如何使用具有特定条件的case语句来求和唯一值

Question

我有一个表，可能有相同的项目，但与不同的尺寸，我想计算这些项目与一个以上的尺寸(例如，神奇衬衫与S，M尺寸将算作"1")，但仍然能够计算出有多少S和M。我有2个结果，我想得到。有关更多详细信息，请参阅下文。

TABLE B   

ITEM_NO ITEM             
=========================
3130C   MARVEL_SHIRT     
1845C   SPONGEBOB_BOXERS 
A900C   CK_COAT          
A988C   RIDER_JEANS      


TABLE C

ITEM_NO SIZE          
===============
3130C   S             
3130C   M             
1845C   M             
A900C   L             
A988C   M     -

我试着数了一下，但它是错误的，因为它计算了有多少不同的大小

select (case substr(item_no, 5, 1)
            when 'C' then 'clothes'
            when 'T' then 'toys'
            else 'misc' 
        end) inv, 
       count(item_no) total 
       ,sum (case when C.size = 'S' then 1 else 0 end) AS small
       ,sum (case when C.size = 'M' then 1 else 0 end) AS med
       ,sum (case when C.size = 'L' then 1 else 0 end) AS large   
       ,count (distinct C.size) AS multiple_sizes
        from B left outer join C on B.item_no = C.item_no 
        group by substr(item_no, 5, 1);

实际结果(不正确)：

INV     TOTAL   Small   Med   Large   Multiple_Sizes
==========================================================
CLOTHES    4       1    3       1       3

期望/预期结果：

INV     TOTAL   Small   Med   Large   Multiple_Sizes
==========================================================
CLOTHES    4       1    3       1       1

下面是这种情况下的另一个可能的预期结果:如果不应该单独计算多个尺码的漫威衬衫(即漫威衬衫有多个尺码，因此不会计算S或M，因为它已经在Multiple_Sizes下计算)？

INV     TOTAL   Small   Med   Large     Multiple_Sizes
==========================================================
CLOTHES    4       0    2     1        1

Answer 1

您可能需要按物料编号(2)和物料类别分组两次(1)：

SELECT SUBSTR(item_no, 5, 1) AS category
     , COUNT(*) AS count_products
     , SUM(count_small) AS small
     , SUM(count_med) AS med
     , SUM(count_large) AS large
     , SUM(CASE WHEN count_small + count_med + count_large > 1 THEN 1 END) AS has_multiple
FROM (
    SELECT b.ITEM_NO
         , COUNT(CASE WHEN c.SIZE = 'S' THEN 1 END) AS count_small
         , COUNT(CASE WHEN c.SIZE = 'M' THEN 1 END) AS count_med
         , COUNT(CASE WHEN c.SIZE = 'L' THEN 1 END) AS count_large
    FROM b
    LEFT JOIN c ON b.item_no = c.item_no
    GROUP BY b.ITEM_NO
) x
GROUP BY SUBSTR(item_no, 5, 1)

| category | count_products | small | med | large | has_multiple |
| C        | 4              | 1     | 3   | 1     | 1            |

其变体是：

SELECT SUBSTR(item_no, 5, 1) AS category
     , COUNT(*) AS count_products
     , SUM(CASE WHEN count_small + count_med + count_large = 1 THEN count_small END) AS small
     , SUM(CASE WHEN count_small + count_med + count_large = 1 THEN count_med   END) AS med
     , SUM(CASE WHEN count_small + count_med + count_large = 1 THEN count_large END) AS large
     , SUM(CASE WHEN count_small + count_med + count_large > 1 THEN 1 END) AS has_multiple
FROM (
    SELECT b.ITEM_NO
         , COUNT(CASE WHEN c.SIZE = 'S' THEN 1 END) AS count_small
         , COUNT(CASE WHEN c.SIZE = 'M' THEN 1 END) AS count_med
         , COUNT(CASE WHEN c.SIZE = 'L' THEN 1 END) AS count_large
    FROM b
    LEFT JOIN c ON b.item_no = c.item_no
    GROUP BY b.ITEM_NO
) x
GROUP BY SUBSTR(item_no, 5, 1)

| category | count_products | small | med | large | has_multiple |
| C        | 4              | 0     | 2   | 1     | 1            |

Answer 2

--creando tabla
create table #temp (itemId int, size nvarchar(1))

--insertando valores
insert into #temp values (1,'S')
insert into #temp values (1,'M')
insert into #temp values (2,'M')
insert into #temp values (3,'L')
insert into #temp values (4,'M')


-- table of Different Item Codes

    select 
        itemId
    into #masDeUnItem
    from
    (select itemId,size from #temp group by itemId,size) t1
    group by itemId 
    having count(1) > 1


-- Variable of Counting different Items
declare @itemsDistintos int 


-- Providing Value to Variable
select @itemsDistintos = count(1) from 
(
    select * from #masDeUnItem
) t1


--Outcome 1
select count(distinct(itemId)) TOTAL
,
sum(case when size = 'S' then 1 else 0 end) SMALL
, sum(case when size = 'M' then 1 else 0 end) MEDIUM
, sum(case when size = 'L' then 1 else 0 end) LARGE
,  @itemsDistintos as Multiple_Sizes
from #temp


--Outcome 2
select count(distinct(a.itemId)) TOTAL
,
sum(case when size = 'S' and b.itemId is null then 1 else 0 end) SMALL
, sum(case when size = 'M' and b.itemId is null then 1 else 0 end) MEDIUM
, sum(case when size = 'L' and b.itemId is null then 1 else 0 end) LARGE
,  @itemsDistintos as Multiple_Sizes
from #temp a 
left join #masDeUnItem b 
on a.itemId = b.itemId