Question

我有一个查询，该查询返回成员，他们的上次访问和最后的付款。我的问题是，没有访问和/或付款，它不会返回会员。

我以前没有包括最后一次访问，然后我用LEFT和RIGHT JOINs而不是INNER进行查询，但是当我添加访问表时，我得到了包括它的索姆帮助，但我们没有注意到我们缺少成员访问或付款中的值为空。

我尝试应用LEFT和RIGHT JOIN没有任何运气。我也尝试添加例如。 “ OR（pt.member_id IS NULL）”也没有成功。

SELECT
    mr.member_id, 
    mr.name, 
    mr.tag, 
    pt.semester, 
    pt.date, 
    vt.date, 
FROM 
    members mr
INNER JOIN 
    payment pt 
ON 
    pt.member_id = mr.member_id 
    INNER JOIN 
        ( SELECT 
            member_id, 
            MAX(payment_id) max_value 
        FROM 
            payment 
        GROUP BY    
            member_id ) pt2 
    ON 
        pt.member_id = pt2.member_id 
    AND 
        pt.payment_id = pt2.max_value   
INNER JOIN 
    visit vt 
ON 
    vt.member_id = mr.member_id 
    INNER JOIN  
        ( SELECT    
            member_id, 
            MAX(date) max_visit_value 
        FROM 
            visit 
        GROUP BY 
            member_id ) vt2 
    ON 
        vt.member_id = vt2.member_id 
    AND 
        vt.date = vt2.max_visit_value

我想得到一个访问和/或付款可以为空的结果。

我希望我有道理，希望有人可以帮助我：）

MySQL 5.6

Answer 1

以下在各处使用左连接的版本可能会为您提供所需的结果：

SELECT
    mr.member_id,
    mr.name,
    mr.tag,
    pt.semester,
    pt.date,
    vt.date,
FROM members mr
LEFT JOIN payment pt
    ON pt.member_id = mr.member_id
LEFT JOIN
(
    SELECT member_id, MAX(payment_id) max_value
    FROM payment
    GROUP BY member_id
) pt2
    ON pt.member_id = pt2.member_id AND pt.payment_id = pt2.max_value
LEFT JOIN visit vt
    ON vt.member_id = mr.member_id
LEFT JOIN
(
    SELECT member_id, MAX(date) max_visit_value
    FROM visit
    GROUP BY member_id
) vt2
    ON vt.member_id = vt2.member_id AND vt.date = vt2.max_visit_value;

Answer 2

相关子查询可能是最简单的解决方案：

SELECT mr.member_id, mr.name, mr.tag, 
       (SELECT pt.semester
        FROM payment pt
        WHERE pt.member_id = mr.member_id
        ORDER BY pt.date DESC
        LIMIT 1
       ) as last_payment_semester
       (SELECT pt.date
        FROM payment pt
        WHERE pt.member_id = mr.member_id
        ORDER BY pt.date DESC
        LIMIT 1
       ) as last_payment_date
       (SELECT MAX(vt.date,)
        FROM visit vt
        WHERE vt.member_id = mr.member_id
        ORDER BY vt.date DESC
        LIMIT 1
       ) as last_visit_date
FROM members mr;

为了提高性能，您希望在payment(member_id, date desc, semester)和visit(member_id, date desc)上建立索引。

诚然，对于日期和学期，必须将基本上相同的子查询重复两次，这有点不雅。

在MySQL 8+中，您可以使用窗口函数：

SELECT mr.member_id, mr.name, mr.tag, pt.semester, pt.date, vt.date
FROM members mr LEFT JOIN
     (SELECT pt.*,
             ROW_NUMBER() OVER (PARTITION BY pt.member_id ORDER BY pt.date DESC) as seqnum
      FROM payment pt
     ) pt
     ON pt.member_id = mr.member_id AND pt.seqnum = 1 LEFT JOIN
     (SELECT pt.*,
             ROW_NUMBER() OVER (PARTITION BY vt.member_id ORDER BY vt.date DESC) as seqnum
      FROM visit vt
     ) vt
     ON vt.member_id = mr.member_id AND vt.seqnum = 1

Answer 3

这里最大的问题是会员可以进行很多付款和多次访问。您只想显示每个最新的。这对于访问很容易，因为您只想显示（最大）日期。但是，对于付款，您还想显示属于最长日期的学期。如果学期像日期一样升序，那么再简单一遍：使用MAX(semester)。如果不是，则必须检索最大日期 row 。

从MySQL 8开始：

SELECT
  mr.member_id, 
  mr.name, 
  mr.tag, 
  pt.semester, 
  pt.date, 
  vt.date 
FROM members mr
LEFT JOIN 
(
  SELECT
    member_id,
    semester,
    date,
    MAX(date) OVER (PARTITION BY member_id) AS last_date
  FROM payment
) pt ON pt.member_id = mr.member_id AND pt.dateb = pt.last_date
LEFT JOIN
(
  SELECT    
    member_id, 
    MAX(date) AS max_visit_value 
  FROM visit 
  GROUP BY member_id
) vt ON vt.member_id = mr.member_id 
ORDER BY mr.member_id;

在早期版本中：

SELECT
  mr.member_id, 
  mr.name, 
  mr.tag, 
  pt.semester, 
  pt.date, 
  vt.date 
FROM members mr
LEFT JOIN 
(
  SELECT
    member_id,
    semester,
    date
  FROM payment
  WHERE (member_id, date) IN 
  (
    SELECT member_id, MAX(date)
    FROM payment
    GROUP BY member_id
  )
) pt ON pt.member_id = mr.member_id
LEFT JOIN
(
  SELECT    
    member_id, 
    MAX(date) AS max_visit_value 
  FROM visit 
  GROUP BY member_id
) vt ON vt.member_id = mr.member_id 
ORDER BY mr.member_id;

Answer 4

如果主表中的记录更多，那么

LEFT JOIN可以为您带来帮助，因为这些ID在其他联接表中不会存在，因此肯定会给您带来空访问/付款。如果不是这种情况，也请尝试通过运行子查询单独进行调试，以检查单独运行时是否存在访问/付款遇到的任何空值。

Answer 5

对Thorsten Kettner的答案进行了一些微调，使其起作用：

谢谢大家：）

SELECT
    mr.member_id,
    mr.name,
    mr.tag,
    pt.semester,
    pt.date,
    vt.date
FROM members mr

LEFT JOIN
(
    SELECT 
        member_id, 
        semester, 
        date        
    FROM payment
    WHERE ( member_id, date ) IN
    (
        SELECT 
            member_id, 
            MAX(date)
        FROM 
            payment
        GROUP BY 
            member_id
    )
 ) pt ON pt.member_id = mr.member_id

 LEFT JOIN
 (
    SELECT
        member_id,
        date,
        door
    FROM visit
    WHERE ( member_id, date ) IN
    (
        SELECT 
            member_id, 
            MAX(date)
        FROM 
            visit
        GROUP BY 
            member_id
    )
 ) vt ON vt.member_id = mr.member_id

SQL连接3个具有空值的表

5 个答案: