如何使用联合所有查询优化派生表？

Question

union all 查询如何优化派生表？

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问题

• 我想删除冗余。
• “团队”表被使用了两次

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   SELECT series_id, team, SUM(Win) As Won, SUM(Loss) as Lost
    FROM
    ( SELECT *,radiant_name as team, 
         CASE WHEN radiant_win = 1 THEN 1 ELSE 0 END as Win, 
         CASE WHEN radiant_win = 1 THEN 0 ELSE 1 END as Loss
      FROM matches
      UNION ALL
      SELECT *,dire_name as team, 
         CASE WHEN radiant_win = 0 THEN 1 ELSE 0 END as Win, 
         CASE WHEN radiant_win = 0 THEN 0 ELSE 1 END as Loss
      FROM matches
    ) as temp
    WHERE series_id = 8313
    GROUP By team
    ORDER By Won, Lost DESC

 

Answer 1

对此进行优化的领域包括：

不要选择 *.只选择您需要的字段。换句话说，这是：

SELECT *,radiant_name as team

可以这样：

SELECT series_id,radiant_name as team

接下来，您的派生表没有过滤，因此它会查询整个表。这个：

WHERE series_id = 8313

应该进入您的子查询 - 两次。

最后，如果您只对一个 series_id 感兴趣，那么选择它是没有意义的。您的查询可能类似于：

SELECT team, SUM(Win) As Won, SUM(Loss) as Lost
FROM
( SELECT radiant_name as team, 
 CASE WHEN radiant_win = 1 THEN 1 ELSE 0 END as Win, 
 CASE WHEN radiant_win = 1 THEN 0 ELSE 1 END as Loss
FROM matches
WHERE series_id = 8313

UNION ALL

 SELECT dire_name as team, 
 CASE WHEN radiant_win = 0 THEN 1 ELSE 0 END as Win, 
 CASE WHEN radiant_win = 0 THEN 0 ELSE 1 END as Loss
 FROM matches
WHERE series_id = 8313

) as temp
GROUP By team
ORDER By Won, Lost DESC

Answer 2

我认为，这些案例陈述是您查询中丑陋的部分。我会通过减去总计数的获胜次数来计算输赢，反之亦然：

select series_id, team, sum(won), sum(lost)
  from
    (
      select series_id, radiant_name as team, sum(radiant_win) as won, count(*)-sum(radiant_win) as lost
        from matches
        group by series_id, radiant_name
      union all
      select series_id, dire_name as team, count(*)-sum(radiant_win) as won, sum(radiant_win) as lost
        from matches
        group by series_id, dire_name
    )
where series_id = 8313
group by series_id, team
order by won, lost desc