我需要加入 4 个表:位置、动物、人、组织和地址。
输入组 ID，即 70
要求:
- 根据收到的日期获取动物的最新位置。
- 从动物表中获取动物名称。
如果该动物有 person_id，则在位置表中，从 Person 表中获取该人的名字和姓氏以及地址信息。
如果位置中有组织 ID，则获取组织名称和地址 来自组织表和地址表。
对于给定的组织 ID 70，如何将这 5 个表连接到一个查询中:
带有数据的示例表结构:
表格位置
id | group_id | person_id | organization_id | fees1 |fees2 |fees3| received_date |animal_id |
23| 70 | 12 | 0 | 10 |10 |0 | 2017-11-11 | 1 |
24| 70 | 1 | 0 | 10 |10 |0 | 2017-10-11 | 1 |
餐 table 动物
id| animal_name |group_id|
1 | demo | 22 | 70
餐 table 上的人
id | first_name | last_name |
1 | Sam | Dam |
表格组织
id | org_name |
77 | test_org |
表地址
id | organization_id | person_id | address1 | country|
45 | 0 | 1 | test address| USA |
group_id 70 的预期输出
输出:
location.id | location.group_id | location.person_id | location.organization_id | fees1 |fees2 |fees3| received_date | animal_id | animal_name | first_name |address1 |
23 | 70 | 1 | 70 | 20 |20 |20 | 2017-11-11 | 1 | demo | Sam | test address |
最佳答案
您可以尝试以下查询
select l.id,l.group_id,l.person_id,l.organization_id,l.fees1,l.fees2,l.fees3,max(l.received_date) as received_date,l.animal_id,p.first_name,aa.address1 from Organization as o inner join location as l on o.id=l.organization_id left join Person as p on l.person_id=p.id inner join Animal as a on l.animal_id=a.id inner join Address as aa on o.id=aa.organization_id where o.id=70 limit 1
或
select l.id,l.group_id,l.person_id,l.organization_id,l.fees1,l.fees2,l.fees3,l.received_date,l.animal_id,p.first_name,aa.address1 from Organization as o inner join location as l on o.id=l.organization_id left join Person as p on l.person_id=p.id inner join Animal as a on l.animal_id=a.id inner join Address as aa on o.id=aa.organization_id where o.id=70 order by l.received_date desc limit 1
关于Mysql 连接 5 个表或存储过程/函数，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/49372826/