Question

下面我可以得到sub_shares的结果，但我不知道如何通过使用sub_shares的结果得到结果总和（sub_shares）：

怎么办？

SELECT * , 
Client.client_chi_name, 
Client.client_eng_name, 
SUM( shares_no ) AS sub_shares
FROM Shareholder
LEFT OUTER JOIN Client ON Shareholder.client_id = Client.client_id
WHERE Shareholder.com_no = 2040628
GROUP BY Shareholder.client_id
ORDER BY SUM( shares_no ) DESC,
Shareholder.date_of_register DESC

表股东

com_no
date_of_register
share_type
class_shares
client_id
shares_no
transferee_id
currency
shares_amount

非常感谢你的帮助和帮助。支持。

Answer 1

SELECT * , 
  Client.client_chi_name, 
  Client.client_eng_name, 
  SUM( shares_no ) AS sub_shares,
  (select sum(shares_no) FROM Shareholder
      LEFT OUTER JOIN Client ON Shareholder.client_id = Client.client_id
      WHERE Shareholder.com_no = 2040628) as sum_of_sum
FROM Shareholder
LEFT OUTER JOIN Client ON Shareholder.client_id = Client.client_id
WHERE Shareholder.com_no = 2040628
GROUP BY Shareholder.client_id
ORDER BY SUM( shares_no ) DESC,
Shareholder.date_of_register DESC

您可以使用subselect

再添加一列

Answer 2

您可以根据需要使用以下代码，

GROUP BY Shareholder.client_id WITH ROLLUP

或

GROUP BY Shareholder.client_id WITH CUBE

有关详细信息，请参阅technet

Answer 3

您可以使用CTE（Commom Table Expression）来完成相同的任务

with cte as
(
SELECT * , 
Client.client_chi_name, 
Client.client_eng_name, 
SUM( shares_no ) AS sub_shares
FROM Shareholder
LEFT OUTER JOIN Client ON Shareholder.client_id = Client.client_id
WHERE Shareholder.com_no = 2040628
GROUP BY Shareholder.client_id
ORDER BY shares_no  DESC,
Shareholder.date_of_register DESC
)

select cte.client_chi_name,cte.client_eng_name,SUM(cte.sub_shares) sub_shares from cte   --you can access more column which is present in * 
GROUP BY client_chi_name,client_eng_name --place extra column yor are accessing

Answer 4

您可以删除GROUP BY并执行以下操作：

SUM( shares_no ) OVER (PARTITION BY Client.client_id) AS sub_shares
SUM( shares_no ) AS total_shares,

您还需要从SELECT DISTINCT开始，否则您将获得重复项。

sql：如何获得总和

4 个答案: