如何获取每行mysql查询的计数？

Question

我有三张桌子

1）review_trans

2）user_comment_trans

第3）helpful_review_trans

表结构如下所示

的 review_trans

 review_id(pk) |   review_desc  |  user_id 
 --------------+----------------+-----------
 1             |  hello world   |   1
 2             |  test2         |   1
 3             |  test3         |   2

的 user_comment_trans

 comment_id  |   review_id  |     user_id
 ------------+--------------+------------
  1          |        1     |         4
  2          |        1     |         2
  3          |        2     |         3

的 helpful_review_trans

 helpful_review_id |  review_id |  user_id
 ------------------+------------+------------
 1                 |     1      |     4 
 2                 |     1      |     2
 3                 |     2      |     3

我希望每次审核都能获得有用的投票和评论。 所以我想要的输出是这样的

的输出

review_id       review_desc      count(helpful_review_id)      count(comment_id)
------------+------------------+----------------------------+--------------------
1           |    hello world   |           2                |           2
2           |    test2         |           1                |           1
3           |    test3         |                            |

我无法获得此记录。我尝试加入表格，但它只显示了一个review_desc 请指教和帮助。

Answer 1

使用左连接计数（不同）

select r.review_id, 
r.review_desc,
count(distinct u.comment_id),
count(distinct h.helpful_review_id)
from review_trans r
left join user_comment_trans u on(r.review_id = u.review_id)
left join helpful_review_trans h on(r.review_id = h.review_id)
group by r.review_id

Demo

Answer 2

SELECT rt.review_id, rt.review_desc, 
       COUNT(DISTINCT hrt.helpful_review_id), COUNT(DISTINCT uct.comment_id)
FROM review_trans rt
LEFT JOIN helpful_review_trans hrt ON rt.review_id = hrt.review_id
LEFT JOIN user_comment_trans uct ON uct.review_id = rt.review_id
GROUP BY rt.review_id

Answer 3

SELECT rt.review_id, rt.review_desc, helpful_count, comment_count
FROM review_trans AS rt
LEFT JOIN (
    SELECT review_id, COUNT(*) AS comment_count
    FROM user_comment_trans
    GROUP BY review_id) AS uct ON rt.review_id = uct.review_id
LEFT JOIN (
    SELECT review_id, COUNT(*) AS helpful_count
    FROM helpful_review_trans
    GROUP BY review_id) AS hrt ON rt.review_id = hrt.review_id

Answer 4

你可以这样做：

SELECT
    review_trans.review_id,
    review_trans.review_desc,
    (
        SELECT
            COUNT(*)
        FROM
            helpful_review_trans
        WHERE
            helpful_review_trans.review_id=review_trans.review_id
    ) AS helpful_review_id,
    (
        SELECT
            COUNT(*)
        FROM
            user_comment_trans
        WHERE
            user_comment_trans.review_id=review_trans.review_id
    ) AS CountComment_id
FROM
    review_trans