这是表格
mysql> desc tags;
+------------+--------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+------------+--------------+------+-----+---------+----------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| content_id | int(11) | NO | | NULL | |
| name | varchar(100) | NO | | NULL | |
+------------+--------------+------+-----+---------+----------------+
3 rows in set (0.00 sec)
一些数据
mysql> select * from tags;
+----+------------+---------------+
| id | content_id | name |
+----+------------+---------------+
| 1 | 36 | banana |
| 2 | 36 | strawberry |
| 3 | 36 | orange |
| 4 | 36 | apple |
| 5 | 36 | watermelon |
| 6 | 37 | qiwi |
| 7 | 37 | apple |
| 8 | 37 | orange |
| 9 | 37 | bed |
| 10 | 38 | grape |
| 11 | 38 | apple |
+----+------------+---------------+
11 rows in set (0.00 sec)
我想要的是获取一个唯一的content_id
列表,其中name
是苹果,而content_id也与ORANGE绑在一起。
所以在这个例子中,如果我想知道哪个content_ids被标记为“apple”和“orange”,则会导致:
+------------+
| content_id |
+------------+
| 36 |
| 37 |
+------------+
因为这些是仅有两个相应标记的content_id。最好的方法是什么?
答案 0 :(得分:1)
您可以left join
content_id
上的表格,其中的名称也是如此:
select tbl.content_id
from tags tbl
left join tags tbl2 on tbl2.content_id = tbl.content_id and tbl2.name = 'orange'
where tbl.name = 'apple' and tbl2.name is not null
group by tbl.content_id
仅返回tbl2.name is not null
子句指示的对应连接的行。
答案 1 :(得分:0)
select content_id
from tags
where `name` in ('banana', 'orange')
group by content_id
having count(distinct `name`) >= 2
答案 2 :(得分:0)
select content_id from fruits
where name in('apple','orange')
group by content_id
having COUNT(*)=2