我在SQL查询中使用sum()寻找帮助:
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
是因为我正在进行“分组依据”,这可以确保同一行不会被重复计数。问题是SUM(conversions.value)对每行的“值”多次计数(由于分组依据)
我基本上想为每个DISTINCT conversions.id做
SUM(conversions.value)
。那可能吗?
最佳答案
我可能是错的,但据我了解
因此,对于每个conversions.id,您最多影响一个link.id。
您的要求有点像做2组的笛卡尔积:
[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,您将获得sizeof([clicks])x sizeof([conversions])行
如前所述,您可以通过以下方式获得请求中唯一身份转化的次数:
count(distinct conversions.id) = sizeof([conversions])
这种独特的方法可以删除笛卡尔积中的所有[clicks]行
但显然
sum(conversions.value) = sum([conversions].value) * sizeof([clicks])
就您而言,
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
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