Question

我有两张这样的表：

人

id | name | sale | commission
1  | abc  | 0    |   0
2  | xyz  | 0    |   0

售：

id | date       | person_id | sale | commission
1  | 2016-05-01 |     1     | 10   |     1
2  | 2016-05-02 |     1     | 10   |     1
3  | 2016-05-03 |     1     | 10   |     1
4  | 2016-05-01 |     2     | 20   |     2
5  | 2016-05-02 |     2     | 20   |     2
6  | 2016-05-01 |     2     | 20   |     2

我想用单个更新查询更新人员表，并将表更改为：

人

id | name | sale | commission
1  | abc  | 30   |   3
2  | xyz  | 60   |   6

我知道我可以像以下一样销售销售，但如何将以下查询结果直接更新到人员表。

SELECT person_id, SUM(sale), SUM(commission) 
FROM sale
GROUP BY person_id;

Answer 1

正如草莓在你提问的评论中所说的那样，在保存这些信息之前，请仔细思考。它被非规范化，变得陈旧。相反，请考虑在报告生成期间使用它。否则，如上所述，你可能会遇到问题。

drop table if exists person;
create table person
(   personId int auto_increment primary key,
    name varchar(100) not null,
    totSales decimal(9,2) not null,
    totComm decimal(9,2)
);
insert person(name,totSales,totComm) values
('Joe',0,0),
('Sally',0,0);
-- just added persons 1 and 2 (auto_inc)

drop table if exists sale;
create table sale
(   saleId int auto_increment primary key,
    saleDate date not null,
    personId int not null,
    sale decimal(9,2) not null,
    commission decimal(9,2) not null,
    index(personId), -- facilitate a snappier "group by" later
    foreign key (personId) references person(personId) -- Ref Integrity
);

insert sale(saleDate,personId,sale,commission) values
('2016-05-01',2,10,1),
('2016-05-01',1,40,4),
('2016-05-02',1,30,3),
('2016-05-07',2,10,1),
('2016-05-07',2,90,9);

-- the following dies on referential integrity, FK, error 1452 as expected
insert sale(saleDate,personId,sale,commission) values ('2016-05-01',4,10,1);

更新声明

update person p 
join  
(   select personId,sum(sale) totSales, sum(commission) totComm 
    from sale 
    group by personId 
) xDerived 
on xDerived.personId=p.personId 
set p.totSales=xDerived.totSales,p.totComm=xDerived.totComm;

结果

select * from person;
+----------+-------+----------+---------+
| personId | name  | totSales | totComm |
+----------+-------+----------+---------+
|        1 | Joe   |    70.00 |    7.00 |
|        2 | Sally |   110.00 |   11.00 |
+----------+-------+----------+---------+
2 rows in set (0.00 sec)

xDerived只是一个别名。无论您是否明确使用别名，所有派生表都需要别名。

Answer 2

 UPDATE person
 SET sale = (
    SELECT SUM(s.sale) FROM sale s
    WHERE s.person_id = person.id
 );

适合我。请参阅：http://ideone.com/F32oUU

使用其他聚合列编辑新版本：

UPDATE person SET 
 sale = (
    SELECT SUM(s.sale) FROM sale s
    WHERE s.person_id = person.id
 ),
 commission = (
    SELECT SUM(s.commission) FROM sale s
    WHERE s.person_id = person.id
 );

http://ideone.com/yo1A9Y

话虽如此，我确信JOIN解决方案更好，并且希望另一位回答者能够发布这样的解决方案。

根据另一个表上的select结果更新表

2 个答案:

更新声明

结果