我有4张桌子：

项目

+----+------+---------+-----+

| id | name | city_id | ... |

+----+------+---------+-----+

属性

+----+------+-----+

| id | name | ... |

+----+------+-----+

item_attribute

+----+---------+--------------+

| id | item_id | attribute_id |

+----+---------+--------------+

市

+----+------+-----+

| id | name | ... |

+----+------+-----+

项目和属性具有多对多关系.

物品仅位于一对多的城市中

题：

我正在使用php(Laravel).如何在一个城市中为一个具有相似属性的项目获取项目列表(带LIMIT)？属性列表永远不等于2个项目.

是否可以使用MySQL查询？

例：

| ItemName | Attributes | City |

+----------+-----------------------+------+

| Alpha | one, two, three, four | NY |

| Beta | five, six, seven | NY |

| Gamma | one, three, seven | NY |

| Delta | one, six, eight | CA |

| Epsilon | two, three, four | NY |

| Zeta | ten, nine | NY |

我想为Alpha选择类似的项目,它们将是：Gamma,Epsilon,因为它们具有相似的属性.

Delta将不会被选中,因为它位于另一个城市.

最佳答案 如果你同时传入了item_id和city_id：

SELECT i.name,

GROUP_CONCAT(a.name) attributes,

c.name

FROM items i

JOIN city c

ON c.id = i.city_id

JOIN item_attribute ia

ON ia.item_id = i.id

AND EXISTS (

SELECT 1

FROM item_attribute ia1

JOIN item_attribute ia2

ON ia2.attribute_id = ia1.attribute_id

AND ia2.item_id = ia.item_id

WHERE ia1.item_id = :item_id /* Pass in item id variable */

)

JOIN attributes a

ON a.id = ia.attribute_id

WHERE i.city_id = :city_id /* Pass in city id variable */

GROUP BY i.name, c.name

如果你只是想传递示例项id :(有点草率,但应该工作)

SELECT i.name,

GROUP_CONCAT(a.name) attributes,

c.name

FROM items base

JOIN items i

ON i.city_id = base.city_id

JOIN city c

ON c.id = i.city_id

JOIN item_attribute ia

ON ia.item_id = i.id

AND EXISTS (

SELECT 1

FROM item_attribute ia1

JOIN item_attribute ia2

ON ia2.attribute_id = ia1.attribute_id

AND ia2.item_id = ia.item_id

WHERE ia1.item_id = base.id

)

JOIN attributes a

ON a.id = ia.attribute_id

WHERE base.id = :item_id /* Pass in item id variable */

GROUP BY i.name, c.name

**更新**

排序：

...

JOIN (

SELECT ia2.item_id, COUNT(*) count

FROM item_attribute ia1

JOIN item_attribute ia2

ON ia2.attribute_id = ia1.attribute_id

AND ia2.item_id = ia1.item_id

/* AND ia2.id != ia1.id /* If you don't want the original item */

WHERE ia1.item_id = base.id

GROUP BY ia2.item_id

) similar

ON similar.id = ia.item_id

...

ORDER BY similar.count DESC