MySQL 调用返回错误组合的结果，或者如果涉及的表为空，则返回全空行

Question

所以我的数据库中有这些表，其中包含以下数据:

building_officer_membership
+--------------------------------+-------------+------------+
| building_officer_membership_id | building_id | officer_id |
+--------------------------------+-------------+------------+
|                              1 |           1 |          1 |
|                              2 |           1 |          2 |
+--------------------------------+-------------+------------+

building
+-------------+-----------------+
| building_id | name            |
+-------------+-----------------+
|           1 | a_nice_building |
+-------------+-----------------+

officer
+------------+------------+-----------+
| officer_id | first_name | last_name |
+------------+------------+-----------+
|          1 | Brandon    | Thompson  |
|          2 | Mark       | Bobby     |
+------------+------------+-----------+

Manager
+------------+---------------+
| manager_id | full_name     |
+------------+---------------+
|          1 | Bill Lumbergh |
|          2 | Bob Page      |
+------------+---------------+

 officer manager membership
+-------------------------------+------------+------------+
| officer_manager_membership_id | officer_id | manager_id |
+-------------------------------+------------+------------+
|                             1 |          1 |          1 |
|                             2 |          2 |          2 |
+-------------------------------+------------+------------+

我有一个 MySQL 调用，它返回与建筑物关联的每个官员(在building_officer_membership 表中)。为了使其信息更丰富，我还在每个结果中列出了这些官员所属的经理。

当我进行 SQL 调用时，

SELECT building_officer_membership.building_officer_membership_id,
building_officer_membership.building_id,
building_officer_membership.officer_id,
GROUP_CONCAT(DISTINCT manager.full_name) as manager_name,
GROUP_CONCAT(DISTINCT manager.manager_id) AS manager_id
from building_officer_membership
JOIN building on building.building_id = building_officer_membership.building_id
JOIN officer on officer.officer_id = building_officer_membership.officer_id
JOIN officer_manager_membership on officer.officer_id = officer_manager_membership.officer_id
LEFT JOIN manager ON officer_manager_membership.manager_id = manager.manager_id

我期望有这样的两个结果:

+--------------------------------+-------------+------------+------------------------+------------+
| building_officer_membership_id | building_id | officer_id | manager_name           | manager_id |
+--------------------------------+-------------+------------+------------------------+------------+
|                              1 |           1 |          1 | Bill Lumbergh          | 1          |
|                              2 |           1 |          2 | Bob Page               | 2          |
+--------------------------------+-------------+------------+------------------------+------------+

相反，我得到了这个:

+--------------------------------+-------------+------------+------------------------+------------+
| building_officer_membership_id | building_id | officer_id | manager_name           | manager_id |
+--------------------------------+-------------+------------+------------------------+------------+
|                              1 |           1 |          1 | Bill Lumbergh,Bob Page | 1,2        |
+--------------------------------+-------------+------------+------------------------+------------+

这是将军官的所有经理组合成一个结果。

如果表为空，并且我进行 SQL 调用，我会得到以下内容，而不是空行:

+--------------------------------+-------------+------------+--------------+------------+
| building_officer_membership_id | building_id | officer_id | manager_name | manager_id |
+--------------------------------+-------------+------------+--------------+------------+
|                           NULL |        NULL |       NULL | NULL         | NULL       |
+--------------------------------+-------------+------------+--------------+------------+

我知道该错误位于 GROUP_CONCAT 部分，因为如果我删除这些部分，

select building_officer_membership.building_officer_membership_id,
    building_officer_membership.building_id,
    building_officer_membership.officer_id
    from building_officer_membership
    JOIN building on building.building_id = building_officer_membership.building_id
    JOIN officer on officer.officer_id = building_officer_membership.officer_id;
+--------------------------------+-------------+------------+
| building_officer_membership_id | building_id | officer_id |
+--------------------------------+-------------+------------+
|                              1 |           1 |          1 |
|                              2 |           1 |          2 |
+--------------------------------+-------------+------------+

我得到了我期望的结果，只是没有经理信息。

所以我不太确定这里发生了什么。我的连接是否太贪婪了还是什么？

我的MySQL版本是5.1.73。由于我无法控制的原因，我无法升级到最新版本。虽然在使用此 SQL 调用的 Web 服务代码中，我可以采取一种解决方法来解决此问题，这会涉及更多 SQL 调用，但如果可能的话，我宁愿一次性执行此操作。

Answer 1

安恩我明白了。真是太蠢了。

选择building_officer_membership.building_officer_membership_id、building_officer_membership.building_id、building_officer_membership.officer_id、GROUP_CONCAT(DISTINCT manager.full_name)作为manager_name、GROUP_CONCAT(DISTINCT manager.manager_id) AS manager_id从building_officer_membership加入building.building_id =building_officer_membership.building_id加入官员Officer.officer_id =building_officer_membership.officer_id 加入Officer_manager_membership 上Officer.officer_id =Officer_manager_membership.officer_id 加入经理ONOfficer_manager_membership.manager_id =Officer.Officer_id

按officer.officer_id分组

TIL group_concat 需要一个 group by 来知道如何组织结果，否则，它将返回尽可能少的结果。