mysql - 选择 GROUP_CONCAT() 中的位置(1 和 2)

Question

entity
---
id  name
---
1   one
2   two
3   three

property
---
id  name
---
1   prop1
2   prop2
3   prop3

entity_property
---
entity_id   property_id
---
1           1
1           2
1           3
2           1

我想要获得至少具有 1 和 2 个(但可以具有更多)属性的实体。

这是我不喜欢的解决方法:

SELECT entity_property.entity_id,
(GROUP_CONCAT(entity_property.property_id)) as props
FROM `entity_property` 
JOIN entity
ON entity_property.entity_id = entity.id
GROUP BY entity.id

它返回:

entity_id props
---
1   1,2,3
2   1

然后我必须用服务器语言将其分解，然后排除。

<小时/>

此查询返回所有实体的行:

SELECT entity.id
FROM entity
WHERE (1 AND 2) IN
    (SELECT property_id
     FROM entity_property
     LEFT JOIN entity 
     ON entity_property.entity_id = entity.id
     WHERE entity_property.entity_id = entity.id)

<小时/>

此查询导致错误:

SELECT entity.id as ent_id
FROM entity
WHERE (1 AND 2) IN
    (SELECT property_id
     FROM entity_property
     LEFT JOIN entity 
     ON entity_property.entity_id = entity.id
     WHERE entity_property.entity_id = ent_id)

Answer 1

您可以使用group by和having获取实体ID:

SELECT ep.entity_id
FROM `entity_property` ep
WHERE ep.property_id IN (1, 2)
GROUP BY ep.entity_id
HAVING COUNT(DISTINCT ep.property_id) = 2;

注释:

• 这不需要实体表。
• 如果 entity_property 中不允许重复对，则不需要 DISTINCT。
• 对于更多属性，您需要更改 WHERE 和 HAVING(其中“2”是您要匹配的内容的数量)。