如何解決ORA-00937：計算百分比時不是單組功能？

Question

我正在嘗試獲取某個區域中可用的itemid的百分比。 使用我的查詢，我得到一個錯誤ORA-00937: not a single-group group function

所有的細節：

我有這兩個表：

ALLITEMS 
--------------- 
ItemId | Areas 
--------------- 
1  | EAST 
2  | EAST 
3  | SOUTH 
4  | WEST 

CURRENTITEMS 
--------------- 
ItemId 
--------------- 
1 
2 
3 

，並希望這樣的結果：

--------------- 
Areas| Percentage 
--------------- 
EAST | 50 --because ItemId 1 and 2 are in currentitems, so 2 items divided by the total 4 in allitems = .5 
SOUTH | 25 --because there is 1 item in currentitems table that are in area SOUTH (so 1/4=.25) 
WEST | 0 --because there are no items in currentitems that are in area WEST 

的DDL：

drop table allitems; 
drop table currentitems; 

Create Table Allitems(ItemId Int,areas Varchar2(20)); 
Create Table Currentitems(ItemId Int); 
Insert Into Allitems(Itemid,Areas) Values(1,'east'); 
Insert Into Allitems(ItemId,areas) Values(2,'east'); 
insert into allitems(ItemId,areas) values(3,'south'); 
insert into allitems(ItemId,areas) values(4,'east'); 

Insert Into Currentitems(ItemId) Values(1); 
Insert Into Currentitems(ItemId) Values(2); 
Insert Into Currentitems(ItemId) Values(3); 

我的查詢：

Select 
areas, 
(
Select 
Count(Currentitems.ItemId)*100/(Select Count(ItemId) From allitems inner_allitems Where inner_allitems.areas = outer_allitems.areas) 
From 
Allitems Inner_Allitems Left Join Currentitems On (Currentitems.Itemid = Inner_Allitems.Itemid) 
Where inner_allitems.areas = outer_allitems.areas 
***group by inner_allitems.areas*** 
***it worked by adding the above group by*** 
) "Percentage Result" 
From 
allitems outer_allitems 
Group By 
areas 

錯誤：

Error at Command Line:81 Column:41 (which is the part `(Select Count(ItemId) From allitems inner_allitems Where inner_allitems.areas = outer_allitems.areas)`) 
Error report: 
SQL Error: ORA-00937: not a single-group group function 

當我運行在SQL Server中完全相同的查詢，它工作正常。 如何在Oracle中修復此問題？

Answer 1

Analytics（分析）是你的朋友：

SELECT DISTINCT 
     areas 
     ,COUNT(currentitems.itemid) 
     OVER (PARTITION BY areas) * 100 
    /COUNT(*) OVER() Percentage 
FROM allitems, currentitems 
WHERE allitems.itemid = currentitems.itemid(+); 

Answer 2

這裏是一個快速第一遍：

select ai.areas, 
     (sum(cnt)/max(tot)) * 100 "Percentage Result" 
    from (select ai.itemid, 
       ai.areas, 
       count(ci.itemid) cnt, 
       count(ai.areas) over() tot 
      from allitems ai 
       left outer join 
       currentitems ci on (ai.itemid = ci.itemid) 
     group by ai.itemid, ai.areas 
     ) ai 
group by ai.areas 

此外，在您的測試數據的itemid的4個需要改變向西。

Answer 3

原始查詢的一個稍微的改變：

Select 
areas, 
(
Select 
Count(*) 
From 
Allitems Inner_Allitems Left Join Currentitems On (Currentitems.Itemid = Inner_Allitems.Itemid) 
Where inner_allitems.areas = outer_allitems.areas 
) *100/(Select Count(*) From allitems) as percentage 
From allitems outer_allitems 
Group By areas 

Answer 4

只爲它赫克，這樣做沒有分析的方法。

由於重複areas，Jeffrey的解決方案需要DISTINCT。 allitems表格實際上是currentitems與假定的areas表格之間的交點表格。在以下查詢中，這由內嵌視圖ai表示。還有另一個內聯視圖tot，它給了我們allitems中的記錄總數。此計數必須包含在GROUP BY子句中，因爲它不是一個聚合投影儀。

SQL> select ai.areas 
    2   , (count(currentitems.itemid)/tot.cnt) * 100 as "%" 
    3 from 
    4  (select count(*) as cnt from allitems) tot 
    5  , (select distinct areas as areas from allitems) ai 
    6  , currentitems 
    7  , allitems 
    8 where allitems.areas = ai.areas 
    9 and allitems.itemid = currentitems.itemid(+) 
10 group by ai.areas, tot.cnt 
11/

AREAS       % 
-------------------- ---------- 
east       50 
south      25 
west       0 

SQL> 

我不知道這種做法是否會表現得比傑弗裏的更好的解決方案：它很可能會表現更差（在分析查詢肯定有較少的一致獲取）。這很有趣，因爲它更清楚地突出了問題。