sql 统计一周的,该怎么处理

sql 统计一周的

create table dd(id varchar(8),time datetime,money money,job varchar(100))
insert into dd select 'bfbumen01','2011-03-14','12454.214','电子商务'
union all
 select 'bfbumen01','2011-03-14','12454.214','电子商务' union all
 select 'bfbumen02','2011-03-15','12454.214','电子商务' union all
 select 'bfbumen03','2011-03-15','12454.214','电子商务' union all
 select 'bfbumen01','2011-03-16','12454.214','电子商务' union all
 select 'bfbumen02','2011-03-16','12454.214','电子商务' union all
 select 'bfbumen01','2011-03-17','12454.214','电子商务' union all
 select 'bfbumen02','2011-03-17','12454.214','电子商务' union all
 select 'bfbumen03','2011-03-17','12454.214','电子商务' union all
 select 'bfbumen01','2011-03-18','12454.214','电子商务' union all
 select 'bfbumen02','2011-03-18','12454.214','电子商务' union all
 select 'bfbumen03','2011-03-19','12454.214','电子商务' union all
 select 'bfbumen02','2011-03-19','12454.214','电子商务' union all
 select 'bfbumen01','2011-03-19','12454.214','电子商务' union all
 select 'bfbumen01','2011-03-20','12454.214','电子商务' union all
 select 'bfbumen01','2011-03-21','12454.214','电子商务' union all
 select 'bfbumen03','2011-03-21','12454.214','电子商务' union all
 select 'bfbumen01','2011-03-22','12454.214','电子商务' union all
 select 'bfbumen02','2011-03-22','12454.214' ,'电子商务'union all
 select 'bfbumen03','2011-03-22','12454.214','电子商务' union all
 select 'bfbumen01','2011-03-18','12454.214','电子商务' union all 
 有很多岗位，id 为员工工号,想得到的结果:以每星期一为一周统计时间，（统计人数用 count(*) 想要的结果（结果只是个例子，不是正确的数据）：
  人数 时间 钱 岗位 
  23 2011-03-14 22152252.54 电子商务
  23 2011-03-21 22152252.54 电子商务
  23 2011-03-28 22152252.54 电子商务
  。。。
 

[解决办法]

SQL code

select count(1), min(time), sum(money), job from dd group by datediff(day,0,time)/7, job
[解决办法]
顺便问下 楼主准备以哪个时间作为统计开始时间
[解决办法]
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SQL code
select count(1), min(time), sum(money), job from dd group by datediff(day,0,time)/7, job