Question

我需要计算在SQL的日期范围内有X个连续缺席的所有员工。

我们有一个缺勤表，每天缺少一名员工记录1个记录，以及一个包含当年工作日的日历表。

tblAbsences
EmployeeID int
AbsenceDate datetime

tblCalendar
WorkDay datetime

有没有人有任何想法如何计算连续缺席？示例：在2009年1月1日至2009年3月1日期间连续3次缺席的所有员工。

Answer 1

这对你有用。在ConsecDates上的GROUP BY找到谁缺席超过X次。

select a.*, 
        (
            select min(b.absenceDate) from tblAbsences b where a.employeeId = b.employeeId 
            and b.absenceDate >= a.absenceDate
            and not exists ( 
                select 1 from tblabsences c where c.employeeId = b.employeeId and dateadd( dd, 1, b.absenceDate) = c.absenceDate  
            )
) ConsecDates
from dbo.tblAbsences a
order by a.AbsenceDate asc

Answer 2

在PostgreSQL中测试; SQL使用发布的示例值。

在提供的表上没有定义主键，下面的代码解决了这个问题。更好的方法是添加主键并优化下面的代码以利用它们：更好的数据质量，更好的性能，更清晰的代码，更快乐的人。

删除tbl前缀为数据库实现提供了更大的灵活性。然后，表，视图和同义词可以互换使用，而不会影响引用数据库对象或违反命名约定的代码。

/* period length as the desired number of consecutive days */
/* time window as the period to be analyzed */
SELECT DISTINCT
 /* Consolidate employees with multiple periods */
 a.employeeid
FROM
 (SELECT
   /* Generate all possible periods */
   pk_c.workday begin_date,
   /* End date for given period length; less one for closed boundaries */
   LEAD(pk_c.workday,3-1,NULL) OVER (ORDER BY pk_c.workday) end_date 
  FROM (SELECT DISTINCT
         /* No calendar PK, remove dupes; if PK, pull in-line view up */
         c.workday
        FROM sandbox.calendar c) pk_c
       ) p
 INNER JOIN sandbox.absences a ON 
  /* Match absences with periods */
  (a.absencedate BETWEEN p.begin_date AND p.end_date)
WHERE
 /* In time window desired; exclude periods extending beyond boundaries */
 (p.begin_date BETWEEN '2009-01-01' AND '2009-03-01'
  AND /* NOT NULL exclusion implied for periods beyond calendar boundaries */
  p.end_date BETWEEN '2009-01-01' AND '2009-03-01')
GROUP BY
 a.employeeid,
 /* Also group period, display only employee */
 p.begin_date
HAVING
 /* Number of absence days to match to the period length */
 /* DISTINCT due to missing absences PK; if PK, drop DISTINCT */
 COUNT(DISTINCT a.absencedate) = 3
;

享受。剥离版本如下：

SELECT DISTINCT
 a.employeeid
FROM
 (SELECT
   pk_c.workday begin_date,
   LEAD(pk_c.workday,3-1,NULL) OVER (ORDER BY pk_c.workday) end_date 
  FROM (SELECT DISTINCT c.workday FROM sandbox.calendar c) pk_c) p
 INNER JOIN sandbox.absences a ON 
  (a.absencedate BETWEEN p.begin_date AND p.end_date)
WHERE
 (p.begin_date BETWEEN '2009-01-01' AND '2009-03-01'
  AND p.end_date BETWEEN '2009-01-01' AND '2009-03-01')
GROUP BY
 a.employeeid, p.begin_date
HAVING
 COUNT(DISTINCT a.absencedate) = 3
;

计算SQL中的连续缺失

2 个答案: