Question

这是我的表格结构：

// users
+----+--------+
| id |  name  |
+----+--------+
| 1  | Jack   |
| 2  | Peter  |
| 3  | John   |
| 4  | Barman |
| 5  | Ali    |
+----+--------+

// vote
+----------+---------------+---------+
| voter_id | owner_post_id | post_id |
+----------+---------------+---------+
| 2        | 3             | 1653    |
| 4        | 2             | 1214    |
| 1        | 1             | 4355    |
| 4        | 2             | 6445    |
| 2        | 2             | 5465    |
| 3        | 2             | 5435    |
+----------+---------------+---------+

这是当前查询：

SELECT t2.id AS user_id,
       t2.name AS user_name,
       t3.id AS voter_id,
       t3.name AS voter_name
FROM vote t1
INNER JOIN users t2
    ON t1.owner_post_id = t2.id
INNER JOIN users t3
    ON t1.voter_id = t3.id
WHERE t1.owner_post_id = 2 AND
      t1.voter_id <> t1.owner_post_id

这是我的当前输出：

+---------+-----------+----------+------------+
| user_id | user_name | voter_id | voter_name |
+---------+-----------+----------+------------+
| 2       | Peter     | 4        | Barman     |
| 2       | Peter     | 4        | Barman     |
| 2       | Peter     | 3        | John       |
+---------+-----------+----------+------------+

现在我想在结果中再添加一列，其中包含每位选民的总投票数。所以这是预期结果：

+---------+-----------+----------+------------+-----------+
| user_id | user_name | voter_id | voter_name | total_num |
+---------+-----------+----------+------------+-----------+
| 2       | Peter     | 4        | Barman     | 2         |
| 2       | Peter     | 3        | John       | 1         |
+---------+-----------+----------+------------+-----------+

我该怎么做？

Answer 1

SELECT t2.id AS user_id,
       t2.name AS user_name,
       t3.id AS voter_id,
       t3.name AS voter_name
       Count(t1.voter_id) AS total_num
FROM vote t1
INNER JOIN users t2
    ON t1.owner_post_id = t2.id
INNER JOIN users t3
    ON t1.voter_id = t3.id
WHERE t1.owner_post_id = 2 AND
      t1.voter_id <> t1.owner_post_id
GROUP BY t2.id AS user_id,
       t2.name,
       t3.id,
       t3.name

修改

SELECT pr1.id AS user_id, pr1.title AS user_name, pr2.id AS liker_id, pr2.title AS liker_name, x.which AS which_table, COUNT(pr1.id) AS total FROM ( SELECT rid, rootid, 'vote' which FROM p_likes UNION ALL SELECT rid, rootid, 'comment' which FROM p_comments UNION ALL SELECT rid, rootid, 'friend' which FROM relations ) x INNER JOIN pagesroot pr1 ON x.rootid = pr1.id INNER JOIN pagesroot pr2 ON x.rid = pr2.id WHERE x.rootid = 1 AND x.rootid <> x.rid GROUP BY pr1.id, pr1.title, pr2.id, pr2.title, x.which ORDER BY x.which

Answer 2

获取用户的所有投票，汇总并计算。然后加入用户表两次;一次为用户，一次为选民。

select
  usr.id as user_id, 
  usr.name as user_name, 
  vtr.id as voter_id, 
  vtr.name as voter_name,
  v.cnt as total_num
from 
(
  select owner_post_id, voter_id, count(*) as cnt
  from vote
  where owner_post_id = 2 and owner_post_id <> voter_id
  group by owner_post_id, voter_id
) v
join user usr on usr.id = v.owner_post_id
join user vtr on usr.id = v.voter_id;

这是应用于您的真实查询的相同方法。只是其他一些表和列名，所以我认为你应该能够自己做到这一点。

select 
  pr1.id as user_id,
  pr1.title as user_name,
  pr2.id as liker_id,
  pr2.title as liker_name,
  x.which as which_table,
  x.cnt as total
from 
(
  select rid, rootid, which, count(*) as cnt
  from
  (
    select rid, rootid, 'vote' which from p_likes
    union all 
    select rid, rootid, 'comment' which from p_comments
    union all 
    select rid, rootid, 'friend' which from relations
  ) all_in_one
  where rootid = 1 and rootid <> rid
  group by rid, rootid, which
) x
join pagesroot pr1 on x.rootid = pr1.id
join pagesroot pr2 on x.rid = pr2.id
order by x.which;

我如何计算每位选民的所有选票数？

2 个答案: