sql 多对多查询结果仅当仅匹配且完全匹配数组中的所有项目时

Posted 2023-02-24

【中文标题】sql 多对多查询结果仅当仅匹配且完全匹配数组中的所有项目时

【英文标题】：sql many-to-many query result only if matches only and exactly all items in array

【发布时间】：2021-03-24 13:57:14

【问题描述】：

我在一个 mysql 数据库上有三个表：

| pieces   | line_up       | instrument    | 
--------------------------------------------
| id_piece | piece_id      | id_instrument |
| title    | instrument_id | instrument    |

现在我想要实现的是：我想查询那些其 line_up 完全由列表中给出的乐器组成的片段，而不是一个少一个。 此 sql 查询将结果减少到仅由 2 种乐器演奏的曲目，但它包括独奏

SELECT id_piece, title FROM pieces WHERE
   (SELECT COUNT(*) FROM line_up WHERE line_up.piece_id = pieces.id_piece)
   =
   (SELECT COUNT(*) FROM line_up
   INNER JOIN instruments ON instruments.id_instrument = line_up.instrument_id
   WHERE line_up.piece_id = pieces.id_piece
   AND instruments.instrument IN ('guitar', 'drums'));

以这些表格为例：

| pieces               |  | line_up                  |  | instruments                |
-----------------------   ---------------------------   ------------------------------
| id_piece | title     |  | piece_id | instrument_id |  | id_instrument | instrument |
-----------------------  ----------------------------   ------------------------------
| 1        | hello     |  | 1        | 1             |  | 1             | guitar     |
| 2        | goodbye   |  | 1        | 2             |  | 2             | drums      |  
| 3        | goodnight |  | 2        | 1             |  ------------------------------
------------------------  | 3        | 2             |
                          ----------------------------

吉他和鼓的唯一实际作品，因此我的查询结果应该是1 | hello。 有什么建议？谢谢！

【参考方案1】：

你可以像这样使用聚合：

select p.id_piece, p.title
from pieces as p
inner join line_up as l on l.piece_id = p.id_piece
inner join instruments as i on l.instrument_id = i.id_instrument
group by p.id_piece
having sum(i.instrument in ('guitar', 'drums')) = 2
   and sum(i.instrument not in ('guitar', 'drums')) = 0
   

如果您没有太多要搜索的工具，另一种选择是having 子句中的字符串聚合：

having group_concat(i.instrument order by i.instrument) = 'drums,guitar'

第二个表达式要求您为查询提供按字母顺序排列的乐器列表。

【讨论】：

应该有一些东西可以计算请求项目的数量来代替 2

@astentx：如果该值列表是在应用程序中构建的，您会希望应用程序知道给出了多少值。

@GMB ..正是如此。非常非常感谢！

有多少乐器会“太多”？我正在考虑 30 种乐器的最大值

【参考方案2】：

另一种方法。按乐器组合匹配。

SELECT title 
FROM pieces
WHERE id_piece IN (  
  SELECT piece_id
  FROM line_up 
  WHERE instrument_id IN (
    SELECT id_instrument
    FROM instruments 
    WHERE instrument IN ('Guitar', 'Drums')
  )
  GROUP BY piece_id
  HAVING group_concat(instrument_id order by instrument_id separator ',') = (
    SELECT group_concat(id_instrument order by id_instrument separator ',')
    FROM instruments
    WHERE instrument IN ('Guitar', 'Drums')
   )
);

【参考方案3】：

您需要计算使用仪器过滤的行数，并将其与您选择的仪器总数和仪器数量进行比较。

db<>fiddle

with pieces as (
  select 1 as id_piece, 'hello' as title union all
  select 2 as id_piece, 'goodbye' union all
  select 3 as id_piece, 'goodnight'
)
, line_up as (
  select 1 as piece_id, 1 as instrument_id union all
  select 1 as piece_id, 2 as instrument_id union all
  select 2 as piece_id, 1 as instrument_id union all
  select 3 as piece_id, 2 as instrument_id
)
, instruments as (
  select 1 as id_instrument, 'guitar' as instrument union all
  select 2 as id_instrument, 'drums' as instrument
)
select p.id_piece, p.title
from pieces as p
  join line_up as l
    on l.piece_id = p.id_piece
  join instruments as i
    on l.instrument_id = i.id_instrument
group by p.id_piece, p.title
having sum(case when i.instrument in ('guitar', 'drums') then 1 else 0 end) = count(1)
and count(1) = (
  /*To select only valid instruments*/
  select count(1)
  from instruments as f
  where f.instrument in ('guitar', 'drums')
)

-----------+--------
| id_piece | title |
|----------+-------|
|        1 | hello |
-----------+--------