我有4张桌子:
项
+----+------+---------+-----+
| id | name | city_id | ... |
+----+------+---------+-----+
属性
+----+------+-----+
| id | name | ... |
+----+------+-----+
item_attribute
+----+---------+--------------+
| id | item_id | attribute_id |
+----+---------+--------------+
城市
+----+------+-----+
| id | name | ... |
+----+------+-----+
项目和属性具有多对多的关系。
项目仅位于一个城市的一对多
问题:
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?
示例:
| 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。
答案 0 :(得分:1)
如果你同时传递了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
答案 1 :(得分:0)
您可以在所有
中执行INNER JOINS
SELECT I.name,I_A.name,city.name FROM attributes as A
INNER JOIN item_attribute as I_A ON I_A.attribute_id = A.id
INNER JOIN city ON I_A.id = city.id
INNER JOIN items as I ON I.id = I_A.item_id
WHERE <Your condition>
要获得以逗号分隔的值,您可以参考here 如果我不明白你的意思,请告诉我。