【发布时间】:2021-11-26 11:26:13
【问题描述】:
请看一下这 3 个表格:
Pets
+----+---------+-------+
| id | petname | owner |
+====+=========+=======+
| 1 | chew | 1 |
+----+---------+-------+
| 2 | yo | 2 |
+----+---------+-------+
| 3 | mah | 3 |
+----+---------+-------+
Owners
+----+-------+-----------+
| id | store | ownername |
+====+=======+===========+
| 1 | 1 | Jonh |
+----+-------+-----------+
| 2 | 2 | Joe |
+----+-------+-----------+
| 3 | 3 | Smith |
+----+-------+-----------+
Stores
+----+------------+
| id | storename |
+====+============+
| 1 | Lite Store |
+----+------------+
| 2 | Mega |
+----+------------+
| 3 | Corner |
+----+------------+
这样有可能得到这个结果吗?
+------------+------------+
| storename | Total Pets |
+============+============+
| Lite Store | 5 |
+------------+------------+
| Mega | 8 |
+------------+------------+
| Corner | 0 |
+------------+------------+
我花了几个小时尝试了很多子查询和连接,但我错过了一些东西,也许是 Union?
在下面我接近了,但仍然很远
SELECT storename, COUNT(distinct stores.storename) as store, count(DISTINCT pets.owner) as petowner from stores inner join owners on owners.id = stores.id inner JOIN pets on pets.owner = stores.id group by stores.id
SELECT stores.storename, COUNT(distinct stores.storename) as store, count(DISTINCT pets.owner) as petowner from stores inner join owners on owners.id = stores.id inner JOIN pets on pets.owner = stores.id group by stores.id
SELECT stores.storename, COUNT(distinct owners.id) as store, count(DISTINCT pets.owner) as petowner from stores inner join owners on owners.id = stores.id inner JOIN pets on pets.owner = stores.id group by stores.id
SELECT COUNT(*),(SELECT COUNT(*) from stores) FROM pets
SELECT COUNT(*),(SELECT DISTINCT(COUNT(*)) from stores),(SELECT DISTINCT(COUNT(*)) FROM owners) FROM pets
SELECT DISTINCT(COUNT(*)), ( select count(DISTINCT(stores.storename)) from stores join owners on stores.id = stores.storename ) FROM pets
select stores.storename, (select count(*) from pets) from stores join owners on stores.id = stores.storename group by storename
select DISTINCT(stores.storename), (select count(*) from pets) from stores join owners on stores.id = stores.storename group by storename
select (count(stores.storename)), (select count(*) from pets) as total from stores join owners on stores.id = stores.storename group by storename
有没有办法得到上面的结果?
任何帮助将不胜感激!
【问题讨论】:
-
在所需的输出中有“total 5 pets”和“total 8 pets”的行,但在 pets 表中只有 3 只宠物。请解释这个 ot 更新源数据集更合适
-
请详细说明如何获得预期结果