Question

假设我们的数据库中有衬衫。所有的衬衫都有尺寸和颜色（以及许多其他选项，但是现在我想用两个来解决这个问题。）

表格如下：

+-------------------------+--------------+--------------+-----------+
| variation_attributes_id | variation_id | attribute_id | option_id |
+-------------------------+--------------+--------------+-----------+
|                       1 |            1 | size         | s         |
|                       2 |            1 | color        | red       |
|                       3 |            2 | size         | m         |
|                       4 |            2 | color        | red       |
|                       5 |            3 | size         | s         |
|                       6 |            3 | color        | green     |
|                       7 |            4 | size         | m         |
|                       8 |            4 | color        | green     |
+-------------------------+--------------+--------------+-----------+

如何选择唯一variation_id。例如，size为S且color为Green时应返回variation_id == 3.

Answer 1

你可以用一个简单的mul.s $f4,$f3,$f0（自连接，这意味着你在特定条件下自己加入表）。假设您的表名为JOIN，它就像这样：

shirts

它将返回所有具有您需要的颜色和大小的variation_id，过滤掉所有其他的variation_id。

此查询可以正常运行，如果您拥有和SELECT DISTINCT s1.variation_id FROM shirts s1 JOIN shirts s2 ON s1.variation_id = s2.variation_id WHERE (s1.attribute_id = 'size' AND s1.option_id='s') AND (s2.attribute_id = 'color' AND s2.option_id='green')

之类的索引，那么它的效果会很好

此外，此查询可让您使用更复杂的参数获取(variation_id, attribute_id, option_id) ，使用variation_ids会更加困难。

例如 - 您需要找到尺寸为＆＃39;的所有HAVING COUNT(*) = ?。或者＆＃39; m＆＃39;和'绿色＆＃39;的颜色或者＆＃39;黄色＆＃39;

variation_ids

Answer 2

您可以在where子句中提供必要条件，并在count子句中having出现{<1}}：

select variation_id
from tbl
where ( attribute_id = 'size' and option_id = 's' )
   or ( attribute_id = 'color' and option_id = 'green' )
group by variation_id
having count(*) = 2

SQLFiddle

Answer 3

您可以使用EXISTS和子查询：

SELECT variation_id
FROM your_table t1
WHERE attribute_id = 'color'
  AND option_id  = 'green'
  AND EXISTS (SELECT 1
              FROM your_table t2
              WHERE t1.variation_id = t2.variation_id
                AND t2.attribute_id = 'size'
                AND t2.option_id = 's');

Answer 4

SELECT T1.`variation_id`
FROM(
(SELECT *
 FROM table_name
WHERE `attribute_id` = 'size') T1 
INNER JOIN
  (SELECT *
     FROM table_name
    WHERE `option_id ` = 's') T2 ON T1.`variation_id` = T2.`variation_id`)

希望这有帮助。

mysql从两个（或更多）行匹配的表中选择，

4 个答案: