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我在订单表上有上周的订单和送货司机的 ID。它看起来有点像这样：

ORDERDATE,    ORDERNO,    DRIVER
23/01/2013,    901398503,    1
23/01/2013,    901332159,    1
23/01/2013,    901334158,    2
24/01/2013,    901338455,    1
25/01/2013,    902907513,    1
25/01/2013,    902338553,    2
25/01/2013,    903936533,    2
27/01/2013,    903944523,    1
27/01/2013,    903981522,    2
27/01/2013,    911334951,    1
28/01/2013,    911338851,    1
28/01/2013,    911339259,    1
28/01/2013,    912332555,    2
28/01/2013,    912336650,    2
29/01/2013,    912337655,    1
29/01/2013,    913969582,    1
29/01/2013,    913973583,    1
29/01/2013,    913982552,    1
29/01/2013,    916379158,    1

我想选择 ORDERDATE、ORDERCOUNT、DRIVER_1_COUNT、DRIVER_2_COUNT。

所以，日期| 总订单 | 司机 1 的总订单 | 司机 2 的总订单

此外，如果 ORDERDATE、ORDERCOUNT、DRIVER_1_COUNT 或 DRIVER_2_COUNT 为 0（或 null），我需要零。

（在 oracle 中）我可以为上周的每一天选择日期，并为每一天选择零订单计数（占位符），如下所示：

select 
 TRUNC(NEXT_DAY(sysdate,'SUNDAY')-7 +i) ORDERDATE,
 0 as ORDERCOUNT   
from
 (select rownum i from all_objects where rownum < 8)

我应该能够使用此输出来确保最终结果中没有缺少日期（在本例中 26 日没有订单）

ORDERDATE,ORDERCOUNT
23/01/2013,0
24/01/2013,0
25/01/2013,0
26/01/2013,0
27/01/2013,0
28/01/2013,0
29/01/2013,0

我需要这个输出：

ORDERDATE,ORDERCOUNT,DRIVER_1_COUNT,DRIVER_2_COUNT
23/01/2013,3,2,1
24/01/2013,1,1,0
25/01/2013,3,1,2
26/01/2013,0,0,0
27/01/2013,3,2,1
28/01/2013,4,2,2
29/01/2013,5,5,0

我可以得到 ORDERDATE & ORDERCOUNT(simple sum) 并与其他查询联合以避免丢失天数，但我也不知道如何为每个驱动程序求和。

提前感谢您的帮助。

埃德

4

3 回答 3

1

在 Oracle 11g 中，您可以这样做：-

SELECT *
FROM orders
PIVOT (
  COUNT( ORDERNO )
  FOR DRIVER IN (1,2,3)
)

如需进一步解释，请参阅11g 中的 pivot 和 unpivot 查询

于 2013-01-31T00:36:27.687 回答
0

首先，您需要总结和交叉表结果：

SELECT ORDERDATE, SUM(ORDERCOUNT) ORDERCOUNT, 
SUM(DECODE(DRIVER,1,ORDERCOUNT,0)) DRIVER_1_COUNT, 
SUM(DECODE(DRIVER,2,ORDERCOUNT,0)) DRIVER_2_COUNT
FROM (
    SELECT ORDERDATE, DRIVER, COUNT(*) ORDERCOUNT
    FROM YourTable
    GROUP BY ORDERDATE, DRIVER
) S
GROUP BY  ORDERDATE

在 Oracle 中可能有更聪明的方法来做到这一点

然后，您需要使用外部连接到您的日期来填写空白：

（请注意，上面的查询在此查询中别名为“T”：）

SELECT D.ORDERDATE, 
NVL(T.ORDERCOUNT,0) ORDERCOUNT, 
NVL(T.DRIVER_1_COUNT,0) DRIVER_1_COUNT, 
NVL(T.DRIVER_1_COUNT,0) DRIVER_2_COUNT
FROM 
(
SELECT ORDERDATE, SUM(ORDERCOUNT) ORDERCOUNT, 
SUM(DECODE(DRIVER,1,ORDERCOUNT,0)) DRIVER_1_COUNT, 
SUM(DECODE(DRIVER,2,ORDERCOUNT,0)) DRIVER_2_COUNT
FROM 
    (
    SELECT ORDERDATE, DRIVER, COUNT(*) ORDERCOUNT
    FROM YourTable
    GROUP BY ORDERDATE, DRIVER
    ) S
GROUP BY  ORDERDATE
) T
RIGHT OUTER JOIN
(
SELECT
TRUNC(NEXT_DAY(sysdate,'SUNDAY')-7 +i) ORDERDATE
FROM (select rownum i from all_objects where rownum < 8)
) D
ON D.ORDERDATE = T.ORDERDATE
于 2013-01-31T00:02:49.653 回答
0

您必须从子查询中进行选择。像这样的东西应该工作。

select orderdate, ordercount, sum(driver1) driver1count, sum(driver2) driver2count
from (
select orderdate
, case when driver = 1 then 1 else 0 end driver1
, case when driver = 2 then 1 else 0 end driver2
, count(*) ordercount

from yourtable

where whatever

group by orderdate
, case when driver = 1 then 1 else 0 end driver1
, case when driver = 2 then 1 else 0 end driver2

) you_need_an_alias_here

group by orderdate, ordercount

order by orderdate
于 2013-01-31T00:06:31.627 回答