不是使用 select case 语句的单组函数

Question

我正在写下面的查询，它将两个 select 查询分开并计算百分比。但是我收到一个错误 not a single-group group function

select CASE WHEN COUNT(*) = 0 THEN 0 ELSE round((r.cnt / o.cnt)*100,3) END from 
    (Select count(*) as cnt from O2_CDR_HEADER WHERE STATUS NOT IN(0,1) and DATE_CREATED > (SYSDATE - 1)) r cross join
    (Select count(*) as cnt from O2_CDR_HEADER WHERE DATE_CREATED > (SYSDATE - 1)) o;

Answer 1

您在同一 SELECT 查询中引用了聚合函数 (COUNT(*)) 和单个列表达式（r.cnt 和 o.cnt）。这是无效的 SQL 除非为相关的单个列添加了 GROUP BY 子句。

如果您可以向 return 阐明您希望此查询的内容（给定示例架构和数据集），那么提供有效的替代方案会更容易。作为一个猜测，我想说你可以简单地用 o.cnt 替换 COUNT(*) 来避免除以 0 的问题。如果此处预期存在其他逻辑，您需要澄清那是什么。

Answer 2

您似乎想要获得不在 0,1 中的状态百分比，如果没有结果则为 0。

也许这就是您想要的第一行？

SELECT CASE WHEN (R.CNT = 0 AND O.CNT = 0) THEN 0 ELSE ROUND((R.CNT *100.0 / O.CNT),3) END

Answer 3

您不需要交叉联接。 Select 计数并稍后进行除法。

select case when ocnt > 0 then round((rcnt / ocnt)*100,3)
       else 0 end
from
(
select 
CASE WHEN STATUS NOT IN(0,1) and DATE_CREATED > (SYSDATE - 1)
 THEN COUNT(*) END as rcnt,
CASE WHEN DATE_CREATED > (SYSDATE - 1)
 THEN COUNT(*) END as ocnt 
from O2_CDR_HEADER
group by status, date_created
) t

Answer 4

这是最适合我的答案

select CASE WHEN (o.cnt = 0) THEN 0 ELSE round((r.cnt / o.cnt)*100,3) END from 
(Select count(*) as cnt from O2_CDR_HEADER WHERE STATUS NOT IN(0,1) and DATE_CREATED > (SYSDATE - 1)) r cross join
(Select count(*) as cnt from O2_CDR_HEADER WHERE DATE_CREATED > (SYSDATE - 1)) o

Answer 5

您不需要使用联接。如果我是你，我会这样做：

select case when count(*) = 0 then 0
            else round(100 * count(case when status not in (0, 1) then 1 end) / count(*), 3)
       end non_0_or_1_status_percentage
from   o2_cdr_header
where  date_created > sysdate - 1;

这是一个简单的演示：

with t as (select 1 status from dual union all
           select 2 status from dual union all
           select 3 status from dual union all
           select 2 status from dual union all
           select 4 status from dual union all
           select 5 status from dual union all
           select 6 status from dual union all
           select 7 status from dual union all
           select 1 status from dual union all
           select 0 status from dual union all
           select 1 status from dual)
select case when count(*) = 0 then 0
            else round(100 * count(case when status not in (0, 1) then 1 end) / count(*), 3)
       end col1
from   t
where 1=0;

      COL1
----------
         0

以防万一您不确定在 case 语句 returns 中过滤计数是否与在 where 子句中过滤时相同，这里有一个演示可以证明这一点：

with t as (select 1 status from dual union all
           select 2 status from dual union all
           select 3 status from dual union all
           select 2 status from dual union all
           select 4 status from dual union all
           select 5 status from dual union all
           select 6 status from dual union all
           select 7 status from dual union all
           select 1 status from dual union all
           select 0 status from dual union all
           select 1 status from dual)
select 'using case statement' how_count_filtered,
       count(case when status not in (0, 1) then 1 end) cnt
from   t
union all
select 'using where clause' how_count_filtered,
       count(*) cnt
from   t
where  status not in (0, 1);

HOW_COUNT_FILTERED          CNT
-------------------- ----------
using case statement          7
using where clause            7

Answer 6

Boneist 的回答很好，但我会这样写：

select coalesce(round(100 * avg(case when status not in (0, 1) then 1.0 else 0
                                end), 3), 0) as non_0_or_1_status_percentage
from   o2_cdr_header
where  date_created > sysdate - 1;

不是使用 select case 语句的单组函数

not a single-group group function using select case statement

sql

oracle

group-by

case-when