Question

我有一个复杂的数据库结构，我正在尝试制定一个SQL查询。首先是表格的结构：

Table ANIMALS:
+---------+--------+
|    id   |  name  |
+---------+--------+
|    1    | Tiger  |
|    2    | Lion   |
|    3    | Cat    |
+---------+--------+


Table ANIMAL_ATTRIBUTES:
+---------------+-----------+
| attribute_id  | animal_id |
+---------------+-----------+
|            10 |         1 |
|            11 |         3 |
|            12 |         3 |
+---------------+-----------+



Table ATTRIBUTE_TEXT:
+--------------+-- ----------+
| attribute_id |    value    |
+--------------+-------------+
|           10 |  black      |
|           11 |  big        |
|           12 |  tail       |
+--------------+-------------+


Table INFORMATION:
+---------------+-----------+
| attribute_id  | filter_id |
+---------------+-----------+
| 10            |    20     |
| 11            |    21     |
| 12            |    22     |
+---------------+-----------+


Table FILTER:
+-----------+-----------------+
| filter_id | name            |
+-----------+-----------------+
|    19     | First           |
|    20     | Second          |
|    21     | Third           |
+-----------+-----------------+

需要检查ATTRIBUTE_TEXT.value是否有相应的FILTER.id，并且应该给出具有此值的ANIMAL（其他字段无关紧要）。到目前为止我得到了这个：

select * 
from FILTER as f join INFORMATION as i ON (f.filter_id = i.filter_id) 
                 join ATTRIBUTE_TEXT as at ON (i.attribute_id = at.attribute_id) 
                 join ANIMAL_ATTRIBUTES as aa ON (at.attribute_id = aa.attribute_id)
                 join ANIMALS as a ON (aa.animal_id = a.id) 
where (f.filter_id = 20 and at.value like '%black%');

应该给我'tIGER'作为animal.name。

问题是我有更多的Filter.id来检查相应的ATTRIBUT_TEXT.value： e.g。

Filter 1:
Filter.id = 20 and ATTRIBUTE_TEXT.value = 'black'
and
Filter 2:
Filter.id = 21 and ATTRIBUTE_TEXT.value = 'big'

只有在两者都正确的情况下才会返回“CAT”

Answer 1

您需要使用OR并使用适当的parantheses：

select * 
from FILTER as f join INFORMATION as i ON (f.filter_id = i.filter_id) 
                 join ATTRIBUTE_TEXT as at ON (i.attribute_id = at.attribute_id) 
                 join ANIMAL_ATTRIBUTES as aa ON (at.attribute_id = aa.attribute_id)
                 join ANIMALS as a ON (aa.animal_id = a.id) 
where 
(f.filter_id = 20 and at.value like '%black%')
OR
(f.filter_id = 21 and at.value like '%big%')

Answer 2

如果你想返回'CAT'并且它必须是'大'和'黑'两者，你应该添加额外的动物属性（INSERT INTO ANIMAL_ATTRIBUTES SELECT 10,3）并使用如下查询：

select aa.animal_id
from FILTER as f join INFORMATION as i ON (f.filter_id = i.filter_id) 
                 join ATTRIBUTE_TEXT as at ON (i.attribute_id = at.attribute_id) 
                 join ANIMAL_ATTRIBUTES as aa ON (at.attribute_id = aa.attribute_id)
                 join ANIMALS as a ON (aa.animal_id = a.id)
where 
(f.filter_id = 20 and at.value like '%black%')
OR
(f.filter_id = 21 and at.value like '%big%')
GROUP BY aa.animal_id
HAVING count(aa.attribute_id)=2

sql制定正确的查询

2 个答案: