不同行的MYSQL sum()

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是因为我正在"分组",这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)

我基本上想要SUM(conversions.value)为每个DISTINCT conversions.id 做.

那可能吗？

Answer 1

我可能错了,但据我所知

• conversions.id是表转换的主键
• stats.id是表统计信息的主键

因此,对于每个转换.您最多只有一个links.id受影响.

您的要求有点像做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 

对于每个链接,您获得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a获得

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

这个独特的设法删除了笛卡尔积中的所有[点击]行

但显然

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

在你的情况下,从那以后

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

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. Jerome的公式产生:

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

Answer 3

有关您看到错误数字的原因的解释,请阅读此内容.

我认为Jerome可以处理导致错误的原因.Bryson的查询会起作用,尽管SELECT中的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