如何计算Oracle中行的月数

Question

我正在使用 ORACLE 来检查 'DATE_PERIOS' 中的条件以及当前时间和上次时间

我的 ORD0011 table:

-------------------------------------------------------------
ORDER_ID  |     BS_NO   |    DATE_PERIOS   |    STATUS  
-------------------------------------------------------------
 3000003       HS00001       4-2021           COMPLETE
 3000003       HS00183       5-2021           COMPLETE
 3000003       HS00776       10-2021          FALSE
 3000003       HS00559       11-2021          COMPLETE
 3000003       HS00221       12-2021          ACTIVE
 3000003       HS00222       1-2022           COMPLETE
--------------------------------------------------------------

当i select 'ORDER_ID' = 3000003时，输出数据如下：

------------------------------
ORDER_ID    |   HS_TIME
------------------------------
3000003           4
------------------------------

这是我的食谱：

如果i select DATE_PERIOS is: 1/2022, 则显示HS_TIME is 4 (3 times COMPLETE before: 4-2021, 5-2021, 11-2021 + 1)

如果i select DATE_PERIOS is: 11/2021, 则显示HS_TIME is 3 (2 times COMPLETE before: 4-2021, 5-2021 + 1)

注意：仅当状态为 COMPLETE 时 +1

如何统计有条件输出结果如上的月数？非常感谢

Answer 1

with cte as(
   select *,row_number()over(partition by ORDER_ID order by ORDER_ID) as seq
   from tb
),
cte2 as(
select *,
       case
         when DATE_PERIOS = '1-2022'
           then (select count(*) + 1 from cte t2 where t1.seq > t2.seq and t2.STATUS = 'COMPLETE')
         else 0 end as HS_TIME_1_2022,
        case
         when DATE_PERIOS = '11-2021'
           then (select count(*) + 1 from cte t2 where t1.seq > t2.seq and t2.STATUS = 'COMPLETE')
         else 0 end as HS_TIME_11_2022
from cte t1)
select ORDER_ID,max(HS_TIME_1_2022)HS_TIME_1_2022,max(HS_TIME_11_2022)HS_TIME_11_2022
from cte2
group by ORDER_ID

下面的输出是在sql-server中得到的，但在Oracle中也可以是运行。

Result

Answer 2

您可以使用以下带有 case 语句的查询来处理状态 = 'COMPLETE' -

SELECT CASE WHEN COUNT(CASE WHEN status = 'COMPLETE' THEN 1 ELSE NULL END) > 0 THEN
                 COUNT(CASE WHEN status = 'COMPLETE' THEN 1 ELSE NULL END) + 1
            ELSE NULL
       END HS_TIME
  FROM tb
 WHERE TO_DATE(DATE_PERIOS, 'MM-YYYY') < TO_DATE('01-2020', 'MM-YYYY');

Demo.

Answer 3

一个选项是有条件地（即CASE表达式）添加1（即SUM函数）如果status 已完成。

WHERE 子句要求 TO_DATE 具有适当的日期格式。否则，您将比较会导致错误结果的字符串；如果 date_perios 以 YYYYMM 格式存储，可能可以；另一方面，也许您想考虑将日期值存储到 DATE 数据类型列中。

示例数据

SQL> with ord0011 (order_id, date_perios, status) as
  2    (select 303, '4-2021' , 'COMPLETE' from dual union all
  3     select 303, '5-2021' , 'COMPLETE' from dual union all
  4     select 303, '10-2021', 'FALSE'    from dual union all
  5     select 303, '11-2021', 'COMPLETE' from dual union all
  6     select 303, '12-2021', 'ACTIVE'   from dual union all
  7     select 303, '1-2022' , 'COMPLETE' from dual
  8    )

查询本身

  9  select order_id,
 10         sum(case when status = 'COMPLETE' then 1 else 0 end) hs_time
 11  from ord0011
 12  where to_date(date_perios, 'mm-yyyy') <= to_date('&par_date', 'mm-yyyy')
 13  group by order_id;
Enter value for par_date: 1-2022

  ORDER_ID    HS_TIME
---------- ----------
       303          4

SQL> /
Enter value for par_date: 11-2021

  ORDER_ID    HS_TIME
---------- ----------
       303          3

SQL>

如何计算Oracle中行的月数

How to count number of months of row in Oracle

sql

oracle

plsql