问：请问，如何计算一个表中的周起始和截止日期并写到表字段中? 我要从一个表向另一个表导入数据，并进行转换，用的是VB 。
我现在有有一个表 主要字段有
time_id int
time_date datetime
year int
week_of_year int
day nvarhar
想要转换成另外一张表
time_id int
time_date datetime
year int
week_of_year nvarchar
原来的表内容是
--------------------
1 2003-07-09 2003 20 星期日
1 2003-07-10 2003 20 星期一
1 2003-07-11 2003 20 星期二
想要变成
--------------------
1 07/09/2003 2003 第20周(7/9-7/17)
1 07/10/2003 2003 第20周(7/9-7/17)
1 07/11/2003 2003 第20周(7/9-7/17)
请问：这个语句应该怎么去写?
解决方法：
| if object_id('tablename') is not null drop table tablename select 1 as time_id, '2003-07-09' as time_date, 2003 as [year], 20 as week_of_year, '星期日' as [day] into tablename union select 1, '2003-07-10', 2003, 20, '星期一' union select 1, '2003-07-11', 2003, 20, '星期二' ------------------------------------------------ select time_id, time_date, [year], '第' + cast(week_of_year as varchar(2)) + '周(' + cast(month(week_begin) as varchar(2)) + '/' + cast(day(week_begin) as varchar(2)) + '-' + cast(month(week_end) as varchar(2)) + '/' + cast(day(week_end) as varchar(2)) as week_of_year from (select *, dateadd(day, 1 - datepart(weekday, time_date), time_date) as week_begin, dateadd(day, 7 - datepart(weekday, time_date), time_date) as week_end from tablename) a /* time_id time_date year week_of_year 1 2003-07-09 2003 第20周(7/6-7/12) 1 2003-07-10 2003 第20周(7/6-7/12) 1 2003-07-11 2003 第20周(7/6-7/12) */ ------------------------------------------------ drop table tablename |
