我有3张桌子:
+-------------+ +-------------+ +-------------+
| hotel | | hot_cat | | category |
+------+------+ +------+------+ +-------------+
| id | name | | hid | cid | | id | name |
+------+------+ +------+------+ +-------------+
| 1 | X | | 1 | 1 | | 1 | cat1 |
+------+------+ +------+------+ +-------------+
| 2 | Y | | 1 | 2 | | 2 | cat2 |
+------+------+ +------+------+ +-------------+
| 3 | Z | | 2 | 2 | | 3 | cat3 |
+------+------+ +------+------+ +-------------+
| 2 | 3 | | 4 | cat4 |
+------+------+ +-------------+
| 2 | 4 |
+------+------+
我想选择类别有些价值的酒店,但是所有其他类别都分配给这家酒店。我有这个问题:
SELECT hot.*,GROUP_CONCAT(cat.name SEPARATOR '<br>') AS cats
FROM hotel hot
LEFT JOIN hot_cat hc ON hc.hid = hot.id
LEFT JOIN category cat ON cat.id = hc.cid
WHERE cat.id = 2
GROUP BY hot.id
所以我明白了:
+------+------+-------------------------+
| id | name | cats |
+------+------+-------------------------+
| 1 | X | 'cat2' |
+------+------+-------------------------+
| 2 | Y | 'cat2' |
+------+------+-------------------------+
我想要实现的目标:
+------+------+-------------------------+
| id | name | cats |
+------+------+-------------------------+
| 1 | X | 'cat1<br>cat2' |
+------+------+-------------------------+
| 2 | Y | 'cat2<br>cat3<br>cat4' |
+------+------+-------------------------+
我也希望它在没有where子句的情况下工作,并获得没有分配类别的酒店:
+------+------+-------------------------+
| id | name | cats |
+------+------+-------------------------+
| 1 | X | 'cat1<br>cat2' |
+------+------+-------------------------+
| 2 | Y | 'cat2<br>cat3<br>cat4' |
+------+------+-------------------------+
| 3 | Z | '' |
+------+------+-------------------------+
答案 0 :(得分:0)
在hot_cat
表格中开始查询,然后加入hotel
。它是表格中的一个额外跳跃,但它解决了这个问题。
SELECT hot.*,GROUP_CONCAT(cat.name SEPARATOR '<br>') AS cats
FROM hot_cat hc1
LEFT JOIN hotel hot ON hc1.hid = hot.id
LEFT JOIN hot_cat hc ON hc.hid = hot.id
LEFT JOIN category cat ON cat.id = hc.cid
WHERE hc1.id = 2
GROUP BY hot.id
相反,如果您不需要WHERE,那么您可以在没有WHERE子句的情况下使用现有查询。