我有这个架构:
mysql> describe suggested_solution_comments;
+-----------------------+----------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+-----------------------+----------------+------+-----+---------+----------------+
| comment_id | int(10) | NO | PRI | NULL | auto_increment |
| problem_id | int(10) | NO | | NULL | |
| suggested_solution_id | int(10) | NO | | NULL | |
| commenter_id | int(10) | NO | | NULL | |
| comment | varchar(10000) | YES | | NULL | |
| solution_part | int(3) | NO | | NULL | |
| date | date | NO | | NULL | |
| guid | varchar(50) | YES | UNI | NULL | |
+-----------------------+----------------+------+-----+---------+----------------+
8 rows in set (0.00 sec)
mysql> describe solution_sections;
+---------------------+---------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+---------------------+---------------+------+-----+---------+----------------+
| solution_section_id | int(10) | NO | PRI | NULL | auto_increment |
| display_order | int(10) | NO | | NULL | |
| section_name | varchar(1000) | YES | | NULL | |
+---------------------+---------------+------+-----+---------+----------------+
我的疑问是:
select s.display_order,
s.section_name,
s.solution_section_id ,
count(c.comment_id) AS comment_count
FROM solution_sections s left outer join suggested_solution_comments c
ON (c.solution_part = s.solution_section_id)
where problem_id = 400
group by s.display_order, s.section_name, s.solution_section_id
order by display_order;
它只返回有计数的行> 0但如果计数为0则不返回那些行。
知道如何让它返回所有行吗? :)
谢谢!
答案 0 :(得分:10)
这是因为where problem_id = 400
删除了没有相应suggested_solution_comments
行的行。将条件从where
过滤器移动到on
子句应该可以解决问题:
select s.display_order, s.section_name, s.solution_section_id ,count(c.comment_id)
AS comment_count
from solution_sections s
left outer join suggested_solution_comments c
ON (c.solution_part = s.solution_section_id) AND problem_id = 400
group by s.display_order, s.section_name, s.solution_section_id
order by display_order;