Sum Union Mysql

Question

我必须找到一个用户出现在两个不同表中的次数...现在我使用了一个联合

我有此查询并且我想要记录<{1}}我的记录

SUM

我的结果：

(SELECT s.proprietario_id as p, COUNT(*) as conta
FROM sn_like_spotted s
GROUP BY s.proprietario_id  
ORDER BY COUNT(*) DESC, s.id ASC
)
UNION ALL
(SELECT s.proprietario_id as p, COUNT(*) as conta2
FROM sn_like_risposta s
GROUP BY s.proprietario_id  
ORDER BY COUNT(*) DESC, s.id ASC
)

我希望p具有最大总和

结果应为

p  conta

1   33
2   15
3   1
4   11
5   2
1   20
2   9
4   2
9   2
18  5
19  7

Answer 1

使用子查询！

SELECT p, SUM(conta) AS conta 
FROM (
  SELECT s.proprietario_id as p, COUNT(*) as conta
  FROM sn_like_spotted s
  GROUP BY s.proprietario_id  
  UNION ALL
  SELECT s.proprietario_id as p, COUNT(*) as conta2
  FROM sn_like_risposta s
  GROUP BY s.proprietario_id  
) AS baseview
GROUP BY p
ORDER BY conta DESC, p ASC

修改

我保留/粘贴您的原始查询，这会产生错误的排序 - 现在已修复。

Answer 2

我不知道mySQL，但是你不想将结果集合在一起，然后再进行分组吗？

概念上类似于

SELECT s.proprietario_id as p, COUNT(*) as conta
from
((SELECT s.proprietario_id as p
FROM sn_like_spotted s

)
UNION ALL
(SELECT s.proprietario_id as p
FROM sn_like_risposta s

))
GROUP BY s.proprietario_id  
ORDER BY COUNT(*) DESC, s.id ASC

Answer 3

如果您希望p使用最大总和：

select p, sum(conta) as conta
from ((SELECT s.proprietario_id as p, COUNT(*) as conta
       FROM sn_like_spotted s
       GROUP BY s.proprietario_id  
      )
      UNION ALL
      (SELECT s.proprietario_id as p, COUNT(*) as conta
       FROM sn_like_risposta s
       GROUP BY s.proprietario_id  
      )
     ) u
GROUP BY p
ORDER BY sum(conta) DESC, s.id ASC
LIMIT 1;

limit 1为您带来最高价值。请注意，子查询中的order by是不必要的，您只需要外部查询中的顺序。

此外，在union中，列以第一个select中的名称命名。别名conta2有点不合适，因为该列名为conta。

顺便说一句，您也可以在子查询中编写没有聚合的查询：

select p, sum(conta) as conta
from ((SELECT s.proprietario_id as p, 1 as conta
       FROM sn_like_spotted s
      )
      UNION ALL
      (SELECT s.proprietario_id as p, 1 as conta
       FROM sn_like_risposta s
      )
     ) u
GROUP BY p
ORDER BY sum(conta) DESC, s.id ASC
LIMIT 1;

在某些情况下，这可能会有更好的表现。