Oracle 11g:如何将oracle表与自身联合？

Question

我有一个表格，如下所示：

date          code                name    score         set
09/09/12     967873         Team A         24            1
09/09/12     967873         Team B         22            1
09/09/12     967873         Team A         21            2 
09/09/12     967873         Team B         16            2
02/04/12     965454         Team X         21            1
02/04/12     965454         Team Y         19            1
02/04/12     965454         Team X         21            2
02/04/12     965454         Team Y         19            2

你猜对了！这是一场排球赛！但是，我希望我的输出在一行中。例如：

date           code               Teams                 Set-1    Set-2     Set-3
09/09/12      967873             Team A VS.Team B       24-22    21-16       -
and so on.... 

**Notice that the game could have a third set as well

我需要某种自我联接，以完善上述格式，以便更容易为用户view...Let我知道，如果你需要更多细节。

谢谢,

Answer 1

查询可能如下所示：

with matches as (
   select "DATE", code, name,
      max(case when "SET" = 1 then score end) score_1,
      max(case when "SET" = 2 then score end) score_2,
      max(case when "SET" = 3 then score end) score_3,
      row_number() over(partition by "DATE", code order by name) team_no
    from games
    group by "DATE", code, name
)
select a."DATE", a.code, a.name || ' vs. ' || b.name teams,
  a.score_1 || '-' || b.score_1 set_1,
  a.score_2 || '-' || b.score_2 set_2,
  a.score_3 || '-' || b.score_3 set_3
from matches a
join matches b on a."DATE" = b."DATE" and a.code = b.code
where a.team_no = 1 and b.team_no = 2;

date和set是非常不合适的列名。

查询分3个步骤进行：

1. 将汇总记录，以便为每个团队创建单行并进行匹配。在这个过程中，分数被分配给三列set_1、set_2、set_3中的一列。
2. 行号被分配给每一行，每次匹配从1开始。结果是一支球队被分配了1，另一支球队被分配了2(列team_no)。
3. 使用匹配(日期和代码)作为连接条件，将结果表连接到自身，左侧连接到编号为1的团队，右侧连接到编号为2的团队。结果是每场比赛都有一行，其中包含两支球队的名字和得分。

Answer 2

首先，按"date", code, "set"对数据进行分组，以便对团队和分数进行LISTAGG。然后将结果旋转到scores列。下面是它的SQL语句：

WITH grouped AS (
  SELECT
    "date", code, "set",
    LISTAGG(name,  ' VS. ') WITHIN GROUP (ORDER BY name) AS teams,
    LISTAGG(score, '-'    ) WITHIN GROUP (ORDER BY name) AS score
  FROM matches
  GROUP BY
    "date", code, "set"
)
,    pivoted AS (
  SELECT
    "date", code, teams,
    nvl("1", '-') AS set1,
    nvl("2", '-') AS set2,
    nvl("3", '-') AS set3
  FROM grouped
  PIVOT (
    MAX(score) FOR "set" IN (1, 2, 3)
  ) p
)
SELECT * FROM pivoted
;

请也看一下这个查询at SQL Fiddle。

Answer 3

我会这样做sg，如果桌子的名字是排球：

SELECT temp.date, temp.code, 
    temp.team1 || ' vs. ' || temp.team2 AS teams, 
    (SELECT v.score FROM volleyball v WHERE v.code = temp.code AND v.name = team1 AND v.set = 1) || '-' || 
        (SELECT v.score FROM volleyball v WHERE v.code = temp.code AND v.name = team2 AND v.set = 1) AS set1, 
    (SELECT v.score FROM volleyball v WHERE v.code = temp.code AND v.name = team1 AND v.set = 2) || '-' || 
        (SELECT v.score FROM volleyball v WHERE v.code = temp.code AND v.name = team2 AND v.set = 2) AS set2, 
    nvl((SELECT v.score FROM volleyball v WHERE v.code = temp.code AND v.name = team1 AND v.set = 3) || '-' || 
        (SELECT v.score FROM volleyball v WHERE v.code = temp.code AND v.name = team2 AND v.set = 3)
        , '-') AS set3 -- optional, if no results, then it will be a '-' 
FROM
    (SELECT v.date, v.code, 
        min(v.name) AS team1,  max(v.name) AS team2 
    FROM volleyball v 
    GROUP BY v.date, v.code) temp; 

这将产生一行摘要。