Question

 Regarding my table as below

WORKING_CALENDAR_TABLE

===================================================
 EMPLOYEE ID    |        DATE        | WORKING DAY (0: Holiday; 1: WORKING DAY)
===================================================
  02661              2017/12/01            1
  02661              2017/12/02            1
  02661              2017/12/03            0
  02661              2017/12/04            0
  02661              2017/12/05            0
  02661              2017/12/06            1
  02661              2017/12/07            1
  02661              2017/12/08            1
  02661              2017/12/09            1

2017/12/10时，我的预期结果如下

===================================================
 EMPLOYEE ID    | CONTINOUS WORKING DAY 
===================================================
   02661                   4

如果我们使用SQL ORACLE，我们是否可以利用SQL ORACLE获得此结果？

Answer 1

这样做的一种方法可以是：

在您的员工不工作的参考日期（2017/12/10）之前的最后一天。
在参考日期之前和之前计算日期之后的行数。强行，每行代表员工的工作日。

以下是包含一些注释的代码：

Select employee_id, count(*) as continuous_days  from mytable 
where employee_id = '02661'  and date > (select max(date) 
where employee_id ='02661' and working_day = 0 and date date < 
'2017/12/10') and date < 
'2017/12/10' group by employee_id

 /* (select max(date) 
where employee_id ='02661' and working_day = 0 and date < 
'2017/12/10') gets the last day where the employee didn't work before the reference date. Each row from a date after max(date) represent a working day for the employee because all of them are going to to have working_day = 1*/

有待改进的地方： 如果员工在参考日期之前没有假期，那么 select max(date) where employee_id ='02661' and working_day = 0 and date date < '2017/12/10'将返回null，因此您可以使用COALESCE来防止错误，如果是null，则会获得一些其他值作为回报。我认为在你的情况下，它可以在很早的时候就足够了。你可以这样使用它：

COALESCE( (select max(date) where employee_id ='02661' and working_day = 0 and date < '2017/12/10'), '1900-01-01')

，完整的查询将是：

select employee_id, count(*) as continuous_days  from mytable 
where employee_id = '02661'  and date > COALESCE( (select max(date) 
where employee_id ='02661' and working_day = 0 and date < 
'2017/12/10'),  '1900-01-01') and date < 
'2017/12/10' group by employee_id

Answer 2

对于给定日期，最通用的解决方案是：

select employee_id, count(*)
from t
where employee_id = '02661' and
      date < date '2017-12-10' and
      date > (select max(t2.date)
              from t t2
              where t2.employee_id = t.employee_id and t2.date < date '2017-12-10'
             );

Answer 3

您可以尝试此查询：

SELECT T.EMPLOYEE_ID,
       COUNT(0) AS CONTINUOUS_DAYS
FROM   WORKING_CALENDAR_TABLE T
WHERE  T.WORK_DATE BETWEEN (SELECT MAX(WORK_DATE) + 1
                            FROM   WORKING_CALENDAR_TABLE I
                            WHERE  I.EMPLOYEE_ID = T.EMPLOYEE_ID
                                   AND I.WORK_DAY = 0)
                            AND DATE '2017-12-10' + 1
GROUP  BY T.EMPLOYEE_ID

如何使用SQL Oracle连续计数

3 个答案: