用于说明问题的表和一些虚拟数据。
存储有关成员的基本信息。
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| member_id | email |
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| 1 | 1@a.com |
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| 2 | 2@a.com |
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| 3 | 3@a.com |
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| 4 | 4@a.com |
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| 5 | 4@a.com |
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为每个成员存储一些额外的元数据
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| member_id | name | surname | company |
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| 1 | A | A | A |
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| 2 | B | B | B |
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| 3 | C | C | C |
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| 4 | D | D | D |
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| 5 | E | E | E |
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系统内的不同类别。
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| cat_id | cat_name |
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| 1 | Cars |
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| 2 | Bikes |
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| 3 | Boats |
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会员必须拥有能够访问类别的许可。
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| id | subid | catid | start_date | end_date | description |
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| 1 | 1 | 1 | 2014-01-01 | 2020-12-31 | Premium |
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| 2 | 1 | 2 | 2014-01-01 | 2015-12-31 | Premium |
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| 3 | 1 | 3 | 2014-01-01 | 2018-12-31 | Premium |
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| 4 | 2 | 1 | 2014-01-01 | 2016-12-31 | Premium |
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| 7 | 3 | 1 | 2014-01-01 | 2014-01-02 | Premium |
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| 8 | 3 | 2 | 2014-01-01 | 2014-01-02 | Premium |
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| 9 | 3 | 3 | 2014-01-01 | 2020-01-31 | Premium |
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| 10 | 5 | 1 | 2014-01-01 | 2014-01-02 | Premium |
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| 11 | 5 | 2 | 2014-01-01 | 2014-01-02 | Premium |
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| 12 | 5 | 3 | 2014-01-01 | 2014-01-02 | Premium |
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会员1拥有1,2和3类的许可。它们都是有效且有效的。 会员2仅拥有类别1的许可。它很活跃。 会员3拥有1,2和3的许可。只有第3类的许可才有效。 会员4没有许可证。 会员5拥有1,2和3类的许可证，但它们都已过期。
我想获取每个会员许可证的记录，以及各自的member_data和类别。许可证必须存在且对于成员的类别有效，以便为该许可证返回数据。
此外，我希望返回的每个许可证都作为一行返回，其中包含以下格式所需的所有数据：
我想输出持有有效许可证的成员，如果他们没有某个类别的许可证但是确实持有另一个类别，则返回其类别的到期日期或没有输出。即：
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| Company | Name | LicenceType | Cars | Bikes | Boats |
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| A |A A | Premium |2020-12-31 | 2015-12-31 | 2018-12-21 |
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| B |B B | Premium |2016-12-31 | | |
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| C |C C | Premium | | | 2020-01-31 |
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SELECT
md.company as Company,
CONCAT(md.name,' ', md.surname) as Name,
l.description as LicenceType,
(CASE WHEN (c.cat_name='Cars') THEN l.end_date ELSE '' END)AS Cars,
(CASE WHEN (c.cat_name='Bikes') THEN l.end_date ELSE '' END)AS Bikes,
(CASE WHEN (c.cat_name='Boats') THEN l.end_date ELSE '' END)AS Boats
FROM
licences as l
JOIN
categories as c ON c.cat_id=l.catid
JOIN
member_data as md ON md.member_id=l.subid
WHERE
l.end_date>='2014-12-17'
AND
(l.description='Premium')
ORDER BY Company ASC
这是当前数据的显示方式：
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| Company | Name | LicenceType | Cars | Bikes | Boats |
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| A |A A | Premium |2020-12-31 | | |
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| A |A A | Premium | | 2015-12-31 | |
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| A |A A | Premium | | | 2018-12-21 |
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| B |B B | Premium |2016-12-31 | | |
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| C |C C | Premium | | | 2020-01-31 |
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问题是，正如您可以看到A公司的记录，显示为三个不同的行。我想按照上面显示的输出格式将三行中的每一行作为一行返回。
我很欣赏有关如何实现这一目标的任何想法。谢谢。
答案 0 :(得分：2)
使用聚合：
SELECT
md.company as Company,
CONCAT(md.name,' ', md.surname) as Name,
l.description as LicenceType,
MAX(CASE WHEN (c.cat_name='Cars') THEN l.end_date ELSE '' END)AS Cars,
MAX(CASE WHEN (c.cat_name='Bikes') THEN l.end_date ELSE '' END)AS Bikes,
MAX(CASE WHEN (c.cat_name='Boats') THEN l.end_date ELSE '' END)AS Boats
FROM
licences as l
JOIN
categories as c ON c.cat_id=l.catid
JOIN
member_data as md ON md.member_id=l.subid
WHERE
l.end_date>='2014-12-17'
AND
(l.description='Premium')
GROUP BY Company, Name, l.description
ORDER BY Company ASC;