从两个表中选择计数不同值

Question

用户表：

+-----------+----------+------------+-------------+
| name      | fb_id    | date       | flipbook    |
+-----------+----------+------------+-------------+
| Tia       | 66783722 | 1975-09-18 |  june 2014  |
| Nikki     | 10438259 | 1972-03-04 |  july 2014  |
| Yamila    | 73370629 | 1972-03-04 | august 2014 | 
+-----------+----------+------------+-------------+

访问表：

+-----------+----------+------------+-------------+
| Name      | fb_id    | Date       |  Flipbook   |
+-----------+----------+------------+-------------+
| Tia       | 66783722 | 1975-09-18 | june 2014   |
| Nikki     | 10438259 | 1972-03-04 | august 2014 |
| Nikki     | 10438259 | 1972-04-04 | october 2014|
+-----------+----------+------------+-------------+

我希望查询从用户表返回所有用户，并计算活动簿的数量，例如：

[1]
name => Tia, 
fb_id = 66783722,
date => 1975-09-18,
count_flip => 1 (june 2014 is in both tables so we count it as 1)

[2]
name => Nikki, 
fb_id = 10438259,
date => 1972-03-04,
count_flip => 3 (because in the first table we have june 2014 and in the second table we have august 2014 and october 2014, so no duplicates)

[3]
name => Yamila, 
fb_id = 73370629,
date => 1972-03-04,
count_flip => 1 (because we have august 2014 in the first table and she is not mentioned in the second one)

我试图做这个查询：

SELECT u.*, (SELECT COUNT(Distinct flipbook) FROM visits v WHERE v.fb_id = u.fb_id) as count_flip FROM users u

但是问题是我缺少了用户表中的用户，但是缺少访问表中的用户。 因此，对于上面的示例，我在查询中不会看到“ Yamila”。

有什么想法如何解决这个问题？

Answer 1

关于逻辑的另一种思考方式是，您希望对users表计数1，然后将visits表中name所有不匹配的活页簿值相加。 这表明相关的子查询：

select u.*,
       (select 1 + count(*)
        from visits v
        where v.name = u.name and v.flipbook <> u.flipbook
       ) as nr
from users u;

编辑：

另一种执行所需操作的方法效率较低（假设您为上述查询使用了正确的索引）：

select u.*, uv.nr
from users u join
     (select u.name, count(distinct flipbook) as nr
      from ((select name, flipbook from users) union all
            (select name, flipbook from visits)
           ) uv
      group by u.name
     ) uv
     on u.name = uv.name;

Answer 2

我想你可以做到这一点：

SELECT users.name, users.fb_id, users.date, COUNT(DISTINCT visits.Flipbook) + 1 AS count_flip
FROM users
LEFT OUTER JOIN visits
ON users.fb_id = visits.fb_id
AND users.flipbook != visits.flipbook
GROUP BY users.name, users.fb_id, users.date

是否进行LEFT OUTER JOIN来为每个具有不同日期的匹配用户获得一行。 然后，仅对结果计数加一（以对用户表中的原始字段进行计数）。

这样可以避免任何子查询。