Question

我已经设法按月计算累积的不同计数但是在尝试多年时我无法得到它。我尝试了分区功能，但我一直遇到各种错误，这是我到目前为止所得到的。你能给我一些指导吗，我是新人。

[SQL小提琴] [1]

PostgreSQL 9.3架构设置：

create table table2 (id serial, fecha_captura date, mn integer, VALUACION1 integer);

    insert into table2 (fecha_captura, mn, VALUACION1)
    values
    ('1996-02-01',2,12345),
    ('1996-02-01',2,12345),
    ('1996-02-01',2,12344),
    ('1996-02-01',2,12343),
    ('1996-03-01',3,12345),
    ('1996-03-01',3,12346),
    ('1996-03-01',3,12347),
    ('1996-03-01',3,12348),
    ('1996-04-01',4,12349),
    ('1996-04-01',4,12350),
    ('1996-04-01',4,12351),
    ('1996-04-01',4,12352),
    ('1997-03-01',4,12353),
    ('1997-03-01',4,12354),
    ('1997-03-01',4,12345),
    ('1997-03-01',4,12352),
    ('1997-04-01',4,12353),
    ('1997-04-01',4,12354),
    ('1997-04-01',4,12345),
    ('1997-04-01',4,12355);

查询1 ：

SELECT Year1,mn,
   COUNT(DISTINCT(VALUACION1)) AS Val_Dist_Mes
  ,SUM(flag) AS Val_Dist_Mes_Nuevas
  ,SUM(SUM(flag)) 
   OVER (partition by Year1 ORDER BY mn,Year1
         ROWS UNBOUNDED PRECEDING) AS Val_Dist_Acum 
FROM
 (
   SELECT
      VALUACION1
     ,EXTRACT(MONTH FROM fecha_captura) AS mn
     ,EXTRACT(YEAR FROM fecha_captura) AS Year1
     ,CASE
         WHEN EXTRACT(MONTH FROM fecha_captura) 
              = MIN (EXTRACT(MONTH FROM fecha_captura))
              OVER (PARTITION BY VALUACION1) 
         THEN 1 
         ELSE 0 
      END AS flag
   FROM table2
   GROUP BY fecha_captura,VALUACION1
 ) AS dt
GROUP BY mn,Year1

[结果] [2] ：

| year1 | mn | val_dist_mes | val_dist_mes_nuevas | val_dist_acum |
|-------|----|--------------|---------------------|---------------|
|  1996 |  2 |            3 |                   3 |             3 |
|  1996 |  3 |            4 |                   3 |             6 |
|  1996 |  4 |            4 |                   3 |             9 |
|  1997 |  3 |            4 |                   3 |             3 |
|  1997 |  4 |            4 |                   1 |             4 |

[预期成果] [2] ：

| year1 | mn | val_dist_mes | val_dist_mes_nuevas | val_dist_acum  |
|-------|----|--------------|---------------------|----------------|
|  1996 |  2 |            3 |                   3 |             3  |
|  1996 |  3 |            4 |                   3 |             6  |
|  1996 |  4 |            4 |                   4 |             10 |
|  1997 |  3 |            4 |                   4 |             4  |
|  1997 |  4 |            4 |                   1 |             5  |

Answer 1

在Year和group by中添加partition by可以解决问题

SELECT AÑO, 
       mn, 
       Count(DISTINCT( valuacion1 ))             AS Val_Dist_Mes, 
       Sum(flag)                                 AS Val_Dist_Mes_Nuevas, 
       Sum(Sum(flag)) 
         OVER ( 
           partition BY AÑO 
           ORDER BY mn rows UNBOUNDED PRECEDING) AS Val_Dist_Acum 
FROM   (SELECT valuacion1, 
               Extract(month FROM fecha_captura) AS mn, 
               Extract(year FROM fecha_captura)  AS AÑO, 
               CASE -- find the first year when a customer placed an order 
                 WHEN Extract(month FROM fecha_captura) = Min ( 
                      Extract(month FROM fecha_captura)) 
                      OVER ( 
                        partition BY valuacion1, Extract(year 
                      FROM fecha_captura)) THEN 1 
                 ELSE 0 
               END                               AS flag 
        FROM   table2 
        GROUP  BY fecha_captura, 
                  valuacion1) AS dt 
GROUP  BY mn, 
          AÑO

SQL FIDDLE DEMO

需要按周期（年，月）计算累积计数

1 个答案: