我有这样的查询：
select display_order , section_name , solution_section_id from solution_sections order by display_order
它是非常基础的，并提供了特定讨论的各个部分。有用。
我想做的是还要在每个部分中显示评论数。因此，我想对评论表进行联接，并对有多少评论进行计数。
这是其他表的架构：
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 | |
+---------------------+---------------+------+-----+---------+----------------+
因此，它必须是solution_section_id和solution_part（尽管它们的名称有些不一致，但它们是外键）的联接，其中problem_id =一些id。
但是，我如何获得对suggested_solution_comments表中返回的评论数的计数？
谢谢！
最佳答案
外部联接已更新：
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)
group by s.display_order, s.section_name, s.solution_section_id
order by display_order