Question

我有两张这样的桌子。

create table teams (
  "ID" Integer NOT NULL ,
  "STADIUM_ID" Integer NOT NULL ,
  "NAME" Varchar2 (50) NOT NULL ,
    primary key ("ID") 
    ) ;

create table matches (
  "ID" Integer NOT NULL ,
  "WINNER_ID" Integer NOT NULL ,
  "OPPONENT_ID" Integer NOT NULL ,
  "WINNERSCORE" Integer,
  "OPPONENTSCORE" Integer,
primary key ("ID","WINNER_ID","OPPONENT_ID") 
) ;

他们有以下数据：

select * from matches;

    ID      WINNER_ID OPPONENT_ID WINNERSCORE OPPONENTSCORE
---------- ---------- ----------- ----------- -------------
     1          5           2           5             2
     2          4           5           1             0
     3          3           2           1             0
     4          3           2           1             0
     5          1           2           2             0
     6          3           1           2             1

select * from teams;

    ID     STADIUM_ID    NAME
---------- ---------- -----------
     1          1        Team1
     2          3        Team2
     3          4        Team3
     4          2        Team4
     5          5        Team5

我需要得到每个团队的目标总和。为此目的，尝试了以下查询并得到以下结果：

select name,
       (select sum(opponentscore)
          from matches
         where opponent_id = teams.id) +
       (select sum(winnerscore) from matches where winner_id = teams.id) sum
  from teams;

NAME                                                      SUM
-------------------------------------------------- ----------
Team1                                                       3
Team2
Team3
Team4
Team5                                                       5

你有什么建议吗？

Answer 1

您需要的是分别计算每个团队的 opponentscore 和 winnerscore ，并将它们与 UNION ALL 结合起来：

select name, sum(score) total_score 
  from
(
 select name, sum(winnerscore) score
   from teams t join matches m on ( t.id = m.winner_id  )
  group by name
 union all
 select name, sum(opponentscore) score
   from teams t join matches m on ( t.id = m.opponent_id  )
  group by name
)
group by name
order by 1;

SQL Fiddle Demo

Answer 2

你应该使用join和group by

  select name, sum(matches.opponentscore) + sum(matches.winnerscore) my_sum
  from matches 
  inner join teams on teams.id = matches.winner_id 
  group by  teams.name

Answer 3

您可以使用表格匹配两次加入表团队：

SELECT name, SUM(wonMatches.WINNERSCORE + lostMatches.OPPONENTSCORE) as goals FROM (teams INNER JOIN matches as wonMatches ON teams.ID = wonMatches.WINNER_ID) INNER JOIN matches as lostMatches ON teams.ID = lostMatches.OPPONENT_ID GROUP BY name

Answer 4

我的解决方案是：更改数据库架构。重新考虑您的应用程序的要求。此架构无法满足用户期望的价值。

从我看到的情况来看，我会说你正在尝试为想要跟踪他们的球队/最喜欢的球员进度的球迷建立一个应用程序，以便他们可以吹嘘。

话虽这么说，最后我会得到那些表格：

风扇
团队（id_team）
播放器（id_player，id_team）
锦标赛（id_tournament）
匹配（id_match，id_tournament，start_on，id_team_home，id_team_visitor）
目标（id_match，id_player，goaled_on）

所以现在，我相信编写查询会更加简单。你只需要加入球队，球员，计算目标和按球队分组。

Answer 5

问题在于NULL s - 当找不到结果时子查询返回NULL，NULL + anything == NULL。

最直接的解决方法是：

select name,
       nvl(
           (select sum(opponentscore) from matches where opponent_id = teams.id),
           0
       )
       +
       nvl(
           (select sum(winnerscore) from matches where winner_id = teams.id),
           0
       ) sum
from teams;

出于性能原因，您可能需要考虑按照其他人的建议使用GROUP BY的联接查询。

获得2列的总和，每列具有独特的条件

5 个答案: