Question

需要快速帮助我有这个小项目，我有这个报告，总共有订单和订单的日期（这实际上是我的另一个问题的延续）所以这就是我一直在做的事情

SELECT  sf.ORDER_QNT, dd.ACTUAL_DATE, dd.MONTH_NUMBER
FROM    sales_fact sf,
        date_dim dd
WHERE   dd.date_id = sf.date_id
AND dd.MONTH_NUMBER = 1;

 ORDER_QNT          ACTUAL_DATE MONTH_NUMBER
 -------------------------------------------
      1100             05/01/13            1
       100             05/01/13            1
       140             06/01/13            1
       110             07/01/13            1
       200             08/01/13            1
       500             08/01/13            1
       230             08/01/13            1
       500             08/01/13            1
       200             08/01/13            1
        53             15/01/13            1
        53             22/01/13            1

现在，我想通过使用last_day函数获得该月的平均值（每天的平均值）。

SELECT  sum(sf.ORDER_QNT)/EXTRACT(DAY FROM LAST_DAY(to_date('05/01/13','dd/mm/rr'))) as AVGPERDAY,
        dd.month_number
FROM    sales_fact sf,
        date_dim dd
WHERE   dd.date_id = sf.date_id
AND     dd.month_number = 1
GROUP BY dd.month_number;

AVGPERDAY    MONTH_NUMBER
--------------------------
113.785714        1

结果很好，但现在当我用dd.actual_date更改日期时，它会给出错误

SELECT  sum(sf.ORDER_QNT)/EXTRACT(DAY FROM LAST_DAY(dd.actual_date)) as AVGPERDAY,
        dd.month_number
FROM    sales_fact sf,
        date_dim dd
WHERE   dd.date_id = sf.date_id
AND     dd.month_number = 1
GROUP BY dd.month_number;


Error at Command Line : 1 Column : 53
Error report -
SQL Error: ORA-00979: not a GROUP BY expression
00979. 00000 -  "not a GROUP BY expression"

任何人都可以帮忙吗？

Answer 1

sum(sf.ORDER_QNT) / decode(round(max(dd.actual_date) - min(dd.actual_date)), 0, 1)

您可以在查询中使用此表达式（但结果可能与LAST_DAY不同）

或者您可以将EXTRACT（DAST FROM LAST_DAY（dd.actual_date））作为GROUP BY的一部分：

SELECT  sum(sf.ORDER_QNT)/EXTRACT(DAY FROM LAST_DAY(dd.actual_date)) as AVGPERDAY,
        dd.month_number
FROM    sales_fact sf,
        date_dim dd
WHERE   dd.date_id = sf.date_id
AND     dd.month_number = 1
GROUP BY dd.month_number, EXTRACT(DAY FROM LAST_DAY(dd.actual_date));

使用extract和last_day函数获取每天的平均值

1 个答案: