[英]not a single-group group function using select case statement
我正在写下面的查询,它划分两个选择查询并计算百分比。 但我收到错误,因为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;
您在同一SELECT查询中引用聚合函数( COUNT(*)
)和单个列表达式( r.cnt
和o.cnt
)。 除非为相关的各个列添加了GROUP BY子句,否则这是无效的SQL。
提供一个有效的替代方案会更容易,您可以澄清该查询要返回的内容(给定示例架构和数据集)。 猜测一下,我想您可以将COUNT(*)
替换为o.cnt
以避免被0除。 如果这里还有其他逻辑,则需要弄清楚这是什么。
您不需要使用联接。 如果我是你,我会做:
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
并且以防万一,您不确定在case语句中对计数进行过滤是否与在where子句中进行过滤时返回的结果相同,以下是一个演示此演示的演示:
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 '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
看起来您希望获得的状态百分比不在0,1或0中(如果没有结果)。
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
您不需要交叉连接。 选择计数,然后进行除法。
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
Boneist的回答很好,但我将其写为:
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;
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
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