在MySQL中创建INTERSECT查询

Question

我有一张桌子

ppv (product_id | property_id | value_id)

和

v (id | property_id | value)

我尝试进行查询以获取products其中：

select product_id from v, ppv as t INNER JOIN ppv as t2 
ON t.product_id = t2.product_id and t1.property_id = t2.property_id
where
( t1.property_id = 1 and v.value > $a and v.value < $b )
and
( t2.property_id = 2 and v.value > $c and v.value < $d )

但是没有任何结果。请帮忙。

所以，

ppv (product_id | property_id | value_id)
 1              1            50
 1              2            2
 1              3            3
 2              1            28
 2              2            2
 2              3            29

下一张表

v( id | property_id | value)
 2          2           3
 3          3           2
 28         1           2600000
 29         3           3
 50         1           2500000

---

**我用这种方式找到了解决这个问题的方法**

SELECT `product_id`, 
       CONVERT(GROUP_CONCAT(v.`property_id`) USING cp1251) as properties
FROM `values` as v INNER JOIN `product_property_values` as ppv
ON ppv.value_id = v.id
WHERE
   (v.`property_id` = 1 and v.value >= 2500000 and v.value <= 3000000 ) 
   or 
   (v.`property_id` = 2 and v.value >= 1 and v.value <= 3 )
GROUP by product_id
HAVING LOCATE('1', properties) and LOCATE('2', properties)

结果：

测试1

v.`property_id` = 1 and v.value >= 2500000 and v.value < 3000000
v.`property_id` = 2 and v.value >= 1 and v.value <= 2

 product_id properties
 Empty

TEST2

v.`property_id` = 1 and v.value >= 2550000 and v.value < 3000000
v.`property_id` = 2 and v.value >= 1 and v.value <= 4

product_id  properties
     2          2,1

TEST3

v.`property_id` = 1 and v.value >= 2000000 and v.value < 3000000
v.`property_id` = 2 and v.value >= 1 and v.value <= 4

product_id  properties
     1          1,2
     2          2,1

没关系！

Answer 1

看起来你实际上并没有加入表v。另外，我不相信有必要两次链接ppv。尝试：

SELECT ppv.product_id
FROM v
INNER JOIN ppv ON v.property_id = ppv.property_id
WHERE
(ppv.property_id = 1 and v.value > $a and v.value < $b )
OR
(ppv.property_id = 2 and v.value > $c and v.value < $d )

Answer 2

第一个问题：什么是t1？

第二个问题：

• t1.property_id = t2.property_id
• t1.property_id = 1
• t2.property_id = 2

这意味着：1 = 2。

---

如果我理解得很好，那么您正在寻找具有以下产品的产品：

• 一个property_id = 1集
• 一个property_id = 2集
• 如果$a存在，则<{1}}的值
• 如果property_id = 1存在，则<{1}}的值
如果$c存在，则
• 低于property_id = 2的值
如果$b存在，则
• 低于property_id = 1的值

我认为您正在寻找此请求：

$d

Answer 3

SELECT `product_id`, 
       CONVERT(GROUP_CONCAT(v.`property_id`) USING cp1251) as properties
FROM `values` as v INNER JOIN `product_property_values` as ppv
ON ppv.value_id = v.id
WHERE
   (v.`property_id` = 1 and v.value >= 2500000 and v.value <= 3000000 ) 
   or 
   (v.`property_id` = 2 and v.value >= 1 and v.value <= 3 )
GROUP by product_id
HAVING LOCATE('1', properties) and LOCATE('2', properties)

结果：

测试1

v.`property_id` = 1 and v.value >= 2500000 and v.value < 3000000
v.`property_id` = 2 and v.value >= 1 and v.value <= 2

 product_id properties
 Empty

TEST2

v.`property_id` = 1 and v.value >= 2550000 and v.value < 3000000
v.`property_id` = 2 and v.value >= 1 and v.value <= 4

product_id  properties
     2          2,1

TEST3

v.`property_id` = 1 and v.value >= 2000000 and v.value < 3000000
v.`property_id` = 2 and v.value >= 1 and v.value <= 4

product_id  properties
     1          1,2
     2          2,1

没关系！