Question

假设我有下表：

+------------+-------------+
| Product_id | customer_id |
+------------+-------------+
| a          | c1          |
| a          | c2          |
| a          | c3          |
| a          | c4          |
| b          | c1          |
| c          | c1          |
| b          | c2          |
| d          | c2          |
+------------+-------------+

我想查找每位客户购买的（a，b，c）产品数量以及每位客户购买的（a，b，d）产品数量。我尝试将COUNT与GROUP BY一起使用，但我只设法找到每个客户FIDDLE的购买数量。我需要使用CASE WHEN还是DECODE？我怎样才能做到这一点？

预期输出类似于：

+-------------+-------------+-------------+
| CUSTOMER_ID | ABC_PRODUCT | ABD_PRODUCT |
+-------------+-------------+-------------+
| c1          |           1 |           0 |
| c2          |           0 |           1 |
| c3          |           0 |           0 |
| c4          |           0 |           0 |
+-------------+-------------+-------------+

Answer 1

您可以使用单个聚合而不使用子查询来执行此操作。关键是使用带有聚合的嵌套case语句来计算每个客户的每个产品。以下内容确定客户是否拥有每个“捆绑”：

SELECT CUSTOMER_ID,
       (case when max(case when product_id = 'a' then 1 else 0 end) +
                  max(case when product_id = 'b' then 1 else 0 end) +
                  max(case when product_id = 'c' then 1 else 0 end) = 3
             then 1
             else 0
        end) as ABC,
       (case when max(case when product_id = 'a' then 1 else 0 end) +
                  max(case when product_id = 'b' then 1 else 0 end) +
                  max(case when product_id = 'd' then 1 else 0 end) = 3
             then 1
             else 0
        end) as ABD           
FROM CUSTOMERS_SALES
GROUP BY CUSTOMER_ID;

现在，您的问题实际上是关于此类购买的数量。所以，我想客户可以两次购买每件商品，你会希望他们计算两次。如果是，那么该数字是任何计数的最小值。你也可以这样做：

SELECT CUSTOMER_ID,
       least(sum(case when product_id = 'a' then 1 else 0 end),
             sum(case when product_id = 'b' then 1 else 0 end),
             sum(case when product_id = 'c' then 1 else 0 end) 
            ) as ABC,
       least(sum(case when product_id = 'a' then 1 else 0 end),
             sum(case when product_id = 'b' then 1 else 0 end),
             sum(case when product_id = 'd' then 1 else 0 end) 
            ) as ABD
FROM CUSTOMERS_SALES
GROUP BY CUSTOMER_ID;

Answer 2

请尝试以下查询，找到拥有产品a, b and c的客户：

SELECT CUSTOMER_ID
FROM CUSTOMERS_SALES
WHERE PRODUCT_ID IN ('a', 'b', 'c') 
GROUP BY CUSTOMER_ID
HAVING COUNT(DISTINCT PRODUCT_ID)=3

要获得点数，请尝试：

SELECT COUNT(*) FROM(
  SELECT CUSTOMER_ID
  FROM CUSTOMERS_SALES
  WHERE PRODUCT_ID IN ('a', 'b', 'd')
  GROUP BY CUSTOMER_ID
  HAVING COUNT(DISTINCT PRODUCT_ID)=3
)x

Answer 3

完全基于@ TechDo的例子：

SELECT DISTINCT CUSTOMER_ID, 
       DECODE((SELECT CUSTOMER_ID
                 FROM CUSTOMERS_SALES CS2
                WHERE PRODUCT_ID IN ('A', 'B', 'C') 
                  AND CS2.CUSTOMER_ID = CS.CUSTOMER_ID
               GROUP BY CUSTOMER_ID
               HAVING COUNT(DISTINCT PRODUCT_ID)=3), NULL, 0, 1) AS ABC_PRODUCT,
       DECODE((SELECT CUSTOMER_ID
                 FROM CUSTOMERS_SALES CS2
                WHERE PRODUCT_ID IN ('A', 'B', 'D') 
                  AND CS2.CUSTOMER_ID = CS.CUSTOMER_ID
               GROUP BY CUSTOMER_ID
               HAVING COUNT(DISTINCT PRODUCT_ID)=3), NULL, 0, 1) AS ABD_PRODUCT
  FROM CUSTOMERS_SALES CS
ORDER BY CUSTOMER_ID

Answer 4

SELECT CUSTOMERS_SALES.CUSTOMER_ID,NVL(MAX(abc.CUSTOMER_ID),0) as ABC_PRODUCT ,NVL(MAX(abd.CUSTOMER_ID),0) as ABD_PRODUCT  FROM CUSTOMERS_SALES 
 LEFT JOIN
(SELECT CUSTOMER_ID 
  FROM CUSTOMERS_SALES
  WHERE PRODUCT_ID IN ('a', 'b', 'd')
  GROUP BY CUSTOMER_ID
  HAVING COUNT(DISTINCT PRODUCT_ID)=3) abd
ON abd.CUSTOMER_ID=CUSTOMERS_SALES.CUSTOMER_ID
 LEFT JOIN
(SELECT CUSTOMER_ID 
  FROM CUSTOMERS_SALES
  WHERE PRODUCT_ID IN ('a', 'b', 'c')
  GROUP BY CUSTOMER_ID
  HAVING COUNT(DISTINCT PRODUCT_ID)=3) abc
ON abc.CUSTOMER_ID=CUSTOMERS_SALES.CUSTOMER_ID
GROUP BY CUSTOMERS_SALES.CUSTOMER_ID
ORDER BY CUSTOMERS_SALES.CUSTOMER_ID;

FIDDLE

基于其他列值的聚合

4 个答案: