我有4张桌子：
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 | |
+---------------------+---------------+------+-----+---------+----------------+
3 rows in set (0.00 sec)
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 | |
+-----------------------+----------------+------+-----+---------+----------------+
mysql> describe users;
+--------------+---------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+--------------+---------------+------+-----+---------+----------------+
| user_id | int(10) | NO | PRI | NULL | auto_increment |
| first_name | varchar(100) | NO | | NULL | |
| last_name | varchar(100) | NO | | NULL | |
| email | varchar(150) | NO | | NULL | |
| user_pass | varchar(40) | NO | | NULL | |
| zip | varchar(100) | NO | | NULL | |
| country | varchar(100) | NO | | NULL | |
| city | varchar(100) | NO | | NULL | |
| state | varchar(100) | NO | | NULL | |
| lat | float(9,6) | YES | | NULL | |
| lng | float(9,6) | YES | | NULL | |
| agreed_terms | tinyint(1) | YES | | NULL | |
| join_date | date | NO | | NULL | |
| last_login | date | NO | | NULL | |
| bio_blurb | varchar(5000) | YES | | NULL | |
+--------------+---------------+------+-----+---------+----------------+
15 rows in set (0.03 sec)
mysql> describe member_photo;
+-------------------+---------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+-------------------+---------------+------+-----+---------+----------------+
| photo_id | int(10) | NO | PRI | NULL | auto_increment |
| member_id | int(10) | NO | | NULL | |
| photo_description | varchar(3000) | YES | | NULL | |
| photo_path | varchar(1000) | NO | | NULL | |
| small_thumb | varchar(1000) | YES | | NULL | |
| mid_thumb | varchar(1000) | YES | | NULL | |
| is_main_photo | tinyint(1) | YES | | NULL | |
+-------------------+---------------+------+-----+---------+----------------+
我有这样的查询：
select comment_id,
commenter_id,
section_name,
comment,
solution_part,
display_order,
solution_section_id,
suggested_solution_id,
DAYOFMONTH(date),
DAYNAME(date),
YEAR(date),
MONTH(date),
first_name,
last_name,
email,
small_thumb,
mid_thumb
from solution_sections
left join suggested_solution_comments on
solution_sections.solution_section_id = suggested_solution_comments.solution_part
left join users on
suggested_solution_comments.commenter_id = users.user_id
left join member_photo on
suggested_solution_comments.commenter_id = member_photo.member_id
where suggested_solution_id = 61 OR
suggested_solution_id IS NULL
order by solution_section_id,
comment_id,
section_name,
comment,
solution_part,
display_order;
它应该做的是从solution_sections表中获取每个section_name，然后查找注释（以及有关谁评论的数据）。有时候没有注释，但它至少应该返回带有section_name的行，并且所有其他内容都为null。
但由于某种原因，它没有。最奇怪的是，如果我给它一个不同的suggested_solution_id来匹配，它将返回solution_sections的所有行。
为什么会发生这样的事情？谢谢!!
我刚刚意识到一件事 - 如果对任何problem_id做了另一个评论，那么这个查询将不会返回该部分的行。
答案 0 :(得分：1)
如果您想显示solution_sections，即使其他所有内容都不存在，您也可以使用“左外部加入”：
select comment_id,commenter_id, section_name, comment, solution_part,
display_order, solution_section_id, suggested_solution_id,
DAYOFMONTH(date), DAYNAME(date), YEAR(date), MONTH(date),
first_name, last_name, email, small_thumb,mid_thumb
from solution_sections
left outer join suggested_solution_comments on solution_sections.solution_section_id = suggested_solution_comments.solution_part
and suggested_solution_id = 61
left outer join users on suggested_solution_comments.commenter_id = users.user_id
left outer join member_photo on suggested_solution_comments.commenter_id = member_photo.member_id
where solution_section_id = ????
order by solution_section_id, comment_id, section_name, comment, solution_part,display_order;
PS。尝试为表使用别名，它更具可读性： - ）
答案 1 :(得分：1)
当不保证子记录退出时，您需要左外连接来查看父表中的所有记录。我还避免在使用外连接时添加where子句。我认为将其连接保留在子选择中更具可读性，并过滤结果..尝试这样的事情：
select * from
(
select sc.comment_id,
sc.commenter_id,
ss.section_name,
sc.comment,
sc.solution_part,
ss.display_order,
ss.solution_section_id,
sc.suggested_solution_id,
DAYOFMONTH(sc.date),
DAYNAME(sc.date),
YEAR(sc.date),
MONTH(sc.date),
u.first_name,
u.last_name,
u.email,
mp.small_thumb,
mp.mid_thumb
from solution_sections ss
left outer join suggested_solution_comments sc on ss.solution_section_id = sc.solution_part
left outer join users u on sc.commenter_id = u.user_id
left outer join member_photo mp on sc.commenter_id = mp.member_id) a
where a.suggested_solution_id = 61 OR
a.suggested_solution_id IS NULL
order by a.solution_section_id,
a.comment_id,
a.section_name,
a.comment,
a.solution_part,
a.display_order;
编辑：
select sc.comment_id,
sc.commenter_id,
ss.section_name,
sc.comment,
sc.solution_part,
ss.display_order,
ss.solution_section_id,
sc.suggested_solution_id,
DAYOFMONTH(sc.date),
DAYNAME(sc.date),
YEAR(sc.date),
MONTH(sc.date),
u.first_name,
u.last_name,
u.email,
mp.small_thumb,
mp.mid_thumb
from solution_sections ss
left outer join suggested_solution_comments sc on ss.solution_section_id = sc.solution_part
AND sc.suggested_solution_id = 61
left outer join users u on sc.commenter_id = u.user_id
left outer join member_photo mp on sc.commenter_id = mp.member_id
order by solution_section_id,
comment_id,
section_name,
comment,
solution_part,
display_order;