Question

我正在使用3个表格：用户（ID，名字，姓氏，类型，禁用）评论（ID，发件人，日期） comment_recipient（COMMENT_ID，USER_ID）

我试图找到给定或收到某个月/年评论的用户的数据。我想要的结果行：名字|姓氏|评论给出了|收到的评论|月|年

单个评论可以有多个收件人（comment_recipient中有许多条目）

我开始关注这个问题，看起来有点类似： PubNub Access Manager to lock down access to your channels

以下是我目前的情况：

SELECT
  firstName AS 'First Name',
  lastName AS 'Last Name',
  given AS 'Comments Given',
  received AS 'Comments Received',
  MONTHNAME(d_yearmonth) AS Month,
  YEAR(d_yearmonth) AS Year
FROM (
  SELECT
    LAST_DAY(c.date) AS d_yearmonth,
    u.firstName,
    u.lastName,
    (SELECT COUNT(*) FROM comment AS c1 WHERE c1.sender=c.sender
    AND LAST_DAY(c1.date)=d_yearmonth) AS given,
    '24' AS received
  FROM comment AS c
  JOIN comment_recipient AS cr ON cr.comment_id=c.id
  JOIN user AS u ON u.id=c.sender
  WHERE u.type<4 AND u.disabled=0
  GROUP BY
    d_yearmonth
) AS s
ORDER BY Year DESC,
MONTH(d_yearmonth) DESC

我只是在使用＆＃39; 24＆＃39;为了收到＆＃39;目前，我正试图一次解决这个问题

我不需要计算每个用户或每个月，所以如果给定和收到它们都会返回0，则可以忽略它们

Answer 1

如果你把它分成两部分，我认为这个问题会更容易解决。

计算用户每月发送的邮件数
计算用户每月收到的邮件数

您可以使用联合“合并”发送和接收的消息。

完整查询可能如下所示：

SELECT user_id,
       lastName,
       firstName,
       SUM(sent) as sent,
       SUM(received) as received,
       year,
       month 

FROM (

  SELECT 
       c.sender as user_id,
       u.lastname as lastName,
       u.firstname as firstName,
       count(*) as sent,
       0 as received,
       YEAR(date) as year, 
       MONTH(date) as month
FROM Comment as C
  JOIN user AS u ON u.id=c.sender
  WHERE u.type<4 AND u.disabled=0
GROUP by user_id, year, month

UNION ALL 

select cr.user_id as user_id, 
       u.lastname as lastName,
       u.firstname as firstName,
       0 as sent,
       count(*) as received,
       YEAR(date) as year, 
       MONTH(date) as month
FROM comment_recipient cr
JOIN comment c ON c.id = cr.comment_id
JOIN user u on u.id = cr.user_id
WHERE u.type<4 AND u.disabled=0
GROUP by user_id, year, month
) A
GROUP by user_id, year, month;

Answer 2

您将从表中生成三个汇总数据集。他们是

一组月末日期
发送的评论摘要。
收到的意见摘要。

然后你可以一起加入他们

您需要为每个子查询提供子查询。从月末日期开始。

                        SELECT DISTINCT LAST_DAY(`date`) month_end
                          FROM comment

您需要发送的评论，按用户ID和月份汇总

                        SELECT COUNT(*) sent, 
                               sender id, 
                               LAST_DAY(`date`) month_end
                          FROM comment
                         GROUP BY sender, LAST_DAY(`date`)

您需要收到的评论，按用户ID和月份汇总。这需要加入评论表，以获取评论的日期。

                        SELECT COUNT(*) received, 
                               cr.user_id id, 
                               LAST_DAY(c.`date`) month_end
                          FROM comment_recipient cr
                          JOIN comment c ON cr.comment_id = c.id
                         GROUP BY cr.user_id, LAST_DAY(c.`date`)

然后，您需要使用用户表将这三个子查询（将它们视为虚拟表）加入。

SELECT u.firstName, u.lastName, 
       ISNULL(s.sent,0) sent,
       ISNULL(r.received,0) received, 
       MONTHNAME(d.month_end), YEAR(d.month_end)
  FROM (
                        SELECT DISTINCT LAST_DAY(`date`) month_end
                          FROM comment
       ) AS d
  JOIN user u ON 1=1
  LEFT JOIN (
                        SELECT COUNT(*) sent, 
                               sender id, 
                               LAST_DAY(`date`) month_end
                          FROM comment
                         GROUP BY sender, LAST_DAY(`date`)
            ) s ON d.month_end=s.month_end AND u.id = s.id
  LEFT JOIN (
                        SELECT COUNT(*) received, 
                               cr.user_id id, 
                               LAST_DAY(c.`date`) month_end
                          FROM comment_recipient cr
                          JOIN comment c ON cr.comment_id = c.id
                         GROUP BY cr.user_id, LAST_DAY(c.`date`)
            ) r ON d.month_end=r.month_end AND u.id = r.id
 WHERE u.type < 4 AND u.disabled = 0
   AND NOT (r.received IS NULL AND s.sent IS NULL) 
   AND d.month_end > CURDATE() - INTERVAL 1 YEAR

获取所需的所有行有点棘手。这就是我们进行ISNULL和LEFT JOIN操作的原因。我们还必须在单独的查询中进行聚合。如果我们只是加入评论和收件人，我们最终会有很多重复。

按月/年查询多个表和组

2 个答案: