不是使用select case語句的單個組的組函數

Question

我在下面的查詢中寫下哪個分隔兩個select查詢並計算百分比。但我得到一個錯誤的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

你並不需要使用加入了答案。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; 

這裏有一個簡單的演示：

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 

---

以防萬一你不知道，這樣做計數的過濾在case語句返回相同的，當你在WHERE子句中過濾器，下面是證明了一個演示它：

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 2

你引用一個聚合函數（COUNT(*)），並在同一個SELECT查詢單個列表達式（r.cnt和o.cnt）。這是無效的SQL，除非爲相關的各個列添加了GROUP BY子句。

提供一個有效的替代方案會更容易，您可以闡明您希望此查詢返回的內容（給出示例模式和一組數據）。作爲一個猜測，我會說你可以簡單地用替換爲o.cnt以避免被0問題分割。如果這裏還有其他一些邏輯出現，那麼你需要澄清這是什麼。

Answer 3

它看起來像你想要得到的狀態百分比不是在0,1或0，如果沒有結果。

也許這是你想要的第一行？

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。選擇計數並稍後再進行分組。

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

這裏是一個完美的作品對我來說

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 6

Boneist的答案是好的，但我會WR將它作爲：

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;