上一期间分区上的SQL和

我有下表，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 |
+----------+-----------+

我想再添加两列：

当月客户汇总表

上个月的客户摘要

以下是期望的结果：

+----------+-----------+----------------------+------------------------+
|   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 |
+----------+-----------+----------------------+------------------------+

我已经设法通过简单的分区函数求和来计算第三列：

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

但是，我找不到一种方法来计算前一个月列的和

感谢您的支持

---

Asaf

我认为使用

lag（）

和聚合子查询可能会更容易。ANSI标准语法为：

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,
             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，这可以写成：

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;

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

---

上个月有点棘手。您的Teradata版本是什么，TD14.10支持

LAST\u值

：

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

---

在这种情况下，我希望有一个0的值。看看这篇文章。好像是同一个话题。我不知道这是Teradata语法。我不相信Teradata已经实现了对LAG的支持。我相信它依赖于OVER（）窗口函数中的前行和后行子句来完成超前和滞后。谢谢Gordon和Rob。我相信罗布是对的。您知道如何使用“行前置”语法实现这一点吗？请参阅下面的@dnoeth response。TD14.10实现了

第一个值

和最后一个值

，这是关于
前置
和
滞后的更通用版本`