我有一个非常旧的数据库,有一个非常糟糕的表结构,我正在尝试创建这些表的更好的结构。为此,我需要匹配两个表,以获取类别的id。
这是我的两张旧桌子:
表类别:
| ID | catname | cat1 | cat2 | cat3 | cat4 |
+----+--------------+--------+-------------+---------+------+
| 1 | bike | bike | NULL | NULL | NULL |
| 2 | accessories | bike | accessories | NULL | NULL |
| 3 | helmets | bike | accessories | helmets | NULL |
| 4 | lights | bike | accessories | lights | NULL |
| 5 | led | bike | accessories | lights | led |
餐桌产品:
| ID | productnr | productname | cat1 | cat2 | cat3 | cat4 |
+----+-------------+---------------+-------+-------------+---------+------+
| 1 | 451157 | productya | bike | accessories | NULL | NULL |
| 2 | 555523 | product11 | bike | accessories | helmets | NULL |
| 3 | 234432 | helmetxqa | bike | accessories | helmets | NULL |
| 4 | 666623 | lightblue | bike | accessories | lights | NULL |
| 5 | 542123 | foobarlight | bike | accessories | lights | led |
首先,我想从products表中去掉cat1、2、3和4列。
所以我得到这样的结果:
| ID | catId | productnr | productname |
+----+---------+------------+---------------+
| 1 | 2 | 451157 | productya |
| 2 | 3 | 555523 | product11 |
| 3 | 3 | 234432 | helmetxqa |
| 4 | 4 | 666623 | lightblue |
| 5 | 5 | 542123 | foobarlight |
有人能告诉我,我应该如何做一个查询,检查4只猫是否都匹配,然后给我猫的id?我试过这种方法,但我认为这是错误的方法,因为每次一个产品只有2或3只猫,我没有得到相关的catId。所以它只适用于定义了所有4种cat的产品。
SELECT
cat.`id`,
prod.`productnr`,
prod.`productname`
FROM
products as prod
LEFT JOIN
categorys as cat
ON
cat.`cat1` = prod.`cat1`
AND
cat.`cat2` = prod.`cat2`
AND
cat.`cat3` = prod.`cat3`
AND
cat.`cat4` = prod.`cat4`
如果有人也对我有用,请告诉我。;-)
谢谢你帮我:)
最佳答案
您可以通过对每个类别组合执行单独的联接来实现这一点。当然,这是一个等级排序,你不想重复。因此,以下检查:
SELECT prod.id, coalesce(c4.id, c3.id, c2.id, c1.id) as catid
prod.`productnr`, prod.`productname`
FROM products prod left join
categorys c4
on c4.cat1 = prod.cat1 and c4.cat2 = prod.cat2 and
c4.cat3 = prod.cat3 and c4.cat4 = prod.cat4 left join
categorys c3
on c3.cat1 = prod.cat1 and c3.cat2 = prod.cat2 and
c3.cat3 = prod.cat3 and c3.cat4 is null and
c4.id is null left join
categorys c2
on c2.cat1 = prod.cat1 and c2.cat2 = prod.cat2 and c2.cat3 is null and
c3.id is null and c4.id is null left join
categorys c1
on c1.cat1 = prod.cat1 and c1.cat2 is null and
c2.id is null and c3.id is null and c4.id is null;
在某些情况下,这可能会产生重复的行(尽管它可以避免这种情况)。如果发生这种情况,那么
group by
可能仍然是必要的。关于mysql - SQL左联接具有多个值,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/26474992/