统计每个会计年度的不同客户数量并在查询结果中显示所有日期

Posted 2023-03-31

【中文标题】统计每个会计年度的不同客户数量并在查询结果中显示所有日期

【英文标题】：Count distinct number of customers per fiscal year and display all dates in query result

【发布时间】：2021-03-17 10:42:48

【问题描述】：

DB-Fiddle

CREATE TABLE customers (
    id SERIAL PRIMARY KEY,
    order_date DATE,
    customerID VARCHAR(255)
);

INSERT INTO customers
(order_date, customerID
)
VALUES 
('2020-01-15', 'Customer_01'),
('2020-02-03', 'Customer_01'),
('2020-02-15', 'Customer_01'),
('2020-03-18', 'Customer_01'),
('2020-03-20', 'Customer_01'),
('2020-04-22', 'Customer_01'),
('2021-01-19', 'Customer_01'),

('2020-01-25', 'Customer_02'),
('2020-02-26', 'Customer_02'),
('2020-11-23', 'Customer_02'),
('2021-01-17', 'Customer_02'),
('2021-02-20', 'Customer_02');

预期结果：

order_date   |      quantity
             |    (fiscal year)
-------------|----------------------------------------------------
2020-01-15   |           1   --> Customer_01 appears the first time between 2019-03 and 2020-02
2020-01-25   |           1   --> Customer_02 appears the first time between 2019-03 and 2020-02
2020-02-03   |           0   
2020-02-15   |           0
2020-02-26   |           0
2020-03-18   |           1   --> Customer_01 appears the first time between 2020-03 and 2021-02
2020-03-20   |           0
2020-04-22   |           0
2020-11-23   |           1   --> Customer_02 appears the first time between 2020-03 and 2021-02
2021-01-17   |           0
2021-01-19   |           0
2021-02-20   |           0

---

在上面的结果中，我想列出所有order dates 并计算每个财政年度的customers 不同的数量。 fiscal year 在日历年之后两个月开始，因此从 March 变为 February。 （例如，从 2020-03 到 2021-02）。

例如Customer_01 在2020-03 财政年度内首次出现在2020-03-18 直到2021-02。 因此，这个order_date 被分配给它1。 如果客户在会计年度内再次出现，则下一个order_date 将被分配给它0。

---

参考MariaDB 中的this question，我能够达到预期的结果，正如您在DB-Fiddle 中看到的那样。

但是，现在我想使用 postgresSQL 获得相同的结果。 因此，到目前为止，我已将查询修改为：

SELECT
order_date,
SUM(rn = 1) AS quantity
FROM 

  (SELECT 
  order_date, 
  row_number() over(PARTITION BY DATE_PART('year', (order_date - INTERVAL '2 month')::date), customerID ORDER BY order_date) rn
  FROM customers
  ) t
  
GROUP BY 1;

但是，现在我在SUM(rn = 1) 部分收到错误function sum(boolean) does not exist。 postgresSQL 中的 SUM(rn = 1) 的等效语法是什么才能达到预期的结果？

【参考方案1】：

这个问题有两个部分。一是确定财政年度。第二个是做不同的计数。

第一个是通过减去两个月并提取日期来解决的。第二个是合乎逻辑的：

select c.*,
       count(distinct customerId) over (partition by fyyyy order by order_date)
from (select c.*, date_trunc('year', order_date - interval '2 month') as fyyyy
      from customers c
     ) c
order by date;

不幸的是，这在 Postgres 中不起作用。但您可以只计算客户第一次出现的时间：

select c.*,
       count(*) filter (where seqnum = 1) over (partition by fyyyy order by order_date)
from (select c.*,
             date_trunc('year', order_date - interval '2 month') as fyyyy,
             row_number() over (partition by customerId, date_trunc('year', order_date - interval '2 month')
                                order by order_date
                               ) as seqnum
      from customers c
     ) c
order by order_date;

Here 是一个 dbfiddle。

【讨论】：

所以没有办法使用 postgresSQL 获得预期的结果？

红移有可能还是会出现同样的问题？

问题被标记为 Postgres，而不是 Redshfit。你已经知道它们是不同的。在任何情况下，您只需将 filter 替换为 sum( (seqnum = 1)::int ) 即可获得 Redshfit。

【参考方案2】：

经过进一步调查，我想出了以下解决方案：

DB-Fiddle

SELECT
order_date,
(CASE WHEN t.rolling_count > 1 THEN 0 ELSE t.rolling_count END) AS quantity
FROM 

  (SELECT 
  order_date, 
  (row_number() over(PARTITION BY DATE_PART('year', (order_date - INTERVAL '2 month')::date), customerID ORDER BY order_date)) AS rolling_count
  FROM customers
  ORDER BY 1
  ) t
  
GROUP BY 1,2
ORDER BY 1;

---

这里比较的是查询的MariaDB：

SELECT
order_date,
(CASE WHEN t.rolling_count > 1 THEN 0 ELSE t.rolling_count END) AS quantity
FROM 

  (SELECT 
  order_date, 
  (row_number() over(PARTITION BY YEAR(order_date - INTERVAL 2 MONTH), customerID ORDER BY order_date)) AS rolling_count
  FROM customers
  ORDER BY 1
  ) t
  
GROUP BY 1
ORDER BY 1;