MySQL 计数唯一列，其中其他列在 Group By 中具有单个值

Question

请看下面的table。我想计算不同的访问者 (visitor_id)，按组分组 (group_id)，但只计算结果始终为该访问者 'declined' 的访问者。

类似

SELECT group_id, COUNT(DISTINCT visitor_id) AS always_declines
FROM customer_actions 
WHERE outcome='declined' [[AND HAS NEVER BEEN IN ('purchased')]]
GROUP BY group_id;

这是我的 table:

的简化版本

SELECT * FROM customer_actions;
+----+------------+-----------+----------+
| id | visitor_id | outcome   | group_id |
+----+------------+-----------+----------+
|  1 |          5 | purchased |        1 |
|  2 |          5 | purchased |        1 |
|  3 |          6 | purchased |        1 |
|  4 |          7 | declined  |        1 |
|  5 |          6 | declined  |        1 |
|  6 |          7 | purchased |        1 |
|  7 |          8 | declined  |        1 |
|  8 |          8 | declined  |        1 |
+----+------------+-----------+----------+
8 rows in set (0.00 sec)

所以基本上，如果它有效，我正在寻找返回的第一行也是唯一一行（在这种情况下）：

group_id = 1

always_declines = 1（对应只拒绝过的访客8）

Answer 1

not exists 运算符应该可以解决问题：

SELECT   group_id, COUNT(DISTINCT visitor_id) AS always_declines
FROM     customer_actions  ca1
WHERE    NOT EXISTS (SELECT *
                     FROM   customer_actions ca2
                     WHERE  ca1.group_id = ca2.group_id AND
                            ca1.visitor_id = ca2.visitor_id AND
                            ca2.outcome != 'declined')
GROUP BY group_id;

Answer 2

解决这个问题的一种方法是作为两个聚合。首先，按组和访客进行聚合，以获取合适的访客。然后计算剩余的行数：

SELECT group_id, count(*) AS always_declines
FROM (SELECT group_id, visitor_id
      FROM customer_actions 
      GROUP BY group_id, visitor_id
      HAVING SUM(outcome <> 'declined') = 0
     ) gv
GROUP BY group_id;

MySQL 计数唯一列，其中其他列在 Group By 中具有单个值

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