Sql 在日期之间存在重叠的情况下，如何限制日期差异的聚合_Sql_Sql Server_Date_Sql Server 2012_Aggregate

Sql 在日期之间存在重叠的情况下，如何限制日期差异的聚合

sql sql-server date sql-server-2012

Sql 在日期之间存在重叠的情况下，如何限制日期差异的聚合,sql,sql-server,date,sql-server-2012,aggregate,Sql,Sql Server,Date,Sql Server 2012,Aggregate,如何在日期之间存在重叠的情况下限制日期差异的聚合如果我有一个像这样的任务表： emp_num from_date to_date 1336 2017-02-07 00:00:00.000 2017-02-08 00:00:00.000 1336 2017-02-15 00:00:00.000 2017-02-16 00:00:00.000 1336 2017-02-21 00:00:00.0

如何在日期之间存在重叠的情况下限制日期差异的聚合

如果我有一个像这样的任务表：

   emp_num   from_date                 to_date
   1336       2017-02-07 00:00:00.000   2017-02-08 00:00:00.000
   1336       2017-02-15 00:00:00.000   2017-02-16 00:00:00.000
   1336       2017-02-21 00:00:00.000   2017-02-23 00:00:00.000
   1336       2017-02-26 00:00:00.000   2017-02-26 00:00:00.000
   1336       2017-02-28 00:00:00.000   2017-03-01 00:00:00.000

我想得到一年中某个月（截止日期-自截止日期）的总和

在本例中，2017年2月的总任务天数应等于9天

我做了以下查询，但由于与下个月重叠，我得到的是

天，而不是

天

SELECT b.emp_num,b.NAME AS FullName ,c.DEPTNAME AS DeptName,d.camp_name AS CampName
,SUM(DATEDIFF(day,a.from_date ,a.to_date)+1) AS Mission_SUM
FROM Mission a INNER JOIN Employee b
ON a.emp_num = b.emp_num
INNER JOIN DEPARTMENT c
ON b.dep_code = c.dep_code
INNER JOIN Branch d
ON d.camp_code = b.camp_code
where ((Year(from_date) =2017 AND Month(from_date)=2) OR (Year(to_date) =2017 AND Month(to_date)=2))
and mission_type = 1
GROUP BY b.emp_num,b.NAME,c.DEPTNAME,d.camp_name
ORDER BY d.camp_name,c.DEPTNAME,b.NAME

我想获得员工2017年2月的总天数？

您可以使用此查询

DECLARE @FirstDate date = '2017-02-01'
DECLARE @EndDate date = EoMonth(@FirstDate)

SELECT b.emp_num,b.NAME AS FullName ,c.DEPTNAME AS DeptName,d.camp_name AS CampName
,SUM(DATEDIFF(day,
            IIF(a.from_date > @FirstDate, a.from_date, @FirstDate) ,
            IIF(a.to_date < @EndDate, a.to_date, @EndDate))+1) AS Mission_SUM
FROM Mission a INNER JOIN Employee b
ON a.emp_num = b.emp_num
INNER JOIN DEPARTMENT c
ON b.dep_code = c.dep_code
INNER JOIN Branch d
ON d.camp_code = b.camp_code
where ((Year(from_date) =2017 AND Month(from_date)=2) OR (Year(to_date) =2017 AND Month(to_date)=2))
and mission_type = 1
GROUP BY b.emp_num,b.NAME,c.DEPTNAME,d.camp_name
ORDER BY d.camp_name,c.DEPTNAME,b.NAME

DECLARE@FirstDate日期='2017-02-01'
声明@EndDate date=EoMonth（@FirstDate）
选择b.emp_num，b.NAME作为全名，c.DEPTNAME作为DEPTNAME，d.camp_NAME作为CampName
，SUM（DATEDIFF）（日），
IIF（a.from_date>@FirstDate，a.from_date，@FirstDate），
IIF（a.截止日期<@EndDate，a.截止日期，@EndDate））+1）作为任务总和
从任务a内部加入员工b
在a.emp_num=b.emp_num上
内部连接部门c
在b.dep_代码=c.dep_代码上
内连接分支d
关于d.camp_代码=b.camp_代码
式中（（年（自日期）=2017年，月（自日期）=2）或（年（至日期）=2017年，月（至日期）=2））
任务类型=1
按b.emp_num、b.NAME、c.DEPTNAME、d.camp_NAME分组
按d.营地名称、c.部门名称、b.名称订购

您可以使用此查询

DECLARE @FirstDate date = '2017-02-01'
DECLARE @EndDate date = EoMonth(@FirstDate)

SELECT b.emp_num,b.NAME AS FullName ,c.DEPTNAME AS DeptName,d.camp_name AS CampName
,SUM(DATEDIFF(day,
            IIF(a.from_date > @FirstDate, a.from_date, @FirstDate) ,
            IIF(a.to_date < @EndDate, a.to_date, @EndDate))+1) AS Mission_SUM
FROM Mission a INNER JOIN Employee b
ON a.emp_num = b.emp_num
INNER JOIN DEPARTMENT c
ON b.dep_code = c.dep_code
INNER JOIN Branch d
ON d.camp_code = b.camp_code
where ((Year(from_date) =2017 AND Month(from_date)=2) OR (Year(to_date) =2017 AND Month(to_date)=2))
and mission_type = 1
GROUP BY b.emp_num,b.NAME,c.DEPTNAME,d.camp_name
ORDER BY d.camp_name,c.DEPTNAME,b.NAME

DECLARE@FirstDate日期='2017-02-01'
声明@EndDate date=EoMonth（@FirstDate）
选择b.emp_num，b.NAME作为全名，c.DEPTNAME作为DEPTNAME，d.camp_NAME作为CampName
，SUM（DATEDIFF）（日），
IIF（a.from_date>@FirstDate，a.from_date，@FirstDate），
IIF（a.截止日期<@EndDate，a.截止日期，@EndDate））+1）作为任务总和
从任务a内部加入员工b
在a.emp_num=b.emp_num上
内部连接部门c
在b.dep_代码=c.dep_代码上
内连接分支d
关于d.camp_代码=b.camp_代码
式中（（年（自日期）=2017年，月（自日期）=2）或（年（至日期）=2017年，月（至日期）=2））
任务类型=1
按b.emp_num、b.NAME、c.DEPTNAME、d.camp_NAME分组
按d.营地名称、c.部门名称、b.名称订购

我不太了解您，但我认为您需要了解到目前为止的差异，因此我创建了一个函数来计算年-月-日，并返回一个表，以便您可以选择所需的值

您可以更新它以获得所需的结果，也可以向我寻求帮助以获得正确的结果：

CREATE FUNCTION [dbo].[GetDateDifferenceInYearsMonthsDays]
(
    @FromDate DATETIME, @ToDate DATETIME
)
RETURNS
 @DateDifference TABLE (
 YEAR INT,  MONTH INT, DAYS INT)
AS
BEGIN
    DECLARE @Years INT, @Months INT, @Days INT, @tmpFromDate DATETIME = @FromDate
    SET @Years = DATEDIFF(yy, @tmpFromDate, @ToDate) - CASE WHEN (MONTH(@FromDate) > MONTH(@ToDate)) OR (MONTH(@FromDate) = MONTH(@ToDate) AND DAY(@FromDate) > DAY(@ToDate)) THEN 1 ELSE 0 END

    SET @tmpFromDate = DATEADD(YEAR, @Years , @FromDate)
    SET @Months =  DATEDIFF(m, @tmpFromDate, @ToDate) - CASE WHEN DAY(@FromDate) > DAY(@ToDate) THEN 1 ELSE 0 END

    SET @tmpFromDate = DATEADD(MONTH, @Months , @tmpFromDate)
    SET @Days =  DATEDIFF(d, @tmpFromDate, @ToDate)


    INSERT INTO @DateDifference
    VALUES(@Years, @Months, @Days)

    RETURN
END

我的意思是，我不太理解你，但我认为你需要得到到目前为止的差异，所以我创建了一个函数来计算今年的月日，它返回一个表，这样你就可以选择你想要的值

您可以更新它以获得所需的结果，也可以向我寻求帮助以获得正确的结果：

CREATE FUNCTION [dbo].[GetDateDifferenceInYearsMonthsDays]
(
    @FromDate DATETIME, @ToDate DATETIME
)
RETURNS
 @DateDifference TABLE (
 YEAR INT,  MONTH INT, DAYS INT)
AS
BEGIN
    DECLARE @Years INT, @Months INT, @Days INT, @tmpFromDate DATETIME = @FromDate
    SET @Years = DATEDIFF(yy, @tmpFromDate, @ToDate) - CASE WHEN (MONTH(@FromDate) > MONTH(@ToDate)) OR (MONTH(@FromDate) = MONTH(@ToDate) AND DAY(@FromDate) > DAY(@ToDate)) THEN 1 ELSE 0 END

    SET @tmpFromDate = DATEADD(YEAR, @Years , @FromDate)
    SET @Months =  DATEDIFF(m, @tmpFromDate, @ToDate) - CASE WHEN DAY(@FromDate) > DAY(@ToDate) THEN 1 ELSE 0 END

    SET @tmpFromDate = DATEADD(MONTH, @Months , @tmpFromDate)
    SET @Days =  DATEDIFF(d, @tmpFromDate, @ToDate)


    INSERT INTO @DateDifference
    VALUES(@Years, @Months, @Days)

    RETURN
END

实现这一点的简单方法如下所示。希望它涵盖所有场景。我只考虑了日期

CREATE TABLE #TMP (FROMDATE DATE, ENDDATE DATE)
INSERT INTO #TMP VALUES('2017-02-07','2017-02-08')
INSERT INTO #TMP VALUES('2017-02-15','2017-02-16')
INSERT INTO #TMP VALUES('2017-02-21','2017-02-23')
INSERT INTO #TMP VALUES('2017-02-26','2017-02-26')
INSERT INTO #TMP VALUES('2017-02-28','2017-03-01')

SELECT SUM(TOTALDAYS) TOTALDAYS
FROM (
        SELECT TOTALDAYS= 
                CASE WHEN MONTH(FROMDATE)=MONTH(ENDDATE) AND YEAR(FROMDATE)=YEAR(ENDDATE) THEN SUM(DATEDIFF(D,FROMDATE,ENDDATE))+COUNT(1) 
                    WHEN  MONTH(FROMDATE)!=MONTH(ENDDATE) THEN SUM(DATEDIFF(D,FROMDATE,EOMONTH(FROMDATE)))+1
                END
        FROM #TMP
        GROUP BY FROMDATE,ENDDATE 
    ) A

DROP TABLE #TMP

实现这一点的简单方法如下所示。希望它涵盖所有场景。我只考虑了日期

CREATE TABLE #TMP (FROMDATE DATE, ENDDATE DATE)
INSERT INTO #TMP VALUES('2017-02-07','2017-02-08')
INSERT INTO #TMP VALUES('2017-02-15','2017-02-16')
INSERT INTO #TMP VALUES('2017-02-21','2017-02-23')
INSERT INTO #TMP VALUES('2017-02-26','2017-02-26')
INSERT INTO #TMP VALUES('2017-02-28','2017-03-01')

SELECT SUM(TOTALDAYS) TOTALDAYS
FROM (
        SELECT TOTALDAYS= 
                CASE WHEN MONTH(FROMDATE)=MONTH(ENDDATE) AND YEAR(FROMDATE)=YEAR(ENDDATE) THEN SUM(DATEDIFF(D,FROMDATE,ENDDATE))+COUNT(1) 
                    WHEN  MONTH(FROMDATE)!=MONTH(ENDDATE) THEN SUM(DATEDIFF(D,FROMDATE,EOMONTH(FROMDATE)))+1
                END
        FROM #TMP
        GROUP BY FROMDATE,ENDDATE 
    ) A

DROP TABLE #TMP