Question

以下是我的架构和问题的基本内容：http://sqlfiddle.com/#!1/72ec9/4/0

请注意，句号表可以指一个可变的时间范围 - 可能是整个赛季，可能是几场比赛或一场比赛。对于给定的团队和年份，所有期间行都代表独有的时间范围。

我写了一个查询，它连接表并使用GROUP BY periods.year来汇总一个季节的分数（参见sqlfiddle）。但是，如果一名教练在同一年有两个职位，GROUP BY将计算两次同一时期的排名。当教练担任两个职位但仍总结一年由多个时期组成的时期时，我怎么能抛弃重复数据呢？如果有更好的方法来制作模式，如果你向我指出，我也会很感激。

Answer 1

潜在问题（加入多个匹配的多个表）将在这个密切相关的答案中解释：

Two SQL LEFT JOINS produce incorrect result

要解决此问题，我首先简化您的查询：

select pe.year
     , sum(pe.wins)       AS wins
     , sum(pe.losses)     AS losses
     , sum(pe.ties)       AS ties
     , array_agg(po.id)   AS position_id
     , array_agg(po.name) AS position_names
from   periods_positions_coaches_linking pp
join   positions po ON po.id = pp.position
join   periods   pe ON pe.id = pp.period
where  pp.coach = 1
group  by pe.year
order  by pe.year;

产生与原始相同的错误的结果，但更简单/更快/更容易阅读。

只要您不使用coach列表中的列，就无需加入表SELECT。我将其完全删除，并将WHERE条件替换为where pp.coach = 1。
您不需要COALESCE。汇总函数NULL中会忽略sum()个值。无需替换0。
使用表别名使其更易于阅读。

接下来，我解决你的问题：

SELECT *
FROM  (
  SELECT pe.year
       , array_agg(DISTINCT po.id)   AS position_id
       , array_agg(DISTINCT po.name) AS position_names
  FROM   periods_positions_coaches_linking pp
  JOIN   positions                         po ON po.id = pp.position
  JOIN   periods                           pe ON pe.id = pp.period
  WHERE  pp.coach = 1
  GROUP  BY pe.year
  ) po
LEFT JOIN (
  SELECT pe.year
       , sum(pe.wins)   AS wins
       , sum(pe.losses) AS losses
       , sum(pe.ties)   AS ties
  FROM  (
     SELECT period
     FROM   periods_positions_coaches_linking
     WHERE  coach = 1
     GROUP  BY period
     ) pp
  JOIN   periods pe ON pe.id = pp.period
  GROUP  BY pe.year
  ) pe USING (year)
ORDER  BY year;

在加入之前单独汇总位置和期间。
在第一个子查询列表中，只需使用DISTINCT进行一次排名。
在第二个子查询
中
- GROUP BY period，因为教练每个时期可以有多个职位。
- JOIN到之后的期间数据，然后汇总以获得总和。

SQL Fiddle.

Answer 2

使用distinct显示here

代码：

select periods.year as year,
sum(coalesce(periods.wins, 0)) as wins,
sum(coalesce(periods.losses, 0)) as losses,
sum(coalesce(periods.ties, 0)) as ties,
array_agg( distinct positions.id) as position_id,
array_agg( distinct positions.name) as position_names

from periods_positions_coaches_linking

join coaches on coaches.id = periods_positions_coaches_linking.coach
join positions on positions.id = periods_positions_coaches_linking.position
join periods on periods.id = periods_positions_coaches_linking.period

where coaches.id = 1

group by periods.year, positions.id
order by periods.year;

Answer 3

在您的情况下，最简单的方法可能是分割位置：

select periods.year as year,
       sum(coalesce(periods.wins, 0))/COUNT(distinct positions.id) as wins,
       sum(coalesce(periods.losses, 0))/COUNT(distinct positions.id) as losses,
       sum(coalesce(periods.ties, 0))/COUNT(distinct positions.id) as ties,
       array_agg(distinct positions.id) as position_id,
       array_agg(distinct positions.name) as position_names
from periods_positions_coaches_linking join
     coaches
     on coaches.id = periods_positions_coaches_linking.coach join
     positions
     on positions.id = periods_positions_coaches_linking.position join
     periods
     on periods.id = periods_positions_coaches_linking.period
where coaches.id = 1
group by periods.year
order by periods.year;

位置数量会增加胜利，损失和关系，因此将其除去会调整计数。

保持行不在GROUP BY中重复计算

3 个答案: