不是使用select case語句的單組分組功能

Question

我正在寫下面的查詢，它划分兩個選擇查詢並計算百分比。 但我收到錯誤，因為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查詢中引用聚合函數（ COUNT(*) ）和單個列表達式（ r.cnt和o.cnt ）。 除非為相關的各個列添加了GROUP BY子句，否則這是無效的SQL。

提供一個有效的替代方案會更容易，您可以澄清該查詢要返回的內容（給定示例架構和數據集）。 猜測一下，我想您可以將COUNT(*)替換為o.cnt以避免被0除。 如果這里還有其他邏輯，則需要弄清楚這是什么。

Answer 2

您不需要使用聯接。 如果我是你，我會做：

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;

這是一個簡單的演示：

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

---

並且以防萬一，您不確定在case語句中對計數進行過濾是否與在where子句中進行過濾時返回的結果相同，以下是一個演示此演示的演示：

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

Answer 3

看起來您希望獲得的狀態百分比不在0,1或0中（如果沒有結果）。

也許這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 4

您不需要交叉連接。 選擇計數，然后進行除法。

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 5

Boneist的回答很好，但我將其寫為：

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;

Answer 6

這是最適合我的答案

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