3

我有一个如下所示的表：

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！这是一场排球比赛！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

我需要某种自我加入来将上述格式改进为更便于用户查看的格式...如果您需要更多详细信息，请告诉我。

谢谢，

4

4 回答 4

7

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是相当不幸的列名。

查询分 3 个步骤进行：

  1. 这些记录被聚合起来，为每支球队和每场比赛创建一行。在该过程中，分数被分配给三个列 set_1、set_2、set_3 之一。
  2. 行号分配给每一行，从 1 开始为每个匹配。结果是一个团队被分配了 1，而另一个团队被分配了 2（列 team_no）。
  3. 结果表连接到自身，左侧为没有的团队。1 和右侧的团队没有。2 使用匹配（日期和代码）作为连接条件。结果是每场比赛一排，两支球队的名字和得分。
于 2012-11-06T19:21:16.570 回答
2

首先，按团队和分数对数据"date", code, "set"进行分组。LISTAGG然后在分数列上旋转结果。这是它的SQL：

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
;

请在 SQL Fiddle 上查看此查询。

于 2012-11-06T20:22:35.510 回答
1

如果表的名称是 VOLLEYBALL，我会这样做 sg：

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;

这将产生一个单行摘要。

于 2012-11-06T19:26:15.563 回答
1

要在不需要连接的情况下返回所需的结果，请尝试：

select "date", 
       code, 
       min_name || ' VS. ' || max_name teams,
       sum(case when "set" = 1 and name = min_name then score end) || '-' || 
          sum(case when "set" = 1 and name = max_name then score end) "Set-1",
       sum(case when "set" = 2 and name = min_name then score end) || '-' || 
          sum(case when "set" = 2 and name = max_name then score end) "Set-2",
       sum(case when "set" = 3 and name = min_name then score end) || '-' || 
          sum(case when "set" = 3 and name = max_name then score end) "Set-3"
from (select g.*,
             min(name) over (partition by "date", code) min_name,
             max(name) over (partition by "date", code) max_name
      from games)
group by "date", code
于 2012-11-06T20:11:53.577 回答