Oracle SQL:基于列值的累积总计
我有两个数据表:Oracle SQL:基于列值的累积总计,sql,oracle,Sql,Oracle,我有两个数据表: 载有体育比赛中得分日期和时间的表格;及 表B列出了比赛日期以及主客场的球队编号 表 Date Time Team Points_scored ----------------------------------------- 20130818 1400 1 2 20130818 1402 1 3 20130818 1407 2 2 20130818 1410 2
- 载有体育比赛中得分日期和时间的表格;及
- 表B列出了比赛日期以及主客场的球队编号
Date Time Team Points_scored
-----------------------------------------
20130818 1400 1 2
20130818 1402 1 3
20130818 1407 2 2
20130818 1410 2 3
20130818 1412 1 2
20130822 1550 4 2
20130822 1552 5 3
20130822 1553 5 2
20130822 1555 5 3
20130822 1559 4 2
表B
Date Home Team Away Team
-----------------------------------------------------
20130818 2 1
20130822 4 5
我想要的是一个查询,提供主客场球队每天的跑步总数,如下所示:
Date Time Home_score Away_score
20130818 1400 0 2
20130818 1402 0 5
20130818 1407 2 5
20130818 1410 5 5
20130818 1412 5 6
20130822 1550 2 0
20130822 1552 2 3
20130822 1553 2 5
20130822 1555 2 8
20130822 1559 4 8
但我甚至不知道从哪里开始。有人有什么想法吗?我正在使用Oracle 11g
非常感谢
以下是创建脚本:
create table tablea (
match_date number,
time number,
team number,
points_scored number);
create table tableb (
match_date number,
home_team number,
away_team number);
insert into tablea values (20130818,1400,1,2);
insert into tablea values (20130818,1402,1,3);
insert into tablea values (20130818,1407,2,2);
insert into tablea values (20130818,1410,2,3);
insert into tablea values (20130818,1412,1,2);
insert into tablea values (20130822,1550,4,2);
insert into tablea values (20130822,1552,5,3);
insert into tablea values (20130822,1553,5,2);
insert into tablea values (20130822,1555,5,3);
insert into tablea values (20130822,1559,4,2);
insert into tableb values (20130818,2,1);
insert into tableb values (20130822,4,5);
commit;
这其中最难的部分不是累积和分析函数。它是使表a和表b之间的连接正确
select b.match_date, a.time,
(case when a.team = b.home_team then a.points_scored else 0 end) as home_points,
(case when a.team = b.away_team then a.points_scored else 0 end) as away_points,
sum(case when a.team = b.home_team then a.points_scored else 0 end) over (partition by a.match_date order by a.time) as cum_home_points,
sum(case when a.team = b.away_team then a.points_scored else 0 end) over (partition by a.match_date order by a.time) as cum_away_points
from TableB b join
TableA a
on a.team in (b.home_team, b.away_team) and b.match_date = a.match_date;
是SQL的把戏
顺便说一下,根据您的数据,
20130818
的最后一个值应该是7
,而不是6
(得2分)。这其中最难的部分不是累积和分析函数。它是使表a和表b之间的连接正确
select b.match_date, a.time,
(case when a.team = b.home_team then a.points_scored else 0 end) as home_points,
(case when a.team = b.away_team then a.points_scored else 0 end) as away_points,
sum(case when a.team = b.home_team then a.points_scored else 0 end) over (partition by a.match_date order by a.time) as cum_home_points,
sum(case when a.team = b.away_team then a.points_scored else 0 end) over (partition by a.match_date order by a.time) as cum_away_points
from TableB b join
TableA a
on a.team in (b.home_team, b.away_team) and b.match_date = a.match_date;
是SQL的把戏
顺便说一下,根据您的数据,
20130818
的最后一个值应该是7
,而不是6
(得2分)。很好的解决方案+回答得很好,但要给达莫留个条子。考虑一下一天可能会有一个以上的游戏。由于您将比赛日期存储为一个数字,请在末尾再添加一个数字,以防万一。很好,戈登,谢谢!工作完全符合要求。抱歉,所需的输出数据不正确。罗伯特,谢谢你提醒我一天中的两场比赛。解决得好,戈登+回答得很好,但要给达莫留个条子。考虑一下一天可能会有一个以上的游戏。由于您将比赛日期存储为一个数字,请在末尾再添加一个数字,以防万一。很好,戈登,谢谢!工作完全符合要求。抱歉,所需的输出数据不正确。罗伯特,谢谢你提醒我一天中的两场比赛。