SQL 连接 3 个具有空值的表

Question

我想查询 return 的成员，他们上次访问和上次付款。我的问题是 return 没有访问 and/or 的会员不会付款。

我之前没有包含上次访问，然后我使用 LEFT 和 RIGHT JOIN 而不是 INNER 进行了查询，但是当我添加访问时 table 我收到了一些帮助以包含它，但我们没有注意到我们在访问或付款中缺少具有空值的成员。

我试过应用 LEFT 和 RIGHT JOIN，但没有任何运气。我也试过添加例如。 "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

我想获得访问 and/or 付款可以为空的结果。

我希望我说得有道理，希望有人能帮助我 :)

MySQL 5.6

Answer 1

以下版本在任何地方都使用左连接可能会给您想要的结果：

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，您需要在 payment(member_id, date desc, semester) 和 visit(member_id, date desc) 上建立索引。

不可否认，必须针对日期和学期重复两次基本相同的子查询有点不雅。

在MySQL 8+中，您可以使用window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

这里最大的问题是一个会员可以有很多次付款和很多次访问。您只想显示每个的最新版本。这对于访问来说很容易，因为您只想显示（最大）日期。但是，对于付款，您还希望显示属于最大日期的学期。如果学期像日期一样递增，那么就又简单了：使用 MAX(semester)。如果不是，那么您必须改为检索最大日期行。

截至 MySQL 8：

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;

早期版本：

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

LEFT JOIN 可以提供帮助，如果主 table 中的记录更多，那肯定会给你 null visits/payments 因为这些 ID 不会出现在其他加入的 table 中.如果不是这种情况，请尝试通过运行子查询进行调试，分别检查当运行时 visits/payments 是否遇到任何空值。

Answer 5

对 Thorsten Kettner 的回答稍作调整使其生效：

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

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