SQL左连接具有多个值

时间:2014-10-20 21:11:02

标签: mysql

我有一个非常糟糕的表结构的旧数据库,我正在尝试创建具有更好结构的这些表。为此,我需要匹配两个表,以获取类别的ID。

这是我的两张旧表:

表格分类:

| 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  |

表产品:

| 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,我想从产品表中删除列cat1,2,3和4.

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  |

有人可以告诉我,我应该如何查询是否所有4只猫都匹配,然后给我猫咪?我已经尝试过这种方式了,但我认为这是错误的方式,因为每次产品只有2或3只猫,我都没有得到相关的catId。因此它仅适用于定义了所有4种猫的产品。

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。 ; - )

感谢您帮助我:)

4 个答案:

答案 0 :(得分:3)

您可以通过为每个类别组合执行单独的连接来实现此目的。当然,这是一个等级排名,你不想要重复。因此,以下是这些检查:

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,则可能仍需要group by

答案 1 :(得分:2)

首先,您需要正确的数据模型。

分类

| ID |  catname     |
+----+--------------+

产品:

| ID |  productnr  |  productname  |
+----+-------------+---------------+

哦,这么神奇的连接表:

|ID | product_id | category_id|
+---+------------+------------+

现在您可以正确地要求您的数据库为您提供数据

SELECT
    cat.`id`,
    prod.`productnr`,
    prod.`productname`

FROM categories as cat

INNER JOIN products2categories p2c
ON p2c.category_id = cat.id

INNER JOIN products prod 
ON p2c.product_id = prod.id

由于我真的不喜欢发放完整的复制粘贴示例,我将让您使用上面的查询来检查如何找到存在于4个类别中的产品。请记住 - 最简单的方法通常始终是正确的方法。

答案 2 :(得分:1)

问题可能是表中的null,因为sql

中的null!= null

要解决这个问题,您可以合并数据

SELECT
    cat.`id`,
    prod.`productnr`,
    prod.`productname`
FROM
    products as prod
LEFT JOIN
    categorys as cat
ON
    coalesce(cat.`cat1`,'No Value') = coalesce(prod.`cat1`,'No Value')
AND
    coalesce(cat.`cat2`,'No Value') = coalesce(prod.`cat2`,'No Value')
AND
   coalesce(cat.`cat3`,'No Value') = coalesce(prod.`cat3`,'No Value')
AND
    coalesce(cat.`cat4`,'No Value') = coalesce(prod.`cat4`,'No Value')

答案 3 :(得分:0)

如果您想在每个id字段加入时仅获得cat次,请尝试使用INNER JOIN代替LEFT JOIN

SELECT
    cat.`id`,
    prod.`productnr`,
    prod.`productname`
FROM
    products as prod
INNER JOIN
    categorys as cat
ON
    cat.`cat1` = prod.`cat1`
AND
    cat.`cat2` = prod.`cat2`
AND
    cat.`cat3` = prod.`cat3`
AND
    cat.`cat4` = prod.`cat4`