使用以下数据:
我如何根据事件中的两列(犯罪\包含,相似\包含)得到每个事件的成本总和 listofincidents
table?
另外,我如何获得过去3个月(1月、2月和3月)的金额?
create table crimeincidents (
id int not null,
name varchar(20),
primary key (id)
);
create table listofincidents (
id int not null,
incidentdate datetime not null,
crime_incidentid int not null,
similar_incidentid int not null,
cost_to_city decimal(8,2),
primary key (id),
FOREIGN KEY (crime_incidentid) REFERENCES crimeincidents(id),
FOREIGN KEY (similar_incidentid) REFERENCES crimeincidents(id)
);
insert into crimeincidents (id,name) values
(1,'Burglary'),
(2,'Theft'),
(3,'Grand theft auto');
insert into listofincidents (id, incidentdate, crime_incidentid,
similar_incidentid, cost_to_city)
values
(1, "2018-01-10 18:48:00", 1, 2, 900),
(2, "2018-02-15 14:48:00", 2, 3, 800),
(3, "2018-02-20 18:10:00", 3, 1, 1500.10),
(4, "2018-03-20 18:48:00", 1, 3, 800.23),
(5, "2018-03-25 18:24:00", 1, 3, 200.00),
(6, "2018-04-15 10:12:00", 1, 2, 400.00);
生成不带每月日期的结果的查询是:
select c.id, c.name, sm.similarIncidentCost, cr.crimeIncidentCost
from crimeincidents c
inner join (
select c.id, sum(s.cost_to_city) similarIncidentCost
from crimeincidents c inner join listofincidents s
on s.similar_incidentid = c.id
group by c.id
) sm on sm.id = c.id
inner join (
select c.id, sum(cr.cost_to_city) crimeIncidentCost
from crimeincidents c inner join listofincidents cr
on cr.crime_incidentid = c.id
group by c.id
) cr on cr.id = c.id;
我想使用过去3个月的数据生成成本。最终结果如下:
1. January | 1500.1 | 1900.23
2. February | 900 | 800
3. March | 1800.23 | 1500.1
1条答案
按热度按时间hwazgwia1#
我想这就是你想要的:
你可以选择:
以更好地满足您的要求。
一开始没注意到你想把“事件”和“类似事件”分开。虽然我觉得很奇怪(因为类似的事件本身也可能有类似的事件),但我还是做了一个问题: