使用SQL Server 2005
表1
ID FromDate ToDate
001 23-02-2009 25-02-2009
001 27-02-2009 29-02-2009
002 12-02-2009, 25-03-2009
...
表2
ID Name Total
001 Raja 30
002 Ravi 22
我想获得这个人格的总天数
尝试查询,
SELECT
table2.Id, table2.name, table2.total,
datediff(day, table1.fromdate, table2.todate)
FROM table1
LEFT OUTER JOIN table2 ON table1.personid = table2.personid
获取输出
ID Name Total Days
001 Raja 30 3
001 Raja 30 3
...,
它应该总计天数,它应该显示在一行中,
注意:假设我选择特定的期间日期意味着它应该仅显示那些天
例如
其中日期为26-02-2009至03-03-2009,应显示
ID Name Total Days
001 Raja 30 3
...,
因为我是在2009年2月25日之后的日期,
Expected Output
ID Name Total Days
001 Raja 30 6
002 Ravi 22 16
如何修改我的查询?
答案 0 :(得分:0)
SELECT table2.Id, table2.name, table2.total,
COALESCE(
(
SELECT SUM(DATEDIFF(day, table1.fromdate, table1.todate) + 1)
FROM table1
WHERE table1.personid = table2.personid
), 0) AS [days]
FROM table2
答案 1 :(得分:0)
我认为GROUP BY查询会更简单:
SELECT table2.Id, table2.name, table2.total,
SUM(DATEDIFF(day, table1.fromdate, table1.todate)) AS Days
FROM table1
left outer join table2 on
table1.personid = table2.personid
GROUP BY table2.Id, table2.name, table2.total
答案 2 :(得分:0)
DATEDIFF给出了两个日期之间的天数差异,因此以相同的方式,1和3之间的差异为2(3 - 1 = 2),DATEDIFF(d)实际上是D2 - D1。因此,为了弥补您想要计算的额外日期,您需要每天DATEADD(ToDate或FromDate)来抵消您的日期:
SELECT table2.id, table2.Name, table2.Total, SUM(DATEDIFF(d, DATEADD(d, -1, table1.FromDate), table1.ToDate))
FROM table1
INNER JOIN table2 ON table1.id = table2.id
GROUP BY table2.id, table2.Name, table2.Total