SQL 前一时期的分区总和

Question

我有以下 table，代表每天的客户：

+----------+-----------+
|   Date   | Customers |
+----------+-----------+
| 1/1/2014 |         4 |
| 1/2/2014 |         7 |
| 1/3/2014 |         5 |
| 1/4/2014 |         5 |
| 1/5/2014 |        10 |
| 2/1/2014 |         7 |
| 2/2/2014 |         4 |
| 2/3/2014 |         1 |
| 2/4/2014 |         5 |
+----------+-----------+

我想添加 2 个额外的列：

当月客户汇总
上个月客户汇总

这是期望的结果：

+----------+-----------+----------------------+------------------------+
|   Date   | Customers | Sum_of_Current_month | Sum_of_Preceding_month |
+----------+-----------+----------------------+------------------------+
| 1/1/2014 |         4 |                   31 |                      0 |
| 1/2/2014 |         7 |                   31 |                      0 |
| 1/3/2014 |         5 |                   31 |                      0 |
| 1/4/2014 |         5 |                   31 |                      0 |
| 1/5/2014 |        10 |                   31 |                      0 |
| 2/1/2014 |         7 |                   17 |                     31 |
| 2/2/2014 |         4 |                   17 |                     31 |
| 2/3/2014 |         1 |                   17 |                     31 |
| 2/4/2014 |         5 |                   17 |                     31 |
+----------+-----------+----------------------+------------------------+

我已经设法通过分区函数的简单求和计算出第 3 列：

Select 
  Date,
  Customers, 
  Sum(Customers) over (Partition by (Month(Date)||year(Date) Order by 1) as Sum_of_Current_month
From table

但是，我找不到计算 Sum_of_preceding_month 列的方法。

感谢您的支持。

阿萨夫

Answer 1

我认为使用 lag() 和聚合子查询可能会更容易。 ANSI 标准语法是：

Select t.*, tt.sumCustomers, tt.prev_sumCustomers
From table t join
     (select extract(year from date) as yyyy, extract(month from date) as mm,
             sum(Customers) as sumCustomers,
             lag(sum(Customers)) over (order by extract(year from date), extract(month from date)
                                      ) as prev_sumCustomers
      from table t
      group by extract(year from date), extract(month from date)
     ) tt
     on extract(year from date) = tt.yyyy and extract(month from date) = t.mm;

在 Teradata 中，这将写为：

Select t.*, tt.sumCustomers, tt.prev_sumCustomers
From table t join
     (select extract(year from date) as yyyy, extract(month from date) as mm,
             sum(Customers) as sumCustomers,
             min(sum(Customers)) over (order by extract(year from date), extract(month from date)
                                       rows between 1 preceding and 1 preceding
                                      ) as prev_sumCustomers
      from table t
      group by extract(year from date), extract(month from date)
     ) tt
     on extract(year from date) = tt.yyyy and extract(month from date) = t.mm;

Answer 2

试试这个：

 SELECT
     [Date],
     [Customers],
     (SELECT SUM(customers) FROM table WHERE MONTH(dte) = MONTH(tbl.dte)),
     ISNULL((SELECT SUM(customers) FROM table WHERE MONTH(dte) = MONTH(DATEADD(MONTH, -1, tbl.dte))), 0)
 FROM table tbl

Answer 3

上个月有点棘手。您的 Teradata 版本是什么，TD14.10 支持 LAST_VALUE:

SELECT 
   dt,
   customers,
   Sum_of_Current_month,
   -- return the previous sum
   COALESCE(LAST_VALUE(x ignore NULLS) 
            OVER (ORDER BY dt 
                  ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING)
           ,0) AS Sum_of_Preceding_month
FROM 
 (
   SELECT 
     dt,
     Customers, 
     SUM(Customers) OVER (PARTITION BY TRUNC(dt,'mon')) AS Sum_of_Current_month,
     CASE -- keep the number only for the last day in month
       WHEN ROW_NUMBER()
            OVER (PARTITION BY TRUNC(dt,'mon')
                  ORDER BY dt)
          = COUNT(*) 
            OVER (PARTITION BY TRUNC(dt,'mon'))
       THEN Sum_of_Current_month
     END AS x
   FROM tab
 ) AS dt

SQL 前一时期的分区总和

SQL sum over partition for preceding period

sql

teradata