选择日期范围内的特定日期
Selecting certain days in a date range
我有下面的查询,其中给出了学生缺勤的天数。 DATEDIFF 和 DATEPART 计算工作日和节假日不应算作缺席日。 studentTable 中的缺席天数分别存储在两个字段中,即 fromDate 和 toDate。所以缺席天数在一个日期范围内。如果学生缺勤一天,则记录为 11/23/2015、11/23/2015。如果学生缺席两天,则 11/23/2015, 11/24/2015.
DECLARE @startDate DATE SET @startDate = '20151121'
DECLARE @endDate DATE SET @endDate = '20151123'
SELECT
a.studentName
,SUM(DATEDIFF(dd, fromDate, toDate)
- (DATEDIFF(wk, fromDate, toDate) * 2)
-CASE WHEN DATEPART(dw, fromDate) = 1 THEN 1 ELSE 0 END
+CASE WHEN DATEPART(dw, toDate) = 1 THEN 1 ELSE 0 END + 1 )- COUNT(h.holiday)
AS totalAbsentDay
FROM studentTable a
LEFT OUTER JOIN holiday h
ON h.holiday < a.toDate and h.holiday > a.fromDate
WHERE a.fromDate = @startDate AND a.toDate = @endDate
GROUP BY a.studentName
这里的问题是,当我尝试声明开始日期和结束日期时,它没有给我正确的缺席天数。
例如,如果学生在 11/23/2015 和 11/26/2015 之间缺席,即缺勤 4 天,并且我声明开始日期为 11/22/2015 和结束日期为 11/27/2015,则查询不会给我 3 的结果。
下面的查询适用于给定的数据库方案,可能不是最佳解决方案,因为使用了三级查询
DECLARE @startDate DATE SET @startDate = '2016-02-05'
DECLARE @endDate DATE SET @endDate = '2016-02-20'
SELECT
studentName,
SUM(AbsentDay) totalAbsentDay
FROM
(
SELECT
a.studentName
,DATEDIFF(dd, fromDate, toDate)
- (DATEDIFF(wk, fromDate, toDate) * 2)
-CASE WHEN DATEPART(dw, fromDate) = 1 THEN 1 ELSE 0 END
+CASE WHEN DATEPART(dw, toDate) = 1 THEN 1 ELSE 0 END + 1 - COUNT(h.holiday)
AS AbsentDay
FROM (
SELECT
studentName,-- Name,
CASE WHEN fromDate<@startDate THEN @startDate ELSE fromDate END fromDate,
CASE WHEN toDate>@endDate THEN @endDate ELSE toDate END toDate
FROM
StudentTable S
WHERE
S.toDate >= @startDate AND s.fromDate <= @endDate
) a
LEFT OUTER JOIN holiday h
ON h.holiday < a.toDate and h.holiday > a.fromDate
GROUP BY studentName, fromDate, toDate
) B
GROUP BY studentName
为了更轻松的查询和更快的执行,请考虑将 studentTable
重新设计为 idStudent
、AbsentDate
..只是一个建议..
我有下面的查询,其中给出了学生缺勤的天数。 DATEDIFF 和 DATEPART 计算工作日和节假日不应算作缺席日。 studentTable 中的缺席天数分别存储在两个字段中,即 fromDate 和 toDate。所以缺席天数在一个日期范围内。如果学生缺勤一天,则记录为 11/23/2015、11/23/2015。如果学生缺席两天,则 11/23/2015, 11/24/2015.
DECLARE @startDate DATE SET @startDate = '20151121'
DECLARE @endDate DATE SET @endDate = '20151123'
SELECT
a.studentName
,SUM(DATEDIFF(dd, fromDate, toDate)
- (DATEDIFF(wk, fromDate, toDate) * 2)
-CASE WHEN DATEPART(dw, fromDate) = 1 THEN 1 ELSE 0 END
+CASE WHEN DATEPART(dw, toDate) = 1 THEN 1 ELSE 0 END + 1 )- COUNT(h.holiday)
AS totalAbsentDay
FROM studentTable a
LEFT OUTER JOIN holiday h
ON h.holiday < a.toDate and h.holiday > a.fromDate
WHERE a.fromDate = @startDate AND a.toDate = @endDate
GROUP BY a.studentName
这里的问题是,当我尝试声明开始日期和结束日期时,它没有给我正确的缺席天数。 例如,如果学生在 11/23/2015 和 11/26/2015 之间缺席,即缺勤 4 天,并且我声明开始日期为 11/22/2015 和结束日期为 11/27/2015,则查询不会给我 3 的结果。
下面的查询适用于给定的数据库方案,可能不是最佳解决方案,因为使用了三级查询
DECLARE @startDate DATE SET @startDate = '2016-02-05'
DECLARE @endDate DATE SET @endDate = '2016-02-20'
SELECT
studentName,
SUM(AbsentDay) totalAbsentDay
FROM
(
SELECT
a.studentName
,DATEDIFF(dd, fromDate, toDate)
- (DATEDIFF(wk, fromDate, toDate) * 2)
-CASE WHEN DATEPART(dw, fromDate) = 1 THEN 1 ELSE 0 END
+CASE WHEN DATEPART(dw, toDate) = 1 THEN 1 ELSE 0 END + 1 - COUNT(h.holiday)
AS AbsentDay
FROM (
SELECT
studentName,-- Name,
CASE WHEN fromDate<@startDate THEN @startDate ELSE fromDate END fromDate,
CASE WHEN toDate>@endDate THEN @endDate ELSE toDate END toDate
FROM
StudentTable S
WHERE
S.toDate >= @startDate AND s.fromDate <= @endDate
) a
LEFT OUTER JOIN holiday h
ON h.holiday < a.toDate and h.holiday > a.fromDate
GROUP BY studentName, fromDate, toDate
) B
GROUP BY studentName
为了更轻松的查询和更快的执行,请考虑将 studentTable
重新设计为 idStudent
、AbsentDate
..只是一个建议..