Question

表销售：

  create table sales (  
    Date date,  
    customer_id integer,  
    product_id integer,  
    units_sold integer,  
    paid_amount integer  
    );

Insert into sales (Date, customer_id, product_id, units_sold, paid_amount)  
VALUES  
('2016-01-01', '1', '1', '5', '45'),  
('2016-01-01', '2', '1', '2', '18'),  
('2016-01-01', '3', '2', '7', '35'),  
('2016-01-07', '1', '3', '3', '45'),  
('2016-01-07', '2', '2', '5', '25'),  
('2016-01-07', '4', '2', '5', '25'),  
('2016-01-10', '1', '4', '5', '30'),  
('2016-01-10', '2', '4', '5', '30'),  
('2016-01-10', '4', '5', '6', '60'),  
('2016-01-10', '4', '3', '9', '135'),  
('2016-01-14', '3', '1', '4', '60'),  
('2016-01-14', '2', '3', '6', '90'),  
('2016-01-14', '2', '3', '6', '90');

每天显示次数最多的产品？

我正在尝试这个查询：但是我没有得到日期的结果第14天剩下的日期没事。

  SELECT * 
  FROM (
     SELECT 
       date, 
       customer_id, 
       product_id,
        paid_amount,
       rank() OVER (
         PARTITION BY date 
         ORDER BY paid_amount 
         DESC
        ) 
FROM sales) s 
WHERE rank = 2;

Answer 1

根据vkp：

  SELECT * 
  FROM (
     SELECT 
       date, 
       customer_id, 
       product_id,
        paid_amount,
       dense_rank() OVER (
         PARTITION BY date 
         ORDER BY paid_amount 
         DESC
        ) 
FROM sales) s 
WHERE rank = 2;

对于上下文，想象一下值列表，rank和dense_rank命令desc：

Value    Rank()    Dense_Rank()
100      1         1
90       2         2
90       2         2
80       4         3
70       5         4

Answer 2

SELECT DISTINCT
   date,
   nth_value(paid_amount, 2)
      OVER (PARTITION BY date
            ORDER BY paid_amount
            ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING
           ) "2nd value"
FROM sales
ORDER BY date;

┌────────────┬───────────┐
│    date    │ 2nd value │
├────────────┼───────────┤
│ 2016-01-01 │        35 │
│ 2016-01-07 │        25 │
│ 2016-01-10 │        30 │
│ 2016-01-14 │        90 │
└────────────┴───────────┘
(4 rows)

每天显示第二大支付金额的产品

2 个答案: