Question

如何使用GROUP BY获取行的总和，并在不使用WITH ROLLUP的情况下形成所有行

我得到的错误是'字段列表'中的未知列'x'

实施例

SELECT SUM(orderValue) as x GROUP BY day;

但那只会给我每天的orderValue，但是我想把所有的顺序加起来好像我没有使用GROUP BY。我知道我可以使用ROLLUP，但我需要在SELECT中使用它。

〔实施例：

SELECT SUM(orderValue) as x, SUM(x) as total GROUP BY day;

ID  orderValue  day
--  ----------  ---
1   400         2015-01-01
2   800         2015-01-01
3   300         2015-01-01
4   400         2015-01-01
5   600         2015-01-01
6   500         2015-01-01
7   400         2015-01-02
8   800         2015-01-02
9   300         2015-01-02
10  400         2015-01-02
11  600         2015-01-02
12  500         2015-01-02

x       total
3000    6000
3000    6000

Answer 1

您可以通过使用带子查询的交叉连接来获取同一个表中的总和

SELECT SUM(orderValue) as x, 
t.total 
FROM table
CROSS JOIN (SELECT SUM(orderValue) total FROM table ) t
GROUP BY day;

DEMO

Answer 2

使用子查询

SELECT SUM(orderValue) as x, (SELECT SUM(orderValue) FROM table) as y from table GROUP BY day;

Answer 3

您可以在ROLLUP语句中使用SELECT：

SELECT day, SUM(orderValue) as x 
FROM my_table
GROUP BY day
WITH ROLLUP;

这将返回类似

的内容

day, x
--------
day1,30
day2,50
null,80

Answer 4

drop table mySales;

create table mySales
( id int not null auto_increment primary key,
  orderValue int not null,
  theDate date
);

insert mySales (orderValue,theDate) values (100,'2014-01-02'),(600,'2015-01-02'),(9,'2014-07-01'),(1400,'2014-07-02');
insert mySales (orderValue,theDate) values (87,'2014-11-02'),(999,'2015-11-30'),(18,'2014-07-01'),(800,'2013-07-02');
insert mySales (orderValue,theDate) values (11,'2014-1-20'),(9,'2015-11-04'),(1,'2014-07-08'),(6,'2013-05-02');
insert mySales (orderValue,theDate) values (100,'2014-01-02'),(600,'2015-01-02'),(9,'2014-07-01'),(1400,'2014-07-02');
insert mySales (orderValue,theDate) values (87,'2014-11-02'),(999,'2015-11-30'),(18,'2014-07-01'),(800,'2013-07-02');
insert mySales (orderValue,theDate) values (11,'2014-1-20'),(9,'2015-11-04'),(1,'2014-07-08'),(6,'2013-05-02');

select theDate,
sum(orderValue) as dateTotal,
( select sum(orderValue) from mySales
) as grandTotal
from mySales
group by theDate
order by theDate;

theDate, dateTotal, grandTotal
2013-05-02, 12, 8080
2013-07-02, 1600, 8080
2014-01-02, 200, 8080
2014-01-20, 22, 8080
2014-07-01, 54, 8080
2014-07-02, 2800, 8080
2014-07-08, 2, 8080
2014-11-02, 174, 8080
2015-01-02, 1200, 8080
2015-11-04, 18, 8080
2015-11-30, 1998, 8080


select sum(orderValue) from mySales;
8080

如何使用SUM和GROUP BY来汇总行和所有？

4 个答案: