在MySQL中结合多个查询结果(按列)

Question

我有4个不同的查询,每个查询返回单独的结果集.我需要将查询结果与使用单个查询相结合.

我的示例查询是:

1. select cls.* from (calls as cls inner join calls_users as clsusr on cls.id=clsusr.call_id) inner join users as usr on usr.id=cls.assigned_user_id where cls.assigned_user_id='seed_max_id'

2. select mtn.* from (meetings as mtn inner join meetings_users as mtnusr on mtn.id=mtnusr.meeting_id) inner join users as usr on usr.id=mtn.assigned_user_id where mtn.assigned_user_id='seed_max_id'

3. select tsk.* from tasks as tsk inner join users as usr on usr.id=tsk.assigned_user_id where tsk.assigned_user_id='seed_max_id'

4. select nts.* from (notes as nts inner join accounts as acnts on acnts.id=nts.parent_id) inner join users as usr on usr.id=acnts.assigned_user_id where acnts.assigned_user_id='seed_max_id'

我尝试了以下方式,但它没有用

Combine: SELECT tbl1.*, tbl2.* 
from (select cls.* from (calls as cls inner join calls_users as clsusr on cls.id=clsusr.call_id) inner join users as usr on usr.id=cls.assigned_user_id where cls.assigned_user_id='seed_max_id') as tbl1 
left  outer join
(select mtn.* from (meetings as mtn inner join meetings_users as mtnusr on mtn.id=mtnusr.meeting_id) inner join users as usr on usr.id=mtn.assigned_user_id where mtn.assigned_user_id='seed_max_id') as tbl2
using(assigned_user_id)

我也试过右外连接和其他内连接我真的卡住了,如果有人知道解决方案那么请帮忙.我需要类似的结果,如何在MySQL中连接两个具有不同行数的表？.

数据样本:

从查询1:

+-------------------------------------------+------------------+-
| Call Name                                 | Call Description |
+-------------------------------------------+------------------+-
| Discuss Review Process                    | NULL             |
| Get More information on the proposed deal | NULL             |
| Left a message                            | NULL             |
| Discuss Review Process                    | NULL             |
+-------------------------------------------+------------------+

从查询2:

+-----------------------+-----------------------------------------------------------
| Meeting Name          | Meeting Description
+-----------------------+-----------------------------------------------------------
| Review needs          | Meeting to discuss project plan and hash out the details o
| Initial discussion    | Meeting to discuss project plan and hash out the details o
| Demo                  | Meeting to discuss project plan and hash out the details o
| Discuss pricing       | Meeting to discuss project plan and hash out the details o
| Review needs          | Meeting to discuss project plan and hash out the details o
+-----------------------+-----------------------------------------------------------

我需要组合如下列:

+-------------------------------------------+------------------+-------------------+-------------------+
| Call Name                                 | Call Description |Meeting Name       |Meeting Description|
+-------------------------------------------+------------------+-------------------+-------------------+
| Discuss Review Process                    | NULL             |Review needs       |Meeting to discuss |
| Get More information on the proposed deal | NULL             |Initial discussion |Meeting to discuss |
| Left a message                            | NULL             |Demo               |Meeting to discuss |
| NULL                                   | NULL             |Discuss pricing    |Meeting to discuss |
| NULL                                      | NULL             |Review needs       |Meeting to discuss |
+-------------------------------------------+------------------+-------------------+-------------------+

Answer 1

您可以做的最好的是UNION或UNION ALL,但这需要它们具有相同的类型和列数.例如:

SELECT 'Customer' AS type, id, name FROM customer
UNION ALL
SELECT 'Supplier', id, name FROM supplier
UNION ALL
SELECT 'Employee', id, full_name FROM employee

列名称不必匹配.第一部分的别名将用于其余部分.

我还要补充一点,而不是:

select cls.* from (calls as cls inner join calls_users as clsusr on cls.id=clsusr.call_id) inner join users as usr on usr.id=cls.assigned_user_id where cls.assigned_user_id='seed_max_id'

你应该删除不必要的子查询,只需:

SELECT c.*
FROM calls c
JOIN calls_users cu ONc.id = cu.call_id
WHERE c.assigned_user_id = 'seed_max_id'

不需要额外的复杂性,上面的内容显然更具可读性.