How to sum up unique values using case statements having certain conditions

I have a table that may have the same item but with different sizes, I would like to count those items with more than one size (e.g. marvel shirt with S, M sizes will count as “1”), but still be able to count how many S and M. I have 2 outcomes I would like to get. Please see below for more detail.

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     -

I tried just counting it but it is incorrect since it counts how many distinct sizes are there

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);

Actual outcome (incorrect):

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

Desired/expected outcome:

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

Below is another possible desired outcome given this scenario: What if those with multiple sizes shouldn’t be counted separately (i.e. Marvel shirt has multiple sizes, thus it won’t count the S or M since it’s already counted under Multiple_Sizes)?

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

Answer

You probably need to group twice (1) by item number (2) by item category:

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            |

And the variation:

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            |