日期:2014-05-18  浏览次数:20584 次

计算每个月的工作小时数。
表calendar:
date type
2012-01-01 1 
2012-01-02 1
2012-01-03 1
2012-01-04 1
2012-01-05 1
......
......
2012-12-27 2
2012-12-28 1
2012-12-29 1
2012-12-30 1
2012-12-31 1
 type: 1表示工作日,2表示周末,3表示节假日;
我想要计算,每个月的工作小时数,每天的工作小时数是6.74小时;从上个月的25号到当月的24号算作当月的工作小时数;
结果为:
月份 当月工作小时数
2012-01 148.28
2012-02 132.64
2012-03 147.30
..........

请问这样的sql怎么写?

------解决方案--------------------
SQL code

--> 测试数据:[calendar]
if object_id('[calendar]') is not null drop table [calendar]
create table [calendar]([date] datetime,[type] int)
insert [calendar]
select '2012-01-01',1 union all
select '2012-01-02',1 union all
select '2012-01-03',1 union all
select '2012-01-04',2 union all
select '2012-01-05',3 union all
select '2012-12-27',2 union all
select '2012-12-28',1 union all
select '2012-12-29',1 union all
select '2012-12-30',1 union all
select '2012-12-31',3

select case when day([date])<=25  
then CONVERT(varchar(7),dateadd(mm,-1,[date]),120) else 
CONVERT(varchar(7),[date],120) end as 月份,
SUM(case when [type]=1 then 6.74 else 0 end) as 当月工作小时数
from [calendar]
group by case when day([date])<=25  
then CONVERT(varchar(7),dateadd(mm,-1,[date]),120) else 
CONVERT(varchar(7),[date],120) end
/*
月份    当月工作小时数
2011-12    20.22
2012-12    20.22
*/

反正就是这么一个思路

------解决方案--------------------
--楼上借用一下数据
if object_id('[calendar]') is not null drop table [calendar]
create table [calendar]([date] datetime,[type] int)
insert [calendar]
select '2012-01-01',1 union all
select '2012-01-02',1 union all
select '2012-01-03',1 union all
select '2012-01-04',2 union all
select '2012-01-05',3 union all
select '2012-12-27',2 union all
select '2012-12-28',1 union all
select '2012-12-29',1 union all
select '2012-12-30',1 union all
select '2012-12-31',3


select convert(nvarchar(7),A.date,23),
(select sum(case [type] when 1 then 6.74 else 0 end) from [calendar] B where convert(nvarchar(7),A.date,23)
=convert(nvarchar(7),[date],23))
 from [calendar] A group by convert(nvarchar(7),[date],23)