total
  FROM qVisitor_Tracking_1 as I, 
      (SELECT COUNT(1) AS site_total
         FROM `qVisitor_Tracking_1` 
        WHERE `referrer_domain` != '' group by site_id) AS T
 WHERE `referrer_domain` != '' 
 GROUP BY `site_id`, referrer_domain


site_id   referrer  count   site total
1         amazon    20      24
1         google    20      24
2         amazon    12      24
2         google    18      24

Answer 1

如果我理解您的要求，您似乎缺少将子查询加入主表的ON子句。

SELECT I.site_id, 
        I.referrer_domain AS referrer_domain, 
        COUNT(*) AS items, 
        T.site_total 
FROM qVisitor_Tracking_1 as I
INNER JOIN 
(
    SELECT site_id, 
            COUNT(1) AS site_total 
    FROM qVisitor_Tracking_1 
    WHERE referrer_domain != '' 
    GROUP BY site_id
) AS T 
ON I.site_id = T.site_id
WHERE referrer_domain != '' 
GROUP BY I.site_id, referrer_domain, T.site_total

Answer 2

在mysql中，您可以使用GROUP BY修饰符 WITH ROLLUP ：

SELECT site_id, referrer_domain COUNT(*) AS items
  FROM qVisitor_Tracking_1
 WHERE referrer_domain != '' 
 GROUP BY site_id, referrer_domain WITH ROLLUP

它应该给出这样的结果：

site_id   referrer  items
1         amazon    7   
1         google    13
1         NULL      20   
2         amazon    6   
2         google    9
2         NULL      15
NULL      NULL      35

请参阅：http://dev.mysql.com/doc/refman/5.0/en/group-by-modifiers.html

Answer 3

你在I和T之间没有直接或间接的联合（在哪个部分）。所以你只有乘法

FROM qVisitor_Tracking_1 as I, 
      (SELECT COUNT(1) AS site_total
         FROM `qVisitor_Tracking_1` 
        WHERE `referrer_domain` != '' group by site_id) AS T
 WHERE `referrer_domain` != ''

MySQL：两个不同分组的Count和Total

3 个答案: