SQL速率（百分比）计算 - 除以零错误

Question

尝试用户NULLIF或IFNULL函数，但仍然接收除零消息。

SELECT 
client_id
,COUNT(distinct CASE WHEN status = 'failed' THEN id END) AS count_FAILS
,COUNT(distinct CASE WHEN status = 'completed' THEN id END) AS count_COMPLETED
,COUNT(distinct CASE WHEN status IN ('failed') THEN id END)
 /CAST(COUNT(CASE WHEN status = 'completed' THEN id END) AS FLOAT) 

FROM journey
GROUP BY 1

示例数据库和查询https://dbfiddle.uk/?rdbms=postgres_10&fiddle=efc0cd25843e852ab7a3aa8fe49e6986

这种查询是否有必要？

谢谢！

Answer 1

这应该工作：

COUNT(distinct CASE WHEN status IN ('failed') THEN id END) * 1.0 /
NULLIF(COUNT(CASE WHEN status = 'completed' THEN id END), 0)

你可以施放到浮子上。我更喜欢使用* 1.0。

我注意到count_completed的定义使用count(distinct)，但这个比例没有。这可能是一个错误。

我更有可能想要整体失败状态的比例，而不是比率。如果这是可以接受的，并且COUNT(DISTINCT)对于任何一个计数都不是必需的，那么这可以简化为：

avg( (status = 'failed')::int )

Answer 2

你可以使用NULLIF：

,COUNT(distinct CASE WHEN status IN ('failed') THEN id END)/
 NULLIF(CAST(COUNT(CASE WHEN status = 'completed' THEN id END) AS FLOAT),0) 

DBFidde Demo

---

我会将整个查询重写为：

SELECT 
  client_id
  ,COUNT(distinct id) FILTER(WHERE status = 'failed') AS count_FAILS
  ,COUNT(distinct id) FILTER(WHERE status = 'completed') AS count_COMPLETED
  ,COUNT(distinct id) FILTER(WHERE status = 'failed')
   /NULLIF(1.0 * COUNT(distinct id) FILTER(WHERE status = 'completed'),0)
FROM journey
GROUP BY client_id;

DBFiddle Demo2