Question

我在SQL查询中使用sum（）寻求帮助：

SELECT links.id, 
       count(DISTINCT stats.id) as clicks, 
       count(DISTINCT conversions.id) as conversions, 
       sum(conversions.value) as conversion_value 
FROM links 
LEFT OUTER JOIN stats ON links.id = stats.parent_id 
LEFT OUTER JOIN conversions ON links.id = conversions.link_id 
GROUP BY links.id 
ORDER BY links.created desc;

我使用DISTINCT，因为我正在“分组”，这可以确保同一行不会被计算多次。

问题是SUM（conversions.value）多次计算每行的“值”（由于分组依据）

我基本上希望为每个DISTINCT conversions.id执行SUM(conversions.value)。

这可能吗？

Answer 1

我可能错了但是从我的理解

conversions.id 是表转化

主键

stats.id 是表统计信息

主键

因此，对于每次转换。您最多只有一个链接。受影响。

你的请求有点像做2套笛卡尔积：

[clicks]
SELECT *
FROM links 
LEFT OUTER JOIN stats ON links.id = stats.parent_id 

[conversions]
SELECT *
FROM links 
LEFT OUTER JOIN conversions ON links.id = conversions.link_id

并为每个链接获得sizeof（[clicks]）x sizeof（[conversions]）行

如您所知，您的请求中的唯一转化次数可以通过

获得

count(distinct conversions.id) = sizeof([conversions])

此独特设法删除笛卡尔积中的所有[点击]行

但显然

sum(conversions.value) = sum([conversions].value) * sizeof([clicks])

在你的情况下，因为

count(*) = sizeof([clicks]) x sizeof([conversions])
count(*) = sizeof([clicks]) x count(distinct conversions.id)

你有

sizeof([clicks]) = count(*)/count(distinct conversions.id)

所以我会用

测试你的请求

SELECT links.id, 
   count(DISTINCT stats.id) as clicks, 
   count(DISTINCT conversions.id) as conversions, 
   sum(conversions.value)*count(DISTINCT conversions.id)/count(*) as conversion_value 
FROM links 
LEFT OUTER JOIN stats ON links.id = stats.parent_id 
LEFT OUTER JOIN conversions ON links.id = conversions.link_id 
GROUP BY links.id 
ORDER BY links.created desc;

让我发布！杰罗姆

Answer 2

Jeromes解决方案实际上是错误的，可能会产生错误的结果!!

sum(conversions.value)*count(DISTINCT conversions.id)/count(*) as conversion_value

让我们假设下表

conversions
id value
1 5
1 5
1 5
2 2
3 1

不同ID的正确值总和为8。杰罗姆的公式产生：

sum(conversions.value) = 18
count(distinct conversions.id) = 3
count(*) = 5
18*3/5 = 9.6 != 8

Answer 3

有关您看到错误号码的原因的解释， read this 。

我认为Jerome可以处理导致错误的原因。 Bryson的查询可行，但在SELECT中使用子查询可能效率低下。

Answer 4

使用以下查询：

SELECT links.id
  , (
    SELECT COUNT(*)
    FROM stats
    WHERE links.id = stats.parent_id
  ) AS clicks
  , conversions.conversions
  , conversions.conversion_value
FROM links
LEFT JOIN (
  SELECT link_id
    , COUNT(id) AS conversions
    , SUM(conversions.value) AS conversion_value
  FROM conversions
  GROUP BY link_id
) AS conversions ON links.id = conversions.link_id
ORDER BY links.created DESC

Answer 5

我使用子查询来执行此操作。它消除了分组问题。因此查询将类似于：

SELECT COUNT(DISTINCT conversions.id)
...
     (SELECT SUM(conversions.value) FROM ....) AS Vals

Answer 6

这样的事情怎么样：

select l.id, count(s.id) clicks, count(c.id) clicks, sum(c.value) conversion_value
from    (SELECT l.id id, l.created created,
               s.id clicks,  
               c.id conversions,  
               max(c.value) conversion_value                    
        FROM links l LEFT
        JOIN stats s ON l.id = s.parent_id LEFT
        JOIN conversions c ON l.id = c.link_id  
        GROUP BY l.id, l.created, s.id, c.id) t
order by t.created

Answer 7

这样就可以了，只需将会话ID除以重复的会话ID计数。

SELECT a.id,
       a.clicks,
       SUM(a.conversion_value/a.conversions) AS conversion_value,
       a.conversions
FROM (SELECT links.id, 
       COUNT(DISTINCT stats.id) AS clicks, 
       COUNT(conversions.id) AS conversions, 
       SUM(conversions.value) AS conversion_value 
      FROM links 
      LEFT OUTER JOIN stats ON links.id = stats.parent_id 
      LEFT OUTER JOIN conversions ON links.id = conversions.link_id 
      GROUP BY conversions.id,links.id
      ORDER BY links.created DESC) AS a
GROUP BY a.id

Answer 8

Select sum(x.value) as conversion_value,count(x.clicks),count(x.conversions)
FROM
(SELECT links.id, 
       count(DISTINCT stats.id) as clicks, 
       count(DISTINCT conversions.id) as conversions,
       conversions.value,       
FROM links 
LEFT OUTER JOIN stats ON links.id = stats.parent_id 
LEFT OUTER JOIN conversions ON links.id = conversions.link_id 
GROUP BY conversions.id) x
GROUP BY x.id 
ORDER BY x.created desc;

我相信这会给您所需的答案。

不同行的MYSQL sum（）

8 个答案: