Question

我有两张桌子：产品和订单

CREATE TABLE product (
   product_id INTEGER,
   product_name CHARACTER VARYING(20),
   category_name CHARACTER VARYING(20)
);

CREATE TABLE  orders (
   order_id  INTEGER,
   date_id INTEGER,
   customer_id INTEGER,
   product_id INTEGER,
   sales INTEGER   
 );

以下是一些示例数据：

INSERT INTO product VALUES (1,'widget','thingamabobs'); 
INSERT INTO product VALUES (2,'sprocket','thingamabobs'); 
INSERT INTO product VALUES (3,'gizmo','doodads'); 

-- 10 orders across 2 dates to 5 customers

INSERT INTO orders VALUES (1,1,1,1,2); 
INSERT INTO orders VALUES (1,1,1,3,2); 
INSERT INTO orders VALUES (2,1,2,2,3); 
INSERT INTO orders VALUES (3,1,3,1,1); 
INSERT INTO orders VALUES (4,1,4,1,2);
INSERT INTO orders VALUES (5,1,5,1,1);
INSERT INTO orders VALUES (6,2,1,1,3); 
INSERT INTO orders VALUES (6,2,1,3,2); 
INSERT INTO orders VALUES (7,2,2,2,3); 
INSERT INTO orders VALUES (8,2,3,1,1); 
INSERT INTO orders VALUES (8,2,3,2,1); 
INSERT INTO orders VALUES (9,2,4,1,1); 
INSERT INTO orders VALUES (10,2,5,1,2);

从上面我需要确定只购买小部件并且在最近一段时间内购买更多小东西的东西。

我写了以下查询：

WITH otherprod AS (SELECT DISTINCT customer_id
FROM orders o
INNER JOIN product p
ON o.product_id=p.product_id 
WHERE category_name='thingamabobs' AND o.product_id<>1 ),
prod AS (SELECT DISTINCT customer_id
FROM orders o
WHERE o.product_id=1 
),
solus AS (SELECT p.customer_id
FROM prod p
LEFT JOIN otherprod op
ON p.customer_id=op.customer_id
),
prodpp AS (SELECT customer_id,SUM(sales) AS ppsales
FROM orders o
WHERE o.product_id=1 
AND date_id<2
GROUP BY customer_id
),
prodtp AS (SELECT customer_id,SUM(sales) AS sales
FROM orders o
WHERE o.product_id=1 
AND date_id>1
GROUP BY customer_id
)
SELECT pp.customer_id
FROM prodpp pp
INNER JOIN prodtp tpp
ON pp.customer_id=tpp.customer_id
INNER JOIN solus s
ON pp.customer_id=s.customer_id
WHERE ppsales<sales;

返回正确的结果：

customer_id

       5
       1

（2行）

但是有更有效的方法吗？

Answer 1

我认为你可以写下这样的查询：

select customer_id,
       sum(case when date_id = 1 then sales else 0 end) as sales_1,
       sum(case when date_id = 2 then sales else 0 end) as sales_2
from (select o.*, p.*,
             min(p.product_name) over (partition by o.customer_id) as minpn,
             max(p.product_name) over (partition by o.customer_id) as maxpn
      from orders o join
           products p
           on o.product_id = p.product_id
      where p.category_name = 'thingamabobs'
     ) op
where minpn = maxpn and minpn = 'widget'
group by custoemr_id
having sales_2 > sales_1;

这是基于您的描述，而不是您的代码。

注意：

子查询计算＆＃34; thingamabobs＆＃34;中每个客户的最小和最大产品。基。
当最小值和最大值相同且它们具有您正在查看的值时，它们是相同的。
我不遵循日期逻辑，从日期转换为期间。也许外部case表达式中应该有更复杂的日期逻辑。

一个类别中的一个项目的独家客户，他们在第2期和第1期间购买更多

customer_id

1 个答案: